Sequence and Series Solved Problems

SEQUENCE AND SERIES SOLVED PROBLEMS Ex.1 Sol. If for an A.P. T3 = 18 and T7 = 30 then find S17 Let first term = a, common difference = d Then T3 = a + 2d = 18 and T7 = a + 6d = 30 Solving these , a= 12, d= 3  Ex.2 Sol. S17 = 17 [2a + (17–1)d] 2 = 17 [24 + 16 x 3] = 612 2 Find The sum of integers in between 1 and 100 which are divisible by 2 or 5 Required sum = ( sum of integers divisible by 2) + (sum of integers divisible by 5) – (sum of integers divisible by 2 and 5) = (2 + 4+ 6 +....+ 100) + ( 5 + 10 + 15 +.....+ 100)– (10 + 20 +....+ 100) = 50 20 [2 × 2 + (50 – 1) x 2] + 2 2 [2 × 5 + (20 – 1) × 10] – 1 + a  a + ...+ a1  a 2 2 3 Sol. Sol. = + a1  a 2 a2  a3 a2  a3 +...+ Ex.5 10 2 If the sum of first 6 terms of a G.P. is nine times of the sum of its first three terms, then find its common ratio     Ex.6 = = = –  1  an d d (n  1)d d an  a1 n 1 a n  a1  =1  1 – r6 = 9 ( 1– r3 ) 1+ r3 = 9 r=2 ( r  1) If pth, qth and rth terms of an A.P. are equal to are respectively x,y,z, then find xy–z. yz–x. zx–y 1 Sol. Let first term of an A.P. be a and c.d. be d and first term of a G.P. be A and c.r. be R, an1  an then a + (p–1) d = ARp–1 = x  a n1  a n a n1  a n  ...(1) a + (q – 1) d = AR =y q – 1 = (y–a)/d ...(2) a + ( r– 1) d= ARr–1 = z a n  a1  a n  a1 [ an = a1 + (n – 1)d] p – 1 = (x– a) /d q–1 Ex.7   corresponding terms of a G.P. and these terms  a 1  a 2  a 2  a 3  ..... a n    –  d   a  a 1 r 3 a 1 r 6 =9 1 r 1 r (Note) = 1 1 1 , , are in A.P.. bc ca ad [dividing by (a + b) (b + c) (c + a)] Let d be the c.d. of the A.P. Now L.H.S. a1  a 2  a2 , b2, c2 are in A.P.  a2 + ab + bc + ca, b2 + bc + ca + ab, c 2 + ca + ab + bc ..... are also in A.P. [adding ab + bc + ca] or (a+c) (a+ b), (b+ c) (a+b), (c+ a) (b+ c) .. are also in A.P. or If a1, a2, a3.....an are in AP where ai > 0 then find the value of 1 If a2,b2, c2 are in A.P. then prove that 1 1 1 , , are in A.P.. bc ca ad Sol. [2 × 10 + (10–1) × 10] = 50 [2 + 49] + 10 [10 + 95] – 5 [20 + 90] = 51 × 50 + 105 × 10 – 110 × 5 = 3050 Ex.3 Ex.4 Sol.  r – 1 = (z–a) / d ...(3)  Given expression = (ARp–1)y–z, (ARq–1) = A0 R (p-1)(y-z)+(q-1)(z-x)+(r-1)+(x-y) = A0R[(x–a)(y-z)+(y-a)(z-x)+(z-a)(x-y)]/d , (ARr–1)x–y z–x =1 If x, y, z are in G.P. and ax = by = cz then prove that logb a = logcb x,y,z are in G.P. y2 = xz ...(i) We have, ax = by= cz = ( say)  x log a = y log b = z log c = log  log  log  log   x = log a , y = log b , z = log c SEQUENCE AND SERIES putting x,y,z in (i) , we get S = 1 + 2x + 3x2 + 4x3 + ....+ n xn–1  xS = x + 2x2+ 3x3 +....+ (n–1) xn–1 + nxn  S – xS = 1 + [x + x2 + ...+ xn–1] –nxn 2 log  log   log     = log a . log c  log b  Ex.8 (log b)2 = log a. log c or loga b = logb c  logb a = logcb  S (1–x) = If a, b, c, d are in G.P., then prove that  S (–1/n) = – n[1–(1+1/n)n] – n (1+1/n)n (a3 + b3)–1, (b3 + c3)–1, (c3 +d3)–1 are in G.P. Sol. Let b = ar, c = ar2 and d = ar3 . Then, 1 3 a b and, 3 = 1 3 3 , 1 a (1  r ) b  c 1 3 c d 3 = 3 3 = 1 3 3 a r (1  r 3 ) 1 If x,y,z are in A.P. and x,y, t are in G.P. then prove that x, x– y, t– z are in G.P. Sol. 2y = x + z or 2xy = x2 + xz  x2 – 2xy = – xz S = n2 terms of a G.P. Sol. a + b + c = 25 and b,c, 18 are in G.P.  c2 = 18 b ...(3) ...(1) Eliminating a and b from (1) ,(2) and (3), gives the following equation ...(2) c2 + 12 c – 288 = 0 or (x2 – 2xy + y2 ) = – xz + xt (c– 12) (c+24) = 0 c = 12, – 24 or (x–y)2 = x (t– z) [Leaving c = – 24 because this is not in x, x–y, t– z are in G.P. between 2 and 18] Ex.10 Find the sum of the series  c = 12, from (3) b = 8 and from (1) a = 5 a – (a + d) + (a + 2d) – (a + 3d) + ... upto Hence a,b,c = 5,8,12 (2n + 1) terms Sol. The given series is an A.G.P. with common ratio Ex.13 Find the sum of n terms of the series S = a–(a + d) + (a + 2d) – (a + 3d)+....+ ( a + 2nd)  – S = – a + (a + d) – (a + 2d) + ...+ (a+ (2n – 1)d) – ( a + 2nd)  2S = a + {– d + d – d + d...upto 2n terms} + (a+2nd)  ...(1) 2b 2, a,b are in A.P.  a = 2 ...(2) (multiplying with x) y2 = xt  The numbers b,c, 18 are three consecutive x,y, t are in G.P.  1 . S = n [1– (1+1/n)n + (1+1/n)n ] = n n 2, a,b are consecutive terms of an A.P. (iii) x,y,z are in A.P.   Ex.12 Find three numbers a,b,c between 2 and 18 such that - (i) Their sum is 25 (ii)The numbers a 3 r 3 (1  r 3 ) Clearly, (a3 + b3)–1,(b3 + c3)–1 and (c3 + d3)–1 are in G.P. with common ratio 1/r3. Ex.9 1 xn – n xn 1 x 8 + 88 + 888 + .... Sol. Sum = = 8 [(10–1) + (100–1) + (1000–1) 9 + .... n terms] 2S = 2a + 2nd S = a + nd Ex.11 Find the sum to n terms of the series 8 [ 9 + 99 + 999 + ...n terms] 9 = 8 [ (10 + 102 +103 + ....+ 10n) – n] 9 2 1  1  1 + 2 1   + 3 1   + .... n n    Sol. Let S be the sum of n terms of the given series and x = 1 + 1/n, Then, = 8 9  10(10 n  1)   n = 8 [10n+1 – 9n – 10]   81  10  1