STUDY PACKAGESubject : Mathematics Topic : Sequence & Progression Available Online : www.MathsBySuhag.com R Index 1. Theory 2. Short Revision 3. Exercise (Ex. 1 + 5 = 6) 4. Assertion & Reason 5. Que. from Compt. Exams 6. 39 Yrs. Que. from IIT-JEE(Advanced) 7. 15 Yrs. Que. from AIEEE (JEE Main) Student’s Name :______________________ Class :______________________ Roll No. :______________________ Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal : 0 903 903 7779, 98930 58881, WhatsApp 9009 260 559 www.TekoClasses.com www.MathsBySuhag.com Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't. page 1 of 26 fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjar e/;e eu dj ';keA iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA jfpr% ekuo /keZ iz.ksrk ln~xq# Jh j.kNksM+nklth egkjkt Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com Sequence & Progression 2n (ii) n 2n (i) Let t n = n put n = 1, 2, 3, 4, .............. we get 8 t 1 = 2, t 2 = 2, t 3 = , t 4 = 4 3 (i) Solution. so the sequence is (ii) Let t n = 2, 2, 3 ( 1)n 3n 8 , 4, ........ 3 3 ( 1)n 3n put n = 1, 2, 3, 4, ...... 2 4 4 2 , , , ,............ 3 9 27 81 Series By adding or substracting the terms of a sequence, we get an expression which is called a seri es. If a1, a2, a3,........an is a sequence, then the expression a 1 + a2 + a3 + ...... + an is a series. Example. (i) 1 + 2 + 3 + 4 + .................... + n (ii) 2 + 4 + 8 + 16 + ................. Progression : It is not necessary that the terms of a sequence always follow a certain pattern or they are described by some explicit formula for the nth term. Those sequences whose terms follow certain patterns are called progressions. An arithmetic progression (A.P.): A.P. is a sequence whose terms increase or decrease by a fixed number. This fixed number is called the common difference. If a is the first term & d the common difference, then A.P. can be written as a, a + d, a + 2 d,....... a + (n 1) d,........ Example – 4, – 1, 2, 5 ........... (i) n th term of an A.P. Let a be the first term and d be the common difference of an A.P., then t n = a + (n – 1) d where d = an – an – 1 Solved Example # 2 If t 54 of an A.P. is – 61 and t 4 = 64, find t 10. Solution. Let a be the first term and d be the common difference so t 54 = a + 53d = – 61 .........(i) and t 4 = a + 3d = 64 .........(ii) equation (i) – (ii) 50d = – 125 5 143 d=– a= 2 2 5 143 so t 10 = + 9 = 49 2 2 Solved Example # 3 Find the number of terms in the sequence 4, 12, 20, ........108. Solution. a = 4, d = 8 so 108 = 4 + (n – 1)8 n = 14 (ii) The sum of first n terms of are A.P. If a is first term and d is common difference then n Sn = [2a + (n – 1) d] 2 n = [a + ] = nt n1 , 2 so the sequence is 2 where is the last term and t n1 is the middle term. 2 th (iii) r term of an A.P. when sum of first r terms is given is t r = sr – Sr – 1. Solved Example # 4 Find the sum of all natural numbers divisible by 5, but less than 100. Solution. All those numbers are 5, 10, 15, 20, ........... 95. 19 Here a = 5 n = 19 = 95 so S= (5 + 95) = 950. 2 "try" & "should" with "I Will". Ineffective People don't. Successful People Replace the words like; "wish", page 2 of 26 for every sequence is the set N of natural numbers, therefore a sequence is represented by its range. If f : N R, then f(n) = t n n N is called a sequence and is denoted by {f(1), f(2), f(3),...............} = {t 1, t 2, t 3, ......................} = {t n } Real Sequence : A sequence whose range is a subset of R is called a real sequence. Examples : (i) 2, 5, 8, 11, ....................... (ii) 4, 1, – 2, – 5, ...................... (iii) 3, –9, 27, – 81, ........................ Types of Sequence : On the basis of the number of terms there are two types of sequence. (i) Finite sequences : A sequence is said to be finite if it has finite number of terms. (ii) Infinite sequences : A sequenceis said to be infinite if it has infinite number of terms. Solved Example # 1 Write down the sequence whose nth term is Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881. FREE Download Study Package from website: www.TekoClasses.com & www.MathsBySuhag.com Sequence : A sequence is a function whose domain is the set N of natural numbers. Since the domain a5. prove that a2. is 27 and the sum of their squares is 293. Ans. a d. . c are in A. a + d..P.. = 2n2 + 3n – 2(n – 1)2 – 3(n – 1) = 4n + 1 so t 50 = 201.P.. 34.e non zero number. Maths : Suhag R. 2 b = a + c & if a.. 5 Solved Example # 9 If a1. Which term of the sequence 2005. (iii) Three numbers in A. 9. Let a1 and a2 be the first terms and d1 and d2 be the common differences of two A. K. is increased. find its 50th term. to find n. FREE Download Study Package from website: www.. a + 5d etc. Find the maximum sum of the A.. are a 2d. 1 are in A.. c.P. 2000. a + d. Find the ratio of their 11 1th terms.P. a 3d. b. 36.. . a.. i page 3 of 26 Teko Classes. 40.P.MathsBySuhag.P. a5 are in A. Solution. a + d so 3a = 27 a=9 Also (a – d)2 + a2 + (a + d) 2 = 293. a4.P. 0 98930 58881. 997. 108. 115. i1 1 1 .Ps. can be taken as a 3d. As a1. decreased.P. b. 997 = 101 + (n – 1) 7 n = 129 129 so S= [101 + 997] = 70821. 1985.P.P. are in ratio . b2 – a2 = c 2 – b2 a2.P. 5 Hence a i = 10.. 14.. 420 Properties of A. can be taken as a d.P.. b2.com Solved Example # 5 Find the sum of all the three digit natural numbers which on division by 7 leaves remainder 3. Successful People Replace the words like.. 1990.P.P.. . 32. Kariya (S. (v) Any term of an A.. . are a 5d.. multiplied or divided by the sA.. a + d.. a + 3d. (i) The common difference can be zero.P. 2. R. a3. a + d. a d. five numbers in A. a3. find them Solution. a2. Sir). ab 1 are in A. a. All these numbers are 101. bc ca 1 1 1 – = ca bc ab ba c b = bc ab Solved Example # 10 If Solution. "try" & "should" with "I Will". Let the numbers be a – d. "wish". For an A. For k = 2. a + d = b + c.. Bhopal Phone : 0 903 903 7779.P. then the resulting sequence is also an A. then find the value of a i1 when a3 = 2. bc ca 1 1 . a + 3d. 4n 27 Sol. (vi) If each term of an A.. Ineffective People don't. 3a2 + 2d2 = 293 d2 = 25 d=±5 therefore numbers are 4. show that t m + t 2n + m = 2 t m + n 3. Solution. ab bc c a c aab 1 – = (c a )(b c ) (a b )(c a) ca . For k = 1..TekoClasses. (iv) The sum of the terms of an A. Solution. (except the first) is equal to half the sum of terms which are equidistant from it.MathsBySuhag. are in A. 1995. 2 7n 1 Solved Example # 6 The sum of n terms of two A. 38.. a + 2d & six terms in A. k < n... a4. four numbers in A.. we have a1 + a5 = a2 + a4 = 2a3. a. equidistant from the beginning & end is constant and equal to the sum of first & last terms.s respectively then n n 1 [2a1 (n 1)d1] a1 d1 2 7n 1 7n 1 2 = = n n 1 4 n 27 4 n 27 [2a1 (n 1)d2 ] a2 d2 2 2 For ratio of 11th terms n 1 = 10 n = 21 2 7(21) 1 so ratio of 11th terms is 4(21) 27 148 = 111 Solved Example # 7 If sum of n terms of a sequence is given by S n = 2n2 + 3n.M..com & www...P.. c2 are in A.. Let t n is nth term of the sequence so t n = sn – sn – 1. with common difference 0.P.P.. 403.... Solved Example # 8 The sum of three numbers in A. positive or negative. a d.. an = (1/2) (an1+ an+1).P.. . contains the first negative term Ans.(ii) If a. an = 1/2 (ank + an+k).com Get Solution of These Packages & Learn by Video Tutorials on www. d are in A. c2 are also in A. Self Practice Problems : 1. a2. an = (1/2) (an2+ an+2) and so on.P. b2. ar.com bc a c ab abc 1 1 1 .. This constant factor is called the common ratio of the series & is obtained by dividing any term by that which immediately proceeds it. 1 1 1 1 Example .. Self Practice Problems : 4. is 7.e. When n rn 0 if r < 1 therefore.P...S = r 1 .. find n. Bhopal Phone : 0 903 903 7779. 2 13 13 n = 2n + 1.. An = a + n1 n1 n1 NOTE : Sum of n A. An. be equal to the A.M.P.MathsBySuhag.Get Solution of These Packages & Learn by Video Tutorials on www. Ineffective People don't. be 1.MathsBySuhag. S n = r 1 na . A2 = a + . A2.. r1 13 .P. is a sequence of numbers whose first term is non zero & each of the succeeding terms is equal to the proceeding terms multiplied by a constant. Solved Example # 15: The first term of an infinite G.. between a & b. "wish". then the middle term is called the A. .. its nth term is 448 and sum of first n terms is 889.M. the 6 sum of these means exceeds their number by unity. .P..M.. Solution. when r < 1. Number of means = 12. between the other two. 0 98930 58881. . between a & b n i. Sir). Let the G. A2.) G. b are any two given numbers & a. is a G.M. Example 2. If n A.P.M. Thus in a G.. K.’s inserted between a & b is equal to n times the single A. .... a b is the A. 14.. b are in A. 4.. . 7rn = 448 r Also Sn = a(r n 1) 7(r n 1) = r 1 r 1 889 = 448r 7 r 1 r=2 Hence T 5 = ar4 = 7(2)4 = 112.P. Ans. Ans..82.M.P.. an even number of A. FREE Download Study Package from website: www.P. c are in A. An are the n A. b.TekoClasses.M. R. between rth and sth term of the A.com Solved Example # 11 If . . 6. then find the fifth term of G. the ratio of successive terms is constant.. 10. so 86 = 2 + (21)d d=4 so the series is 2.... Here 2 is the first term and 86 is the 22nd term of A.P.. Therefore a. 5. Maths : Suhag R. 82.M.M. Kariya (S..P. are in A.. are also in A. 1 r Solved Example # 14: If the first term of G. between 2 and 86. so if a. .P. "try" & "should" with "I Will".. a b c If three terms are in A... are in A. Solved Example # 13 Insert 20 A. r 1 (ii) Sum of the first n terms i. with a as the first term & r as common ratio. Successful People Replace the words like. Let a and b be two numbers and 2n A. r 2. (a) n Arithmetic Means Between Two Numbers: If a. 3 9 27 81 th n1 (i) n term = a r a rn 1 . between pth and qth terms of an A. Solution.. 14.. ar 2. a b c a b c bc a c ab abc Solution... r 1 a (iii) Sum of an infinite G.M. 8. of a and b.... For what value of n. n = 0 an bn Geometric Progression (G. a n 1 b n 1 .. 16 . n = 11 Solved Example # 12 Between two numbers whose sum is 6.. n (b a ) ba 2 (b a ) A1 = a + . a b c 1 1 1 divide each by a + b + c .. Solution.. Find the series. are in A.com & www..P..M.M.P.. is 1 and any term is equal to the sum of all the succeeding terms. b is A. 10. Solution. A 1.P. Find the number of means. then prove that p + q = r + s.e. ar3.P. a b c Add 2 to each term bc a c ab abc ..s are inserted between a and b then 2n (a + b) = 2n + 1. are in A.s are inserted between 20 and 80 such that first means : last mean = 1 : 3. 86 required means are 6.’s between a & b... then .. Given a = 7 the first term t n = arn – 1 = 7(r)n – 1= 448. If A..P. . ar4.M.P.. .P.P. given a b 6 6 n = 6.. r. Ar = nA where A is the single A...): Teko Classes. then A1.s is inserted.. r3..P. page 4 of 26 Arithmetic Mean (Mean or Average) (A...M. of a & c. Given .. .. ar 2..ar k in case we have to take 2k + 1 terms in a G.. a2... 486. FREE Download Study Package from website: www.. find the sum of 8 2 4 (i) first 20 terms of the series (ii) infinite terms of the series..P.... Kariya (S.P. are two G. .. 10 100 1000 10000 3 3 1 10 = 1 = 9 = 3.... a2b2..... Solved Example # 17: Find three numbers in G.. 3 = 0... 9.are in A.. S1 4.. 1 a Solution.a...P..(i) r r 3 and a = 216 a=6 so from (i) 6r2 – 13r + 6 = 0.. is also a G.P.P. a3...P. 2 3 Solved Example # 18: Find the product of 11 terms in G. can be taken as .... a2.. . r k r k 1 r k 2 a a (iii) Any four consecutive terms of a G. Ineffective People don't...233333 = 1..TekoClasses.P.. 3 3 3 3 = + + + + . where each ai > 0... Let x = 0. . consist of 2n terms. prove that a q – r br – p cp – q = 1.. and its converse is also true. . a.. If the sum of the terms occupying the odd places is S1 and that of the terms S2 occupying the even places is S 2 then find the common ratio of the progression....0003 + . and b1. a2.MathsBySuhag. b2.. – 3. (i) S20 = = . express 0... a3. 2 Properties of G..P..... a4.. ..... ar......... having sum 19 and product 216.... a3. 3 Ans...P.. qth ...: Using the property a1a11 = a2a10 = a3a9 = . loga3.. q Solution.. .... Ans..... .P. in general if a1.P’s with common ratio r 1 and r2 respectively then the sequence a1b1. "try" & "should" with "I Will".. 2 1 r 1 1 1 Hence series is 1. ar so a 1 r = 19 .2 + + + + .P....... ...... then a1an = a1an – 1 = a3 an – 2 = . Hence the three numbers are 4. rth terms of a G.003 + 0.. a (ii) Any three consecutive terms of a G. Solution...... 0 98930 58881.. .. ... an are in G. 3 2 r= .P....2 3 as form. Maths : Suhag R. b...03 + 0..com 1 r2 r= . in general we take r a a a ..... 2 1 1 2 2 Self Practice Problems : 1.. 2.. 3....... b3.. is 4. be a... c respectively.P. b....P. with common ratio r1 r2. whose 6 th is 5. . 1 20 1 2 1 2 20 1 Solution. 2.....003 + 0... ...P..P. the resulting sequence is also a G.. then log a1. (vi) If a1. K. Find the G... (ii) S = 19 1 1 = 2.ar 2k – 1 in case we have to take 2k terms in a G...... and the sum of their cubes is 192. = a62 = 25 Hence product of terms = 511 p Solved Example # 19:Using G.. = 0. loga2... Sir).3 + 0..... c are in G... in general we take r r a a a .... .. (i) If a. .. 2 4 8 1 1 1 Solved Example # 16:Let S = 1 + + + + .3333 . if the common ratio of G. 6.P. 18. n th term is 486 and sum of first n terms is 728. is 3..2 + 0. ... r r 2k 1 r 2k 3 (iv) If each term of a G..P.... 162...0003 + ... an – 1 ..... A G. be multiplied or divided or raised to power by the some nonzero quantity... 10 2 10 3 10 4 Successful People Replace the words like.com given condition r = . page 5 of 26 Teko Classes.... a3b3. The sum of infinite number of terms of a G.. a ..P.. Let the three numbers be .....03 + 0. 6...P.P.... 54. can be taken as 3 .. ar3. b2 = ac.. ar. "wish"... find the series.. 6..... (v) If a1...2 3 = 1. ar.... Bhopal Phone : 0 903 903 7779. If the pth .. Ans.P... ar.MathsBySuhag.. ..... 1 10 Let y = 1........ R.. 3 3 3 = 1.are in G..com & www.. 3 and 1.Get Solution of These Packages & Learn by Video Tutorials on www. 18. b. 1 = an H n a1 a 2 Solved Example # 22: If m th term of H. "wish".] 9 7 = [(10 – 1) + (102 – 1) + (103 – 1) + . Ineffective People don't. 1/an is an A.... b: If a.. 1 b n1 Solution... Let S = 7 + 77 + 777 + . 7 = [9 + 99 + 999 + .M between 2/3 and 2/13.. b (n 1)(a b) ac c bc NOTE : (i) If a. 1 Solution.. + upto n terms] 9 7 = [10 + 102 + 103 + .): If a.. Sir).P. 1 30 30 1 10 Solved Example # 20 Evaluate 7 + 77 + 777 + .P. Here we do not have the formula for the sum of the n terms of a H. b.com & www. G r = (G)n where G is the single G...s between a & b is equal to the nth power of the single G.. a ab = a bc (ii) If a.MathsBySuhag...com . b is the H... H of these numbers is given by 1 1 1 1 . For H... a2 . a3. between a & c. Maths : Suhag R.... Bhopal Phone : 0 903 903 7779.) : A sequence is said to H.. an is an H. 0 98930 58881. b = 1 + 10 + 102 + 103 + 104 and c = 1 + 105 + 1010 + . G n are n G.): If a.M... is n. c > 0..P.. Common ratio of the series is given by r = = (243)1/5 = 3 a Hence four G..upto n terms. Solution..P... If the sequence a1. 20. so d = 2 2 = 1. Self Practice Problems : 1.. an are ‘n’ non-zero numbers then H.. R. G 2. the nth term is t n = a ab 2ac ab ..: Given Tm = n or a (m 1) d = n. c are in H. then 1/a1.s are 6.M..P.M...M. a (m n d) d 1 m n 1 mn Solved Example # 23: Insert 4 H. 162.M. Ans.P.. "try" & "should" with "I Will". + 1050. where a is the first term and d is the common difference of the corresponding A..e. while nth term is m. . a > 0..M.. upto n terms.P. a2. between a & b n i. If a1. the products are in A. 5 1 2 3 5 = +1= or H1 = H1 5 2 2 Successful People Replace the words like. Harmonic Progression (H. G n.... The sum of three numbers in G...P.. (a) nGeometric Means Between a..P.MathsBySuhag...+ 10n – n] 9 n 7 10 (10 ) 1 n 7 = = [10n + 1 – 9n – 10] 9 9 81 Geometric Means (Mean Proportional) (G. in 70...P.P... b are in G..P.. b. 1 1 mn 1 so a + (m – 1)d = and a + (n – 1) d = (m – n)d = or d = n m mn mn 1 (m 1) 1 so a= – = n mn mn 1 mn mn Hence T (m + n) = = = . Teko Classes.. If a. 54. c are in G.P. Then G 1....s between a & b.1 2.. c are in A.P.M. G 1.TekoClasses.. if the two extremes be multiplied each by 4 and the mean by 5.. 1/a2.. G n = a(b/a)n/n+1 NOTE : The product of n G.. therefore b = a c .M.com Get Solution of These Packages & Learn by Video Tutorials on www.. Let d be the common difference of corresponding A. Find the numbers..P..2 + = . G 1 = a(b/a)1/n+1.. K. find its (m + n) th term... G 2 = a(b/a)2/n+1. b² = ac.. G 3. r1 Solved Example # 21 Insert 4 G.M. if the reciprocals of its terms are in A.. between a & c. 13 3 Solution.. FREE Download Study Package from website: www. b.. b = or = ..s between 2 and 486. 10.P...M.2 + = 1.. whose first term is a and second term is b...P. a ab = b bc Harmonic Mean (H. c are in H.. then prove that 55 (i) ‘a’ is a composite number (ii) a = bc... c are in G. between a & b. 40 111 .... G 2..... Kariya (S.. then b = 2ac/[a + c]..page 6 of 26 3 1 37 10 2 = 1. b is the G.. & converse. b......M.P. b are two given numbers & a... If a = .. Using the relation A.. (ii) A.. Let x be the first term and d be the common difference of the corresponding A.M. c be in H. ab = 36 also a + b = 15 Hence the two numbers are 3 and 12... If H be the harmonic mean of a and b then find the value of + – 1. Let the numbers be a and b. "try" & "should" with "I Will".e. 2. r terms of a H.e. H are respectively A.M. H.. now using the relation G2 = A.Relation between means : (i) If A..an)1/n and their H. page 7 of 26 1 3 7 2 = +2= or H2 = H2 2 2 7 1 2 3 9 = + 3 = or H = 3 H3 9 2 2 1 3 11 2 = +4= or H4 = .....s as 12 to 13. H4 2 2 11 th th th Solved Example # 24: If p ......(ii) b 1 = x + (r – 1) d . exceeds the H. If a..P..M... 0 98930 58881.M. 3 xyz 1 1 1 3 x y z Solved Example # 27 Solution.M.TekoClasses.P. Ineffective People don't... then we define their a a 2 a 3 ...If the H. A... show that a : a – b = a + c : a – c....M.. between a & b both being unequal & positive then. a2..... b.M.. = 1 1 1 It can be shown that . the show that ab + bc + cd = 3ad Teko Classes. Kariya (S. Bhopal Phone : 0 903 903 7779. Self Practice Problems : 1..(iii) c (i) . If a..M... c respectively....... G² = AH i... c. G...(vi) (iv) + (v) + (vi) gives bc (q – r) + ac(r – p) + ab (p – q) = 0. K. G. 1 so = x + (p – 1)d . z prove that (x + y + z) x y z 9 Using the relation A. = 1 ..MathsBySuhag....M.M. b... prove that the quantities are in ratio 4 to 9... 3 b c a b c a 3 1 1 1 For non-zero x. G.M. an A. H H 3. H..... R.M.M. a1 a 2 an A. 3 6 = G G 2 5 3 9 = G2 + G– G=6 10 5 i..M. H are in G. = (a1 a2 a3 .. 3 Solved Example # 25:The A. Let a1.. between two quantities is to their G.. b.H.(ii) ab(p – q)d = b – a . y.. prove that (q – r)bc + (r – p) ac + (p – q) ab = 0 Solution. 1 1 1 (x + y + z) x y z 9 Successful People Replace the words like.P.com .... by and the G..M.com Get Solution of These Packages & Learn by Video Tutorials on www.. their n n G.. Maths : Suhag R. 0 2a 2b 4.. . FREE Download Study Package from website: www.. q .(i) a 1 = x + (q – 1) d .M. 5 Solution..(iv) (ii) . .. "wish".M... by 2 6 . find the numbers.com & www..(i) ac (r – p) d = a – c .M... and equality holds at either places iff a1 = a2 = a3 = . be a.. > 0 prove that + + 3 b c a Solution.. G.(v) (iii) ...MathsBySuhag.P.M. b.. G... H.(iii) bc (q – r)d = c – b ..M..P.an be n positive real numbers... we have 1 a b c a b c 3 a b c b c a . H....M. c..= an a b c Solved Example # 26 If a... a3. Sir)...M. G. d are in H. of two numbers exceeds the G. Ans. . are in A.M. Sir). b.P. 1 + a1 2 a1 1 + an 2 an (1 + a1) (1 + a2) . x 2.....M. Prove that ABC is an equilateral triangle iff tan A + tan B + tan C = 3 3 If a.. ...MathsBySuhag... then show that 1 1 1 1 1 (i) a + d > b + c (ii) + >2 ab cd bd ac ad then 4. 2n–1 we have 1 2 2 2 . 0 98930 58881.TekoClasses...MathsBySuhag.... K. Bhopal Phone : 0 903 903 7779.. b.Get Solution of These Packages & Learn by Video Tutorials on www. & 1. c.M.(1 + an) 2n a1a 2a 3 .. an = 1 Hence (1 + a1) (1 + a2) .(1 + an) 2n i 1 Using A. b...g. Let S=1+ + 2 + 3 + . 23...... is called the AritH. 2n1 > (1. Here 1. y.... 3. "wish". G.. 1 r 1 r 1 r2 a dr n Sum To Infinity: If r < 1 & n then Limit . (1 + an) 2n...M.. n r = 0 S = 1 r 1 r2 Solved Example # 33 Find the sum of the series 7 10 4 1+ + 2 + 3 + . FREE Download Study Package from website: www.P... + + .P.. i.P.....com n Sol.. Prove that nn > 1 .(2n – 1) 3.. G. 5 .. Ex... c > 0 prove that [(1 + a) (1 + b) (1 + c)]7 > 77 a4 b4 c4 Arithmetico-Geometric Series: A series each term of which is formed by multiplying the corresponding term of an A.....2n–1)1/n n Equality does not hold as all the numbers are not equal..... to n terms. Ineffective People don't.P.M. Solved Example # 29 If n > 0 prove that 2n > 1 + n 2n1 Solution.. > H......... # 30 Solution.... x 3.. 5..(ii) n 1 5 5 5 5 5 5n (i) – (ii) 3 3 3 3n 2 4 3 S=1+ + 2 + 3 + . 1 + 3x + 5x 2 + 7x 3 +.. page 8 of 26 1 + a2 2 a 2 Teko Classes. Maths : Suhag R...a n 1/ n As a1 a2 a3 ... an cn > (an cn)1/2 2 an + cn > 2 (ac)n/2 ..M... + .eticoGeometric Series.. then prove that (1 + a1) (1 + a2) (1 + a3) . d be four distinct positive quantities in G.......P... ac > b (ac)n/2 > bn .com & www.M.... Using the relation A. Sum of n terms of an ArithmeticoGeometric Series: Let Sn = a + (a + d) r + (a + 2 d) r² +.. & G.(i) Also G.... z subject to the condition xy + yz + zx = 12.. Ex... r 1...e.com Solution. G. 22..(ii) hence from (i) and (ii) an + cn > 2bn Self Practice Problems : 1..... Using the relation A. 5 5 5 7 10 3n 2 4 Solution. If a.... c are real and distinct then show that a2 (1 + b2) + b2 (1 + c2) + c2 (1 + a2) > 6abc 2.2 22 23 . c are in H.. on the numbers 1.M. + n 1 – . and they are distinct and positive then prove that an + cn > 2bn Solution.......... are in G........ Kariya (S.. e. 3 .... + [a + (n 1)d] rn1 d r 1 r n 1 a a (n 1) d r n . b..... 2... 1 ( n1) n n 2 1 > n 2 2 2 1 n n n 2 –1>n (n 1) 2 2 ( n1) 2 2 2 >1+n Find the greatest value of xyz for positive value of x.(i) 5 5 5 5n1 4 7 3n 5 3n 2 1 1 S= + 2 + 3 + . xy yz zx (x2 y2 z2)1/3 4 (x y z)2/3 xyz 8 3 Solved Example # 32 If a...... If a.. "try" & "should" with "I Will".. 5.. Let an and cn be two numbers Sol.M. x. 5 5 5 5 5 5n n 1 3 1 1 5 5 3n 2 4 S =1+ – 1 5 5n 1 5 then Sn = Successful People Replace the words like. R... # 28: If ai > 0 i N such that a i 1 .... .....P......n times = nk.. n (ii) k ar = k r1 r1 n k = k + k + k...... u3 – u2....... + (un – un–1) Where the series (u2 – u1) + (u3 – u2) + ..... .(i) S= u1 + u2 + .. 6 12 METHOD OF DIFFERENCE Type – 1 Let u1....... is either an A.MathsBySuhag.... where k is a constant.. or a G.....+ n = r1 n (v) ar .. + an 2) j i j 1 Solved Example # 37: Find the sum of the series to n terms whose general term is 2n + 1.TekoClasses................. Solution..(ii) (i) – (ii) un = u1 + (u2 – u1) + (u3 – u2) + .........5 n S = – ... then nth term un of this sequence is obtained as follows S = u1 + u2 + u3 + ..(ii) (i) ..... Evaluate 1.(ii) 1 (i) . Evaluate 1 + 3x + 6x 2 + 10x 3 + . Ineffective People don't.... .(i) xS = x + 2x 2 + 3x 3 + ....MathsBySuhag.... R...... 0 98930 58881.......... 5n1 16 4 Solved Example # 35: Evaluate 1 + 2x + 3x 2 + 4x 3 + .... Solution.. + (un – un–1) is = 1 2 Successful People Replace the words like.... n2 Important Results n (i) n (ar ± br ) = r1 n n ar ± r1 b r. 2100 Ans.... Maths : Suhag R. u2. (1 x )3 3... 16 ... 2 n Solved Example # 38:T k = k2 + 2k then find T k . 6 n Solved Example # 39: Find the value of the expression i j 1 i 1 j 1k 1 n Solution.Get Solution of These Packages & Learn by Video Tutorials on www.. upto infinity where | x | < 1. . be a sequence..+ n2 = r1 n (n 1) (2n 1) (vi) 6 n r3 = 13 + 23 + 33 +. to infinite terms for | br | < 1.. 1 S= (1 br )(1 r ) Self Practice Problems : 1.(ii) (1 – x) S = 1 + x + x 2 + x 3 + . Let S = 1 + 2x + 3x 2 + 4x 3 + ................ Sir). + 100.... n Tk = k 1 n k2 + k 1 k 2 k 1 n (n 1) (2n 1) 2(2n 1) = + 6 2 1 = n (n 1) (2n 1) + 2n + 1 – 2.....+ an )2 – (a12 + a22 + ..(i) rS = r + (1 + b) r 2 + . ......... Bhopal Phone : 0 903 903 7779. + un–1 + un ..com & www..........(ii) (1 – r)S = 1 + br + b2r2 + b3r3 + .. u3 .. Ans.... Sn = T n = (2n + 1) = 2n + 1 2(n 1) n = +n = n2 + 2n or n(n + 2).... 99.....22 + 3....... r1 n (iii) r² = 12 + 22 + 32 +....+ n3 = r1 n (n 1) 2 n 2 (n 1) 2 4 n (vii) 2 a a i = (a1 + a2 + . Solution..P...... + un ......com 1 3n 2 3 3 – × n 1 – 4 4 5 5n (12n 7) 12n 7 35 7 = – 4....2 + 2.(iv) r1 r = 1 + 2 + 3 +.. page 9 of 26 Teko Classes. FREE Download Study Package from website: www. Let S = 1 + (1 + b)r + (1 + b + b2) r2 +.... "wish".... "try" & "should" with "I Will".. such that u2 – u1.. .. or S = (1 x )2 Solved Example # 36 Evaluate 1 + (1 + b) r + (1 + b + b2) r2 + ... 1 1 1 + 2 1 + 3 1 n n Sum to n terms of the series 2 + ...... k 1 n Solution..... Ans..2101 + 2.: i n j 1 = i i 1 j 1 i 1 j 1k 1 n j = i (i 1) 2 i1 n n i2 i i 1 i 1 1 n (n 1) (2n 1) n (n 1) = 6 2 2 n (n 1) (n 2) n (n 1) = [2n + 1 + 3] = . upto infinite term where | x | < 1. Kariya (S....23 + . 1 2....... K..com =1+ . + Tn ..com either in A..... 4 ..... Maths : Suhag R.. Solution.. 2 . Kariya (S...... + T n–1 + Tn .... 3 ... 3 .. 3]. 5 + .....2 + 2.P.. 2 .. Now S = T r r 1 = T1 + T2 + T3 + ..com & www..+ Tn 1 1 1 [1 . Let S = 3 + 7 + 13 + 21 + .3 + 3........... FREE Download Study Package from website: www.....MathsBySuhag..MathsBySuhag. Bhopal Phone : 0 903 903 7779. 4] 3 3 3 1 1 Tn = [n(n+1) (n + 2) – (n – 1)n (n + 1)] S = n (n + 1) (n + 2) 3 3 Hence sum of series is f(n) – f(0)........ To express tr = f(r) – f(r–1) multiply and divide tr by [(r + 2) – (r – 1)] r so Tr = (r + 1) [(r + 2) – (r – 1)] 3 1 = [r (r + 1) (r + 2) – (r – 1) r (r + 1)]... 3 + 2 .............. then we can find un and hence sum of this series as S = ... + Tn 1 1 1 = x 1 x 1 (n 1)x = n (1 x )[1 (n 1)x] 4 5 6 Solved Example # 44 Sun to n terms of the series 1 ... 2 .. Let Tr be the general term of the series 1 1 [1 (r 1)x] (1 rx ) Tr = So Tr = x (1 rx)(1 (r 1)x ) (1 rx )(1 (r 1)x) T1 = 1 1 1 1 rx 1 ( r 1 ) x x Tr = f(r) – f(r + 1) S = Tr = T1 + T2 + T3 + . This can be explained with the help of examples given below... + (Tn – Tn–1) n 1 =3+ [8 + (n – 2)2] 2 = 3 + (n – 1) (n + 2) = n2 + n + 1 Hence S = (n2 + n + 1) = n2 + n + 1 n(n 1)(2n 1) n 2 n(n 1) = + +n = (n + 3n + 5) 6 3 2 Solved Example # 41 Find the sum to n-terms 1 + 4 + 10 + 22 + . 3 .P. "wish"..... (1 x )(1 2x ) (1 2x )(1 3 x ) (1 3 x )(1 4 x ) Solution.. 3 1 Let f(r) = r (r + 1) (r + 2) 3 n so Tr = [f(r) – f(r – 1)]...Get Solution of These Packages & Learn by Video Tutorials on www...(i) S= 1 + 4 + 10 + .. Let Tr = = = r 3 r(r 1)(r 2) 1 3 + (r 1)(r 2) r(r 1)(r 2) 1 1 1 3 1 = + r 1 r 2 2 r(r 1) (r 1)(r 2) 1 1 1 3 1 S= + 2 n 2 2 2 (n 1)(n 2) Successful People Replace the words like.. R. 0 98930 58881.. + Tn ...... 2 1 – 2n = 3................ Solution... Solution.. + Tn – Tn–1 ) 2n 1 1 Tn = 1 + 3 2 1 Tn = 3 .. Solution..... Solved Example # 42 Find the sum to n-terms of the series 1... Let S = 1 + 4 + 10 + 22 + ... "try" & "should" with "I Will".TekoClasses... T2 = [2 . Ineffective People don't.. + Tn–1 + Tn . 4 – 1 . 4 .(i) S= 3 + 7 + 13 + . 2n–1 – 2 So S = Tn = 3 2n–1 – 2 2n 1 = 3 ..2n – 2n – 3 Type – 2 If possible express rth term as difference of two terms as tr = f(r) – f(r ± 1). 3 – 0]..... 4 + 3 ..... K... 5 – 2 ... T3 = [3 .. Sir)..(ii) (i) – (ii) Tn = 3 + 4 + 6 + 8 + ..com k u r Solved Example # 40 Find the sum to n-terms 3 + 7 + 13 + 21 + ..(ii) (i) – (ii) Tn = 1 + (3 + 6 + 12 + . 1 1 1 Solved Example # 43 Sum to n terms of the series + + + . page 10 of 26 r 1 Teko Classes.......4 + .. Let Tr be the general term of the series So Tr = r(r + 1). or in G. ... Solved Example # 45 Find the nth term and the sum of n term of the series 2............. 6 .+Tn ... 1 1 1 4 3 (2n 1)(2n 3) (iii) 1 ... n (n + 1) (n + 8) (n + 9) 4 (iv) 4 + 14 + 30 + 52 + 82 + 114 + .MathsBySuhag....(ii) (i) – (ii) Tn = 2 + 10 + 24 + 44 + 70 + 102 + .............. 1.... Ineffective People don't.. + (Tn – Tn–1) ...... 3 . page 11 of 26 Teko Classes....MathsBySuhag....com = ....9 Ans........... Ans.... 36..(i) S= 9 + 16 + 29 + 54 + 103 + ... 252 Solution.. Let S = 9 + 16 + 29 + 54 + 103 + .... un – un–1 = Tn. Sum to n terms the following series 1 2 3 1 1 2 2n (i) Ans.(iv) n–2 n–2 12 24 . + (Tn–1–Tn–2) + (Tn – Tn–1) ..(ii) (i) – (ii) Tn = T1 + (T2 – T1) + (T3 – T2) + .... 10 + 3 .. n(n + 1)2 (v) 2 + 5 + 12 + 31 + 86 + ... So un = T1 + T2 + ..7......... Ans.....(i) S= 2 + 12 + 36 + 80 + 150 + 252 + ... + T n ..........TekoClasses.. FREE Download Study Package from website: www..+Tn–1 + Tn ..... Bhopal Phone : 0 903 903 7779. + (Tn–1–Tn–2) + (Tn – Tn–1) .. 3 + 3 3 + 3 1 2 3 n1 1 1 2 (ii) 1 1 1 + + + .5. + + 2n = 6(2n – 1) + 2 2 2 1 Self Practice Problems : 1.....com 3 1 5 1 5 – = – [2n + 5] 1 2 ( n 1 ) 2 ( n 1 )(n 2) 4 n2 4 Note : It is not always necessary that the series of first order of differences i..... 0 98930 58881............... Hence sum of this series is S = n3 + n2 2 2 n(n 1)(2n 1) n (n 1) n (n 1) = + = (3n2 + 7n + 2) 6 12 4 1 n (n + 1) (n + 2) (3n + 1) 12 Solved Example # 46: Find the general term and sum of n terms of the series 9...... R... in such case let u1 = T1 . u3 – u2 ... 11 + ...com & www.......(iii) Tn = 2 + 10 + 24 + 44 + 70 + 102 + ............ u3 – u2 = T3 .. Sir)....... then un = u1 + Tr Otherwise in the similar way we find series of higher order of differences and the nth term of the series... is always either in A......... Ans. u2 – u1 = T2 .... 29...(i) un = T1 + T2 + ..P.... or in G...... 3 3 + ... 54..P.....+ Tn–1 + Tn . 12. 103 Sol. the series (T2 – T1) + (T3 – T2) + ... n = [4 + (n – 1) 6] = n [3n – 1] = Tn – Tn–1 = 3n2 – n 2 general term of given series is Tn – Tn–1 = 3n2 – n = n3 + n2. or in G.... Let S = 2 + 12 + 36 + 80 + 150 + 252 + .. Maths : Suhag R... ...P........ 5 . (iii) – (iv) Tn – Tn–1 = 9 + (–2) + 6 = 7 + 6 [2 – 1] = 6(2) + 1. "try" & "should" with "I Will".... + (Tn – Tn–1) .........+ Tn ..... u2 – u1..(iv) (iii) – (iv) Tn – Tn–1 = 2 + 8 + 14 + 20 + 26 + . "wish". 5 3..P. + Tn–1 +Tn ......... With the help of following example this can be explained........Get Solution of These Packages & Learn by Video Tutorials on www.... 3n n2 n 1 2 Successful People Replace the words like. 9 + 2 .... + (Tn – Tn–1) is series of second order of differences and when it is either in A.(iii) Tn = 9 + 7 + 13 + 25 + 49 + .. + (Tn – Tn–1) Now....... K...e... 7.7 5........ un – un–1. Kariya (S.. 150..... 80........ ( n 2 ) terms General term is Tn = 6(2)n–1 + n + 2 Also sum S = Tn = 62n–1 + n + 2 n (n 1) n(n 5) (2n 1) =6......(ii) (i) – (ii) Tn = 9 + 7 + 13 + 25 + 49 + ... 16. (vi) tr = Sr Sr1 (vii) If a . b. Sum of an infinite GP when r < 1 when n rn 0 if r < 1 therefore.S = 1r If each term of a GP be multiplied or divided by the same non-zero quantity. a .. a. A2 = a + . c are in AP 2 b = a + c.com & www. (iii) The common difference can be zero. ar3. Kariya (S. c are in AP.. c MEANS ARITHMETIC MEAN :If three terms are in AP then the middle term is called the AM between the other two. c are in GP.. n n . b (n 1)(a b) If a.e. a n AM for any n positive number a1. nth term of this AP tn = a + (n – 1)d. a + 5d etc. Successful People Replace the words like. a – 3d. ..com T h e s u m o f t h e f i r s t n t e r m s o f t h e A P i s g i v e n b y ..... a2. "try" & "should" with "I Will". "wish".MathsBySuhag. ar. An = a + nd .. ar3 & so on.. b.. the nth term is tn = ab .. a2.ARITHMETIC MEANS BETWEEN TWO NUMBERS : If a. is a GP with a as the first term & r as common ratio. b are any two given numbers & a. then AP can be written as a. An.g. (iv) The sum of the two terms of an AP equidistant from the beginning & end is constant and equal to the sum of first & last terms. K. c are in HP b = 2ac or ac ab a = bc. .. If the sequence a1. . multiplied or divided by the same non zero number. b is AM of a & c . any 4 consecutive terms of a GP can be taken as a/r3.. e. Maths : Suhag R. ar4. a – d. Sir).. a/r. 1/a2.. . . GEOMETRIC PROGRESSION (GP) : GP is a sequence of numbers whose first term is non zero & each of the succeeding terms is equal to the proceeding terms multiplied by a constant . NOTES :(i) If each term of an A. where d = an – an-1. a.. Ar = nA where A is the single AM between a & b. so if a. if r 1 .. (vi) If a. a + 3d . Sn = ..... an is an HP then 1/a1. A = 1 2 3 .Get Solution of These Packages & Learn by Video Tutorials on www. a + 2 d. .. . (ii) Three numbers in AP can be taken as a – d .... 2 2 where l is the last term. S n = n n [2 a + (n – 1)d] = [a + l]. Ineffective People don't.. (i) nth term = a rn –1 (ii) Sum of the Ist n terms i. Thus in a GP the ratio of successive terms is constant.. four numbers in AP can be taken as a – 3d... is a sequence. R.e.. This fixed number is called the common difference... page 12 of 26 DEFINITION : A sequence is a set of terms in a definite order with a rule for obtaining the terms. 1/2 . A1 = a + n (b a ) ba 2 (b a ) . a + (n – 1)d. A1.TekoClasses..... c are in GP b2 = ac. . a + d. If a is the first term & d the common difference.. (v) Any 3 consecutive terms of a GP can be taken as a/r.. . . . Here we do not have the formula for the sum of the n terms of an HP... a3.MathsBySuhag. a + d.P. a r n 1 r 1 a ( | r | 1) . This constant factor is called the COMMON RATIO of the series & is obtained by dividing any term by that which immediately proceeds it. five numbers in AP are a – 2d . positive or negative. a a a . r1 GEOMETRIC MEANS : If a. a + d . b. where d = ba n 1 NOTE : Sum of n AM’s inserted between a & b is equal to n times the single AM between a & b n i.. Teko Classes. . = a + 2 d .. An = a + n1 n1 n1 =a+d. . is increased... For HP whose first term is a & second term (iii) (iv) is b. AN ARITHMETIC PROGRESSION (AP) :AP is a sequence whose terms increase or decrease by a fixed number. Bhopal Phone : 0 903 903 7779.. 0 98930 58881.. An are the n AM’s between a & b . b .. 1/n . HARMONIC PROGRESSION (HP) :A sequence is said to HP if the reciprocals of its terms are in AP.. . a + 3d... a – d. b. ar .... . a – d . a + d. A2. .. the resulting sequence is also a GP. Therefore a.com SHORT REVESION (SEQUENCES AND SERIES) . an is . ar2. decreased. (v) Any term of an AP (except the first) is equal to half the sum of terms which are equidistant from it. ar. a + 2d & six terms in AP are a – 5d. A2. ... FREE Download Study Package from website: www. then the resulting sequence is also an AP. a + d. . 1/3 . b are in AP then A1. b is the GM between a & c.. 1/an is an AP & converse. . 1 . Note that A. S = 1 r dr 1 r2 .. ARITHMETICO-GEOMETRIC SERIES : A series each term of which is formed by multiplying the corresponding term of an AP & GP is called the Arithmetico-Geometric Series. Remember that to find the sum of n terms of a series each term of which is composed of r factors in AP.. we “write down the nth term. are in GP. .. . 7 + 77 + 777 + . n-GEOMETRIC MEANS BETWEEN a..8 If the pth. Q.. Q. HARMONIC MEAN :If a..com Get Solution of These Packages & Learn by Video Tutorials on www.. Ineffective People don't.... Sir). G....n i. 1 + 3x + 5x2 + 7x3 + . Determine the value of the constant by applying the initial conditions”. b are two given numbers & a. (i) G² = AH (ii) A > G > H (G > 0)...2 Q. T2... r 1 Gr = (G)n where G is the single GM between a & b. G1 = a(b/a)1/n+1.. . x3 . "wish".5 If the 10th term of an HP is 21 & 21st term of the same HP is 10. THEOREM : If A. Gn = a(b/a)n/n+1 = ar .(ii) k ar = k ar. Maths : Suhag R...... b : If a... .. = ar² .. Q.MathsBySuhag... Then G1.. + [a + (n 1)d] rn1 SUM TO INFINITY : a n If r < 1 & n then Limit n r = 0 .9 If one AM ‘a’ & two GM’s p & q be inserted between any two given numbers then show that p3+ q3 = 2 apq ..3 Q. constitute an AP/GP... a > 0. divide by the number of factors thus increased and by the common difference and add a constant. qth & rth terms of an AP are in GP .. HM between a & b both being unequal & positive then.. Q.. 5. EXERCISE–1 Q...com & www. 3. Kariya (S. b.. Gn are n GMs between a & b . where r = (b/a)1/n+1 NOTE : The product of n GMs between a & b is equal to the nth power of the single GM between a & b Teko Classes... Tn are the terms of a sequence then some times the terms T2 T1. H are respectively AM.. pq Successful People Replace the words like.) n 30 2 (sum of the cubes of the first n natural numbers) (n + 1) (2n + 1) (3n² + 3n 1) METHOD OF DIFFERENCE : If T1.g. nth term of the series is determined & the sum to n terms of the sequence can easily be obtained. Here 1. Gn. Express the recurring decimal 0. b is the HM between a & c.. .. G1.6 Find the sum of the n terms of the sequence Q. are in AP & 1... GM. c are in HP. x.. = arn. up to n terms) = 2n ln 2 + 2 There are n AM’s between 1 & 31 such that 7th mean : (n 1)th mean = 5 : 9. affix the next factor at the end.. .TekoClasses..e. n (n 1) ln 3 Show that ln (4 × 12 × 36 × 108 × .. to n terms..com .. G2. G. c > 0. Bhopal Phone : 0 903 903 7779.. "try" & "should" with "I Will". FREE Download Study Package from website: www. T3 T2 .MathsBySuhag. G2 = a(b/a)2/n+1. G3 . r1 r1 r1 r1 r1 n (iii) k = nk . where k is a constant. therefore b = a c . page 13 of 26 b² = ac.. x2. Find the sum of the series . r1 RESULTS n (i) r= r1 n (n 1) 2 n (ii) r² = r1 n (iii) r1 n (iv) n (n 1) (2n 1) 6 (sum of the squares of the first n natural numbers) n n 2 (n 1) 2 r r3 = r 1 4 r4 = r1 (sum of the first n natural nos. the first factors of several terms being in the same AP.. e..1 1 2 4 1 1 1 2 2 1 2 2 4 3 2 1 3 34 . .7 The first term of an arithmetic progression is 1 and the sum of the first nine terms equal to 369.4 Q. Find the seventh term of the geometric progression. The first and the ninth term of a geometric progression coincide with the first and the ninth term of the arithmetic progression. then b = 2ac/[a + c]. Standart appearance of an Arithmetico-Geometric Series is Let Sn = a + (a + d) r + (a + 2 d) r² + . SIGMA NOTATIONS n THEOREMS :(i) n n n n (ar ± br) = ar ± br...1 576 as a rational number using concept of infinite geometric series.... T3.. .... H constitute a GP. ... K. b are in GP. R. . then find the 210th term. then find the value of n.. G2. Show that the common ratio of the GP is q r . 0 98930 58881. 8 an a r (1 r 2 n ) . If the first number is the same as the fourth .25 For or 0 < < /2. 3. The airthmetic mean A & the geometric mean G satisfy the relation 2 A + G² = 27... Q. z = cos2n sin2n then : Prove that (i) xyz = xy + z n0 (ii) xyz = x + y + z EXERCISE–2 Q.. Show that the sum of their nth terms is (m/2) (mn m + n + 1). A computer solved several problems in succession.. c . "try" & "should" with "I Will".. 127 min to solve all the problems except for the last one.5 Q. b .23 Sum the following series to n terms and to infinity : (iii) r1 i r (r + 1) (r + 2) (r + 3) r1 (iv) n n (a) 1 1..13 In an AP of which ‘a’ is the Ist term.. 6. 2 .8 rs 2r 3s where rs is zero if r s & rs is one if r = s.5 min to solve all the problems except for the first two? If the sum of m terms of an AP is equal to the sum of either the next n terms or the next p terms of the same AP prove that (m + n) [(1/m) (1/p)] = (m + p) [(1/m) (1/n)] (n p) If the roots of 10x3 cx2 54x 27 = 0 are in harmonic progression.. ... Maths : Suhag R. 1. 3 ...24 Find the value of the sum (b) n 1 1 1 . If their sum is S. 7 .22 The harmonic mean of two numbers is 4. Q. 8.. P the product & R the sum of the reciprocals of a GP .13 n 1 (i) s 1 j 1... which are in GP is S² . the first three are in GP & the last three are in AP . find the four numbers.. 18 are consecutive terms of a GP . if : x= n0 n0 cos2n . R..P. Sir). are a and b. . 0 98930 58881. 4)..com n Q. find the value of P 2 R . FREE Download Study Package from website: www. Q.12 The first and last terms of an A. 6n Q.18 In a set of four numbers. m . i 1 j1 k 1 Q. (ii) 6 + 13 + 22 + 33 + .MathsBySuhag. A new series is formed by multiplying each of the first 2n terms by the next term. b are consecutive terms of an AP & (iii) the numbers b . c . If there be m AP’s beginning with unity whose common difference is 1 . "wish". c respectively.3 1.4 The series of natural numbers is divided into groups (1). page 14 of 26 Teko Classes.. the same last term & the same number of terms .MathsBySuhag. 3) .11 If S be the sum . show that the sum of the next q terms is a (p + q) q/(p 1)..15 An AP & an HP have the same first term. Q.... Q. S Q.14(a) The interior angles of a polygon are in AP.20 Find the nth term and the sum to n terms of the sequence : (i) 1 + 5 + 13 + 29 + 61 + .1 Q.. prove that the product of the rth term from the beginning in one series & the rth term from the end in the other is independent of r. Ineffective People don't. b.. Find the numbers. If Sn represents the sum to n terms of a GP whose first term & common ratio are a & r respectively. Show that the sum of the numbers in the nth group is (n 1)3 + n3 .7 Q..5 . Find the two numbers. 2 3 Q. 4 4.. Q.. The sum of the squares of three distinct real numbers . Find the ratio of their nth term. show that x .10 .5 min to solve all the problems except for the first. Determine the number of sides of the polygon if its largest interior angle is 172°. (i) their sum is 25 (ii) the numbers 2. r1 n (ii) 4 r2 1 Q.6 Q. qth & rth terms a. The smallest angle is 120° & the common difference is 5°. and 31. 7. show that ² (1/3 . Q..17 Given that ax = by = cz = du & a ..Get Solution of These Packages & Learn by Video Tutorials on www. Q.19 Find the sum of the first n terms of the sequence : 1 2 1 1 3 1 1 4 1 1 . b .. How many problems were suggested to the computer if it spent 63.. (2.3.3 Q. u are in HP .6 4.10 7 . y = sin2n .6.16 Find three numbers a . K. The time it took the computer to solve each successive problem was the same number of times smaller than the time it took to solve the preceding problem. Show that a (b c) log a + b (c a) log b + c (a b) log c = 0. (5. then find c & all the roots. c between 2 & 18 such that .. 1) (1 .. . n n n Q... y ...com Q. Successful People Replace the words like.... 4 . with common difference 6. 9).. then prove that S1 + S3 + S5 + . Show that the sum of the new series is (4n 2 1)(a 2 b 2 ) (4n 2 2)ab .21 The AM of two numbers exceeds their GM by 15 & HM by 27 .2 Q.com & www... (b) The interior angles of a convex polygon form an arithmetic progression with a common difference of 4°. + S2n-1 = Q.. Kariya (S. d are in GP.. 1 r (1 r ) 2 (1 r ) A geometrical & harmonic progression have the same pth. if the sum of the Ist p terms is equal to zero .. a. 7 4 .. & so on.10 The sum of n terms of two arithmetic series are in the ratio of (7 n + 1) : (4 n + 27) . z . Find the number of sides of the polygon.. There are altogether (2n + 1) terms. Bhopal Phone : 0 903 903 7779.TekoClasses. 2 2 2 23 2 4 If there are n quantities in GP with common ratio r & Sm denotes the sum of the first m terms. b is an AP or HP .+ a2n S3 = a2n+1 + a2n+2 + . Sir).. c be in GP & logc a... e are in HP then: (i) Prove that a. z . n3 (n 1)3 + .. c... and hence solve the equation x3 – 12x2 + 39x –28 = 0. show that the difference between the first & the last mean is equal to ac(a c) ... FREE Download Study Package from website: www. (iii) If a = 2 & e = 18 . Q. x2... Prove that : = 1 2 1 1 1 .. c are the first three terms of a geometric series. Find the number.. 3... The real numbers x1.. Q. e are in AP. + an S2 = an+1 + an+2 + .. a1 a n a1 a 2 a3 an Q... S1 = a1 + a2 + a3 + .14 If n is a root of the equation x² (1 ac) x (a² + c²) (1 + ac) = 0 & if n HM’s are inserted between a & c.. "wish". and let r & s be the roots of the equation x2 18x + B = 0 .. Kariya (S...3 3.13 The sequence a1.... "try" & "should" with "I Will"........... d .. a2.24 In a GP the ratio of the sum of the first eleven terms to the sum of the last eleven terms is 1/8 and the ratio of the sum of all the terms without the first nine to the sum of all the terms without the last nine is 2... 1 a 1 1 a 1 1 a 2 1 a 1 1 a 2 1 a 3 Q..15 (a) The value of x + y + z is 15 if a . an = an-1 + Q. . c . ..... y ..22 If a1 = 1 & for n > 1 .. 0 98930 58881.. x . 1 2x 3 x2 .... (1/x)+(1/y)+(1/z) is 5/3 if a ...2 Q. an be an AP ... Prove that the sequence S1 .7 7.P..9(a) Let a1.10 Let Let a1 ...MathsBySuhag. d are in GP & c. z ....P....21 Q. 1] x = 1+ 3a + 6a² + 10a3 + ... a2.. x .. e be 5 numbers such that a.... page 15 of 26 (b) Show that in any arithmetic progression a1...23 ... c. Q. logb c..... 2. then show that the common difference of the AP must be 3/2.. a... c are in HP.. be an A.. c.9 ..P . ...... K..4 Q.. a2. [ JEE ’96.... b ..... + ( 1)n 1 l3 = ______ . Now if we subtract four from the hundred's digit of the initial number and leave the other digits unchanged. + a²2K 1 a²2K = [K/(2 K 1)] (a1² a²2K) ... Find the intervals in which and lie .... If p < q < r < s are in arithmatic progression.16 An AP .Get Solution of These Packages & Learn by Video Tutorials on www.. b are in HP ..com 1 1 1 1 ............ b.18 Q.5 For any odd integer n 1.. a1² a2² + a3² a4² + .... Q.. k 1 Q.. If we subtract 792 from it...... x 1 (x 1) (x 2) (x 1) (x 2) (x 3) a1 a2 a3 ..19 Q. we get a number written by the same digits in the reverse order . find all possible values of b . a3 . y . b1 & c1 are in GP... c. Q..1 Q. Bhopal Phone : 0 903 903 7779. d. b. in terms of x & y. an .... Find the number of terms in the GP. e are in GP .. (ii) Prove that e = (2 b a)²/a ..... show that a1 .12 If a. 3] Let p & q be roots of the equation x2 2x + A = 0.. and B = ______... Ineffective People don't. EXERCISE–3 Q.... b/b1 ...... a98 satisfies the relation an+1 = an + 1 for n = 1.. Show that their (n + 2)th terms will be 2 n 2 a 2 n2 n1 . then show that 12 < a75 < 15.3 Q. is an arithmetic progression whose common difference is n2 times the common difference of the given progression.. x ..17 Prove that the sum of the infinite series Q. x3 satisfying the equation x3 x² + x + = 0 are in A..... loga b be in AP . in GP if b n ba b 2n a 2 n 1. c are in AP ......MathsBySuhag... [JEE ’96... d. S3 .P.. ..... z ....11 If a.. If the harmonic mean of a & b is 12 and that of b Successful People Replace the words like.... a2 . d. show that the sum of the products of these m terms taken two & two together is [r/(r + 1)] [Sm] [Sm 1] .. then A = ______..... Q. b..com Q.25 Given a three digit number whose digits are three successive terms of a G.TekoClasses..P.. Q.. R........ If a . ..... . Evaluate a 2k .. a < 1 y = 1+ 4b + 10b² + 20b3 + .. If ax2 + 2bx + c = 0 & a1x2 + 2 b1x + c1 = 0 have a common root & a/a1 .. we get a number whose digits are successive terms of an A. Show that e = ab²/(2a b)² .. b are in AP while the value of . c/c1 are in AP..97 and has 49 the sum equal to 4949.. an+1 ... b < 1. Find the values of a & b assuming them to be positive integer .... d are in GP & c..+ a3n .. a GP & a HP have ‘a’ & ‘b’ for their first two terms ... Teko Classes. Find a & b . b. Maths : Suhag R. (b) The values of xyz is 15/2 or 18/5 according as the series a . find S = 1+ 3ab + 5(ab)² + .. a3.. a1 a n a 2 a n 1 a 3 a n 2 a n a1 ... Find the condition that the roots of the equation x3 – px2 + qx – r = 0 may be in A... a3 .. b...23 Sum to n terms : (i) (ii) 1 a n 1 .. b.5 5.com & www.20 Q.......... . S2 .... y ....... be positive real numbers in G. b2. A2 . (B) in A.P. r= 4 3 (C) a = .. then n equals (A) 10 (B) 12 (C) 11 (D) 13 (c) Let the positive numbers a..12 Solve the following equations for x and y log2x + log4x + log16x + . Q. 6 ] Select the correct alternative(s).P. are in : 1 n x 1 n y 1 n z (A) AP (B) HP (C) GP (D) none of the above (c) Prove that a triangle ABC is equilateral if & only if tan A + tan B + tan C = 3 3 ...com & www.. Bhopal Phone : 0 903 903 7779.. Kariya (S. .P. . –32 (B) –2.. [ JEE '98.. 61.. b. Hn.. be in A. G2.... find the first five terms of the series.. A2 ... a2 .P. 5 out of 60 ] H1 H 2 H1 H 2 9ab (A) (B) (C) Successful People Replace the words like... for r = 1.. H1... Q... a2... FREE Download Study Package from website: www../H. Mains..M. .com Q. & are in H. z > 1 are in GP. b be positive real numbers. a .. c.....P.... then Tmn equals : n m 1 1 1 (A) (B) mn m n (b) If x = 1. page 16 of 26 Tm = Teko Classes.... Screening.P.7(a) The harmonic mean of the roots of the equation 5 2 x2 4 5 x + 8 + 2 5 = 0 is (A) 2 (B) 4 (C) 6 (D) 8 (b) Let a1. b are in H. b. of G1. y > 1.MathsBySuhag.. are (A) –2. and a2... find all possible values of the common ratio. .. . If the inverse of its common ratio is an integer. abd... a2 ..11 The sum of roots of the equation ax2 + bx + c = 0 is equal to the sum of squares of their reciprocals. If the sum is 4 and the second term is 3/4. Ineffective People don't... ..... 2. For each n..... c are in A.MathsBySuhag. n and the first terms of the series. a3. .. then : 7 . 3 Q.. bcd are (A) NOT in A.. 3 (D) –6.. 59.. a2. "wish".. the arithmetic mean... . "try" & "should" with "I Will". 2 + 2 + 8 ] Let Tr be the rth term of an AP...10 Given that . .? [ REE 2001. Find whether bc2.P. d are positive real numbers such that a + b + c + d = 2. be respectively.. = y 5 9 13.. n we have 1 1 & Tn = .6 (a) ..Gn in terms of A1... r = 4 (B) a = 2 .. A1 . 5 out of 100 ] Q. b are in A. Find an expression for the G.14(a) Suppose a.. acd..com & c is 36.. show that G1 G 2 A A2 ( 2a b ) ( a 2 b ) 1 [ JEE 2002 . 4 out of 100 ] Q./G.13(a) Let be the roots of x2 – x + p = 0 and be the roots of x2 – 4x + q = 0.. then M = (a + b) (c + d) satisfies the relation : (A) 0 M 1 (B) 1 M 2 (C) 2 M 3 (D) 3 M 4 (c) The fourth power of the common difference of an arithmetic progression with integer entries added to the product of any four consecutive terms of it .P. are roots of the equation... d be in A. [ REE 2000. or H.P. (4y 1) = 4log4x [ REE 2001. 2. b. (2y 1) Q.. (C) in G...Get Solution of These Packages & Learn by Video Tutorials on www.P. Then abc. [ JEE 2000. If a .... 5 out of 100 ] 1 3 5..... . H2. 3 out of 100 ] Q.. 57. then the value of 2 a is 1 1 1 1 1 1 (D) 2 2 2 2 2 3 2 3 (b) Let a..P. r = 2 (A) a = 3 7 1 2 3 8 1 (D) a = 3 ... a10.. –32 (b) If the sum of the first 2n terms of the A. [ REE '98.an.P.P.P. If for some positive integers m. 0 98930 58881. 3 (C) –6. h2. Sir)...... H1 .P.. Mains . Gn..P.P.. r = (b) If a.. Maths : Suhag R.P.. & h1..P.P.. K. A x2 4 x + 1 = 0 and . R. Prove that the resulting sum is the square of an integer. If a1 = h1 = 2 & a10 = h10 = 3 then a4 h7 is: (A) 2 (B) 3 (C) 5 (D) 6 Q.P.Hn..TekoClasses.. 8. [JEE 2001... b are in G. then (C) 1 (D) 0 1 1 1 . 1 + 1 + 1 out of 35 ] (d) Let a1. c2 are in G. find values of A and B. 3... If are in G. ca2 and ab2 in A. B x2 6 x + 1 = 0. let An. h10 be in H. geometric mean and harmonic mean of a1..An. ... 5...8 The sum of an infinite geometric series is 162 and the sum of its first n terms is 160. such that .. the roots of the equation.P.. (D) H.9(a) Consider an infinite geometric series with first term 'a' and common ratio r . and a . then the integral values of p and q respectively. a1 . c. H2 .. G... If a < b < c and a + b + c = ... is equal to the sum of the first n terms of the A.. [JEE 2005 (Mains)... If p is positive. is 121 ( 3 + 1) (C) 243 ( 3 + 1) (D) 243 ( 3 – 1) 2 If a1. b 1 is equal to: r 1 2 (A) 1 (B) 3/4 (C) 4/3 (D) none If a.P.. then prove that either a = b = c or a. 2 Q.. 1 1 p (1 p) 2 -.. and h is the harmonic mean of a 2 a 2n 1 an a n 1 a1 a 2n a and b. form a G. then g g + g g + .M. . 14. Maths : Suhag R. R. then the sum of the series (sin d) [cosec a1 cosec a2 + cosec a2 cosec a3 + . then value of 1 + + + . a/2. ) a 2. Find n. and the runs scored in (b) If total number of runs scored in n matches is 4 the kth match are given by k·2n+1– k. G & H are respectively the A..TekoClasses... & H. a3...M. 3.MathsBySuhag. & c is given by: (A) x 3 3 Ax 2 + 3 G 3x G 3 = 0 (B) x 3 3 Ax 2 + 3 (G 3/H)x G 3 = 0 3 2 3 3 (C) x + 3 Ax + 3 (G /H) x G = 0 (D) x 3 3 Ax 2 3 (G 3/H) x + G 3 = 0 4.. and a.. 12. of positive terms.. b are in A. (C) H. then a. c2 are in H. g2.P..P. FREE Download Study Package from website: www.. "wish". with common difference d 0. If 10.. + cosec an–1 cosec an ] (A) sec a1 – sec an (B) cosec a1 – cosec an (C) cot a1 – cot an (D) tan a1 – tan an Sum of the series S = 12 – 22 + 32 – 42 + . – 20022 + 20032 is (A) 2007006 (B) 1005004 (C) 2000506 (D) none of these 1 5 1 1 3 2n 1 If Hn = 1 + + + .. then prove that [(1 + a) (1 + b) (1 + c)]7 > 77 a4 b4 c4..P. 0 98930 58881. 5] (C) [5. Q..15 If a.P. a2n .P. then 2 + 2 + 2 +. (A) 2/12 (B) 2/24 The sum to 10 terms of the series (A) 11..... if = b2 – 4ac and + ..P.P. + is 3 3 2 n 2 n (A) 2n – Hn (B) 2n + Hn (C) Hn – 2n (D) Hn + n 1 1 1 1 The sum of the series log 4 + log 4 + log 4 + . is: 1 p (1 p) 2 (1 p) 3 1 2 1 1 1 1 1 + + +. 2 + 2.... a2 . c are positive real numbers... Kariya (S. are the roots of ax2 + bx + c = 0.. 13.. b.MathsBySuhag. 2n n (B) 2nh (C) nh (D) h h One side of an equilateral triangle is 24 cm. b. b.. 8. b.. then 'a' must lie in the interval: (A) [1.P.. 25x + 25–x form an A. b are in G... the numbers 51+x + 51–x.P. are in G. 3 + 3 are in G.+ . & c. G. an are in A.... g3.P..2 Q. + log 4 is 2 4 8 2n (A) 121 ( 6 + (A) 1 n (n + 1) 2 2) (B) (B) page 17 of 26 c Q. Bhopal Phone : 0 903 903 7779.18(a) In the quadratic equation ax2 + bx + c = 0. where 1 k n. (A) 1/2 (B) 3/4 (C) 1 (D) none of these In a G. b. is (A) 2 cos 18° (B) sin 18° (C) cos 18° (D) 2 sin 18° 9..16 The first term of an infinite geometric progression is x and its sum is 5. . 5] (B) [2. c are in A.. Sir). upto = .g2n .. 1n 1 and Bn = 1 – An. (D) none of these If A. b2 .P.. g1. Then the sum of the perimeters of all the triangles is (A) 144 cm (B) 212 cm (C) 288 cm (D) none of these 7.. then (A) 0 (B) b = 0 (C) c = 0 (D) = 0 n 1 n+1 (2 – n – 2) where n > 1.. K.. = 2 2 2 6 1 1 2 3 3 5 2 + (C) 2/8 6 + 18 + (D) none of these 54 + ... The midpoints of its sides are joined to form another triangle whose mid points are in turn joined to form still another triangle. any term is equal to the sum of the next two terms...M. Teko Classes. c 1 1 1 If x > 1 and . This process continues indefinitely. 6] EXERCISE–4 Part : (A) Only one correct option 1.. b... 2] 1 n (n + 1) (2n + 1) 12 1 (C) n (n 1) (D) 1 n (n + 1) 4 Successful People Replace the words like.... (B) G.P. "try" & "should" with "I Will". [JEE 2006.. The sum r2 5. where . of three positive numbers a. then find the minimum natural 4 4 4 4 number n0 such that Bn > An.com & www. a2. then the equation whose roots are a....19 3 n 3 3 3 3 If A n . + g g is equal to 2 2n 1 1 2n n n 1 6.P.. a2. The common ratio of the G. a1...com Get Solution of These Packages & Learn by Video Tutorials on www.. c are in x x x (A) A. If x R.17 If a. n > n0.com . then the sum to infinity of the series... 12] (D) [12... Ineffective People don't. Then (A) 0 x 10 (B) 0 < x < 10 (C) –10 < x < 0 (D) x > 10 Q. a3.. the value of r for which the inequality b3 > 4b2 – 3b1 holds is given by (A) r > 3 (B) r < 1 (C) r = 3..MathsBySuhag. an are positive real numbers whose product is a fixed number c.. If n is a root of the equation x² (1 ac) x (a² + c²) (1 + ac) = 0 & if n HM’s are inserted between a & c. The sum of the squares of three distinct real numbers. are in H. (2.2 1. S2. p and common ratios Successful People Replace the words like.....P. a + b c are in A.. If a. K. 2 2 2 bc ca a b (A) 24.. 7.TekoClasses... S2.. c are in A.. If b1. (c + a) abc . c are in H.. 3.. (i) (a2 b2). 10. ( x an 1 ) x( x a1 ) .P.. Find the two progressions..... Ineffective People don't. . their cubes respectively. their squares. 5.... c2 (a + b) are also in A.P. The tangent of the smallest angle is 2 2 5 1 5 1 (A) (B) (C) (D) 51 5 1 2 2 Sum to n terms of the series S = 12 + 2(2)2 + 32 + 2(42) + 52 + 2(62) + .com 15.an (A) n n n n {an } and {bn } are two sequences given by an = ( x )1/ 2 + ( y )1/ 2 and bn = ( x )1/ 2 – ( y )1/ 2 for all n N. d are in G. Bhopal Phone : 0 903 903 7779...(ii) b + c a. 7). a2.M. 3). (ii) 1 2 a b 3.P. The value of a1a2a3. c are positive real numbers Find the sum in the nth group of sequence.. . is 1 1 (A) n (n + 1)2 when n is even (B) n2 (n + 1) when n is odd 2 2 1 2 1 (C) n (n + 2) when n is odd (D) n(n + 2)2 when n is even. Let S1.. then r 1 21. (b + c) .. 7.. 11. 9).P. which are in GP is S². are in H. (4. then show that: (i) a2 (b + c).. 2 (iii) (a + b) ... Using the relation A.. Get Solution of These Packages & Learn by Video Tutorials on www..P. 2 .com Teko Classes.a3 a1a 2 . In a circle of radius R a square is inscribed...P. 1 2 c d2 are in G. b2..P... upto ) loge 4} satisfies the quadratic equation y 2 – 20y + 64 = 0 the find the value of x. 4. then nq – p2 is (A) independent of ‘a’ (B) independent of ‘n’ (C) dependent on ‘a’ (D) none of these x x( x a1 ) x( x a1 )( x a 2 ) Sum of n terms of the series 1 + + + + .M. S3 are the sums of first n natural numbers.17. "wish". Maths : Suhag R.(x a n1 ) x( x a1 ). If a.P. If p and q are respectively the sum and the sum of the squares of n successive integers beginning with ‘a’. b2 (c + a). c + a b. if 0 < < (ii) (x 2y + y2z + z2x) (xy2 + yz2 + zx 2) > 9x 2 y2 z2.. b. 1) (1. c – 1 are in A... 4 4 If a. page 18 of 26 16. ... (2.. 8. (5.. (ii) (1).. If their sum is S. "try" & "should" with "I Will".. .P...MathsBySuhag. + an – 1 + 2an is [IIT . (D) c/a is an integer The sides of a right triangle form a G. . then the minimum value of a1 + a2 + a3 + .an 1 a1a 2 .. with the first terms 1. ..... FREE Download Study Package from website: www.. If S1 . 2. 22.com & www... (b2 c2).Sp denote the sum of an infinite G.2002. 9.... Find the limit of the sum of areas of all the circles and the limit of the sum of areas of all the squares as n . (A) a + c = b + d (B) e = 0 (C) a. The first term of the AP is equal to the common ratio of the GP & the first term of the GP is equal to the common difference of the AP. 4). prove that (i) tan + cot 2 . Sir). b. 8. c. If a1. if a.. 1 2 b c 2 . b3 (bi > 0) are three successive terms of a G.P..P.... 0 98930 58881.P. b.an is equal to xy xy xy (A) x – y (B) (C) (D) bn bn bn 19. 23.. .. 6... c are in G. show that ² (1/3.. then a circle is inscribed in the square.. (B) = + bc b ba bca ca b a bc b b b a b c (C) a . up to 2 + 3 + (13) (13) (13)4 13 If 0 < x < and the expression exp {(1 + cos x + cos2 x + cos3 x + cos4 x + . then: 2 1 1 a b c . b – 2/3.. ( x a n ) (B) (C) (D) none of these a1a 2 ...P.P. . 9. (8. (D) . b.. 15).. is a1 a1a 2 a1a 2a 3 ( x a1 )( x a 2 ). show that the difference between the first & the last mean is equal to ac(a c). G.. (i) 1. 55 555 5555 5 Find the sum of the series + + . 5. 6. with common ratio r.. (c2 d2) are in G. R. Kariya (S... 3] (A) n(2c)1/n (B) (n + 1) c1/n (C) 2nc1/n (D) (n + 1)(2c)1/n Part : (B) May have more than one options correct 18.. 6. 3). n 20. then S3 (1 8S1 ) is equal to S 22 (A) 1 (B) 3 (C) 9 (D) 10.5 (D) r = 5... prove that : EXERCISE–5 2. The sum of the first ten terms of an AP is 155 & the sum of first two terms of a GP is 9. a new square in the circle and so on for n times. If r(r 1) (2r + 3) = an 4 + bn3 + cn2 + dn + e.. (1 a n ) EXERCISE–3 Q 1. 1 Q 3.9 + 2 2 +. FREE Download Study Package from website: www.. b..13 2499 Q 15. find values of A and B. n (n + 1)/2 (n² + n + 1) Q 7.(2n )(2n 2) 2 Q 24... a = 5 . If 's' be the sum of 'n' positive unequal quantities a. 14. (3. 7 Q 23. 20.com 12. [IIT. 0 98930 58881. (iii) b = 4 . 2n+2 4 3n (ii) n² + 4n + 1 . b = 8 . 120 .. 4 ...3. 3/2 . 8) Q 19.7 24 + 2 5 . Show that S 1 + S2 + . d = 9 OR b = 2 . .. 27 Q 10. 8. satisfy the system of equations u + 2v + 3w = 6. then prove that 18. (1/6) n (n + 1) (2n + 13) + n Q 21..2000...5 16 + 2 2 3 . 35/222 Q 6.. b = 9 OR b = 1 . (ii) (a + b + c)3 > 27abc.7 2 + 1 7 . Ineffective People don't.. 216..6... 648 Q 7.. + Sp = . 3 Q 23. (a) 1 xn (x 1) (x 2) .com & www.. "try" & "should" with "I Will". n = 38 (b) 1 Q 25. roots are 1..9 Let a.. . 4] If a. 8 problems . b. (a) 9 . If a. page 19 of 26 1/2. Maths : Suhag R. (b) a = 1 .. 1/(p + 1) respectively. v. S = (7/81){10n+1 9n 10} Q 5. (1/27) Q 5..com 1 p(p + 3) 2 Circles are inscribed in the acute angle so that every neighbouring circles touch each other. (a) B (b) D Successful People Replace the words like. 30 Q 22. 4. Given that are roots of the equation. b & c are in arithmetic progression and a 2. w.. (a) C (b) B Q3. 4] ANSWER KEY EXERCISE–1 Q 1....Get Solution of These Packages & Learn by Video Tutorials on www.. s s s n2 . 10] The fourth power of the common difference of an arithmetic progression with integer entries added to the product of any four consecutive terms of it. (14 n 6)/(8 n + 23) Q 11. c.. 1/3. Sum the following series to n terms and to infinity: n (i) r (r + 1) (r + 2) (r + 3) r1 1 (ii) 2 4 11 1 32 + 2 2 12 2 4 + 3 2 13 3 4 +. K. (a) a = 1 . (i) sn = (1/24) [1/{6(3n + 1) (3n + 4) }] . 3. (x n) Q 24.. Prove that the resulting sum is the square of an integer. c d be real numbers in G.1999.(2n 1)(2n 1) (iv) Sn = 2 . 72.. 24.. (b) 12 Q 16..P. µ = 14 Q 4. R... (iii) (a + b + c)3 > 27 (a + b – c) (c + a – b) (b + c – a) 16. 7 . a = 9 .. The airthmetic mean between m and n and the geometric mean between a and b are each equal to 13. Sir). mn 15. (iii) 1 3. B x 2 6 x + 1 = 0.. S = 1 2.5. Kariya (S. 1 Q 14.. 931 1 (1 a 1 ) (1 a 2 ) .5 + 1 5. 2 . 127. c = 6 . c are positive real numbers then prove that (i) b2c2 + c2a2 + a2b2 > abc (a + b + c). 3/5) Q 12.. d = 18 Q. n² Q 20. either a = b = c or a.4.MathsBySuhag. ma nb : find the m and n in terms of a and b. c = 12 Q 18. 19. If u.. 2 [IIT – 2003.8 C = 9 .. (1/3) . (a) (6/5) (6n 1) (b) [n (n + 1) (n + 2)]/6 (iii) n/(2n + 1) EXERCISE–2 Q 6. 4u + 5v + 6w = 12 6u + 9v = 4 then show that the roots of the equation 1 1 1 x 2 + [(b – c)2 + (c – a)2 + (d – b)2] x + u + v + w = 0 and u v w 20x 2 + 10 (a – d)2 x – 9 = 0 are reciprocals of each other.MathsBySuhag.. 77 Q 6.TekoClasses.19 2p3 – 9pq + 27r = 0. c = 6 . (i) 2n+1 3 .... (8 . 1 4 (2n 1) (n + 1)² Q 2.P. b = 3 or vice versa Q. b.. Bhopal Phone : 0 903 903 7779. Teko Classes. b2 & c2 are in harmonic progression. such that . If the radius of the first circle is R then find the sum of the radii of the first n circles in terms of R and . 6 . the roots of the equation. b & c are in geometric progression. & are in H..5 minutes Q. "wish".. S = 1 ab (1 ab) 2 Where a = 1 x–1/3 & b = 1 y–1/4 Q 4. then . A x 2 4 x + 1 = 0 and . s = 1/24 (ii) (1/5) n (n + 1) (n + 2) (n + 3) (n + 4) 1 1. [IIT . sa s b sc n 1 17. ...... A ABCD 8.. C 12. A 11. A = 3 . (ii) n (n 1) 2 ( n 2 n 1) ..com Q 12.. (i) 2n 2 (2n + 2n 1 1) (ii) (n 1)3 + n3 6.. 2 3 3 EXERCISE–5 4. ABCD 24..16 B Q. A 17.. Maths : Suhag R.. Bhopal Phone : 0 903 903 7779. A..MathsBySuhag.. (a) D Q.) 9. m = 4 R2 7. (a) D (b) A Q 10. Ineffective People don't. An H1. (b) C. a = 144/180 OR r = ± 1/3 . (i) (1/5) n (n + 1) (n + 2) (n + 3) (n + 4) (iii) page 20 of 26 Q 11. R. 20..TekoClasses. (3 + 6 + 12 +. 18. 19. x = 2 2 and y = 3 1 Q 13. n R 1 sin 2 1 sin 2 1 12. a = 160 Q 9. B 5.. B 4... BC 22. r = ± 1/9 .. A 13.). (b) n = 7 EXERCISE–4 1. B = 8 2b a 2a b . H n 2 n Q14. A 21.n= a b a b 17. ABCD 7. "wish". s = 1 2 4 n ( n 3) n + 3 (2 n 3) 9 (2 n 3)2 Successful People Replace the words like. FREE Download Study Package from website: www.MathsBySuhag. D 2. A = 3.. n = 2 . Teko Classes. C 16.. K. A 3.. 1 sin 2 2 sin 2 65 36 13. 14.P. Q... n = 1 . AB 23.Get Solution of These Packages & Learn by Video Tutorials on www. . A 14.. 2 ..19 n0 = 5 C A 10. D 15. (d) A1 ... n = 4 .. 0 98930 58881. a = 108 OR r = 1/81 . (c) D . "try" & "should" with "I Will". B = 8 .18 (a) C. H2 . B 8. (2/3 + 25/3 + 625/6 +. D C 9. A2 . 2 R2..com & www... Sir). (a) A. Kariya (S.com Q 8.. A 6.