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SOLVED EXAMPLE, Ex.1, , Sol., , If for an A.P. T3 = 18 and T7 = 30 then S17 is equal, to(A) 612, (B) 622, (C) 306, (D) None of these, Let first term = a, common difference = d, Then T3 = a + 2d = 18 and T7 = a + 6d = 30, , Sol., , Solving these, a= 12, d= 3, S17 =, =, , (C) 3050, (D) 3500, 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, , =, , 17, [2a + (17–1)d], 2, , 10, [2×10+(10–1) × 10], 2, 50 [2 + 49] + 10 [10 + 95] – 5 [20 + 90], 51 × 50 + 105 × 10 – 110 × 5 = 3050, , [2×5 + (20 –1) ×10] –, , 17, [24 + 16 × 3] = 612, 2, , =, =, , Ans. [A], , Ans. [C], Ex.2, , Sol., , The first, second and middle terms of an AP are a,, b, c respectively. Their sum is2( c a ), (A), ba, , 2c ( c a ), (B), +c, ba, , 2c ( b a ), (C), ca, , 2 b (c a ), (D), ba, , Ex.4, , If a1, a2, a3.....an are in AP where ai > 0 i then, the value of, , 1, a1 a 2, , , , middle term = c a + nd = c, , a + n (b – a) = c n =, Sum =, =, , 1 ca , 1, 2 , 2 ba , , Ans. [B], Ex.3, , a1 a 2, a1 a 2, , +, , 2c ( c a ), 1 2 (c a ), , 1 {2c} =, =, +c, , ba, 2 ba, , , The sum of integers in between 1 and 100 which, are divisible by 2 or 5 is(A) 3100, (B) 3600, , a n 1 a n, , =, , a1 a n, , n 1, a1 a n, , Let d be the c.d. of the A.P. Now L.H.S., , ca, ba, , , , ca , (b a ), 2a 2 , b, , a, , , , , , 1, , 1, , (D), , a1 a n, , =, , 2n 1, [2a + (2n +1–1)d], 2, , + ...+, , (B), , n, , (C), Sol., , a2 a3, , a1 a n, , d = common difference = b – a, It is given that the middle term is c. This, means that there are an odd number of terms, in the AP. Let there be (2n+1) terms in the, AP. Then (n+1)th term is the middle term., , 1, , 1, , (A), , We have first term = a, second term = b, , +, , +, , a2 a3, a2 a3, , +..., , a n 1 a n, , (Note), , a n 1 a n, , =, , a 1 a 2 a 2 a 3 ... a n , , –, , , d, , , , =, , –, , d, , (n 1) d, , =, d, , =, , ( a1 a n ), , a, , n, , a1, , , , =, , 1, d, , (a n a 1 ), a n a1, , [ an = a1 + (n – 1)d], , n 1, a n a1, PROGRESSIONS, , Ans. [D], , 44
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0, , =, Ex.5, , If a2,b2, c2 are in A.P. then, , Ex.8, , 1, 1, 1, ,, ,, are inad, bc ca, , (A) A.P., (C) H.P., Sol., , Sol., , If x, y, z are in G.P. and ax = by = cz then(A) logb a = logac, (B) logc b = logac, (D) None of these, , x, y, z are in G.P. y = xz, 2, , y, , z, , x log a = y log b = z log c = log , log , log , log , ,y=, ,z=, log b, log a, log c, , x =, , + ab + bc ..... are also in A.P., [adding ab + bc + ca], , putting x, y, z in (i) , we get, , (a +c) (a+ b), (b+ c) (a+b), (c+ a) (b+ c) .., are also in A.P., , 2, , log , log log , , =, ., log a log c, log b , , 1, 1, 1, ,, ,, are in A.P., bc ca ad, , (log b)2 = log a. log c, , [dividing by (a + b) (b + c) (c + a)], , or, , loga b = logb c logb a = logcb, , Ans. [A], Ex.6, , Sol., , Ex.7, , Sol., , Ans. [C], , If the sum of first 6 terms of a G.P. is nine times, of the sum of its first three terms, then its, common ratio is(A) 1, (B) 3/2 (C) 2, (D) – 2, , Ex.9, , a (1 r 6 ), a (1 r 3 ), =9, 1 r, 1 r, 1 – r6 = 9 (1– r3 ), 1+ r3 = 9, r = 2, , Sol., , If a, b, c, d are in G.P., then, (a3 + b3)–1, (b3 + c3)–1, (c3 +d3)–1 are in(A) A.P., (B) G.P., (C) H.P., (D) None of these, 2, Let b = ar, c = ar and d = ar3 . Then,, 1, , ( r 1), , a b, 3, , and,, , Ans. [C], , If pth, qth and rth terms of an A.P. are equal to, corresponding terms of a G.P. and these terms are, respectively x, y, z, then, xy–z. yz–x. zx–y equals(A) 0, (B) 1, (C) 2, (D) None of these, 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, then, a + (p –1) d = ARp–1 = x, p – 1 = (x – a) /d, ...(1), q–1, a + (q – 1) d = AR = y, q – 1 = (y – a)/d, ...(2), r–1, a + (r – 1) d= AR = z, r – 1 = (z–a) / d, ...(3), Given expression, = (ARp–1)y–z, (ARq–1) z–x, (ARr–1)x–y, = 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, , ...(i), , We have, a = b = c = (say), x, , a2 + ab + bc + ca, b2 + bc + ca + ab, c2 + ca, , or, , AR =1, , (C) logb a = logcb, , (B) G.P., (D) None of these, , a2, b2, c2 are in A.P., , or, , [By (1), (2) and (3)], Ans. [B], , 0, , 3, , =, , 1, 1, 1, ,, =, a 3 (1 r 3 ) b 3 c 3 a 3 r 3 (1 r 3 ), , 1, c d, 3, , 3, , =, , 1, a r (1 r 3 ), 3 3, , Clearly, (a3 + b3)–1,(b3 + c3)–1 and (c3 + d3)–1, are in G.P. with common ratio 1/r 3. Ans. [B], Ex.10, , If a, b, c, d and p are distinct real numbers such, that (a2 + b2 + c2) p2 – 2p (ab + bc + cd) +, (b2 + c2 + d2) 0 then a, b, c, d are in (A) A.P., (B) G.P., , Sol., , (C) H.P., (D) None of these, Here the given condition, (a2 + b2 + c2) p2 – 2p(ab + bc + ca)+ b2 + c2 + d2 0, (ap – b)2 + (bp – c)2 + (cp– d)2 0, Since the squares can not be negative, ap – b = 0, bp – c = 0, cp – d = 0, , , 1 a b c, = = =, p b c d, , , , a, b, c, d are in G.P., PROGRESSIONS, , Ans. [B], 45
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2S = 2a + 2nd S = a + nd, Ex.11, , Sol., , If pth, qth and rth terms of H.P. are u, v, w, respectively, then the value of the expression, (q – r)vw + (r – p)wu + (p – q)uv is(A) 1, (B) 0, (C) – 2, (D) – 1, Let H.P. be, 1, 1, 1, +, +, + ....., a, a d a 2d, , u=, , Ex.14, , 2, , Sol., , 1, 1, ,v=, ,, a (p 1)d, a (q 1)d, , , , xS = x + 2x2+ 3x3 +....+ (n–1) xn–1 + nxn, , , , S – xS = 1 + [x + x2 + ...+ xn–1] –nxn, , 1, 1, a + (p – 1)d = , a + (q– 1) d =, ,, u, v, , , , S (1– x) =, , , , S (–1/n) = – n[1– (1+1/n)n] – n (1+1/n)n, , , , 1, . S = n [1– (1+1/n)n + (1+1/n)n ]= n, n, , , , S = n2, , 1, 1, +...... =, (q – r) +, (r – p) + ...., u, v, , (q – r) vw + ..... = 0, , Ex.13, , Sol., , (A) n2, (B) n (n+1), 2, (C) n (1+1/n), (D) None of these, Let S be the sum of n terms of the given series, and x = 1 + 1/n, Then,, S = 1 + 2x + 3x2 + 4x3 + ....+ n xn–1, , 1, w=, a (r 1)d, , (q – r) {a + (p – 1)d} + (r – p) {a + (q –1) d}, , Sol., , The sum to n terms of the series, 1, 1, 1 + 2 1 + 3 1 + .... is given byn, , , n, , 1, a + (r – 1)d =, w, , Ex.12, , Ans. [C], , Ans. [B], , If x,y,z are in A.P. and x,y, t are in G.P. then x,, x– y, t– z are in (A) G.P., (B) A.P., (D) H.P., (D) A.P. and G.P. both, x,y,z are in A.P., 2y = x + z, or, 2xy = x2 + xz, (multiplying with x), 2, x – 2xy = – xz, ...(1), x,y, t are in G.P., y2 = xt, ...(2), 2, 2, or (x – 2xy + y ) = – xz + xt, or (x – y)2 = x (t – z), x, x – y, t– z are in G.P., Ans. [A], The sum of the series, a – (a + d) + (a + 2d) – (a + 3d) + ... upto (2n + 1) terms is(A) – nd, (B) a + 2 nd, (C) a + nd, (D) 2nd, The given series is an A.G.P. with common ratio, S = a – (a + d) + (a + 2d) – (a + 3d)+....+ (a + 2nd), , Ex.15, , Sol., , 2S = a + {– d + d – d + d...upto 2n terms} +, (a+2nd), , Ans. [A], , 1+ 2.2 + 3.22 + 4.23 + ....+ 100.299 equals(A) 99.2100, (B) 100.2100, (C) 1 + 99.2100, (D) None of these, Let, S = 1+ 2.2 + 3.22 + 4.23 +...+100.299 .....(1), 2S = 2+2.22 +3.23 +....+ 99.299 +100.2100, .....(2), Subtracting (2) from (1), we get, – S = (1+ 2+ 22 + 23 +.....+ 299) – 100.2100, , 2100 1, 2 1, 100, 100, = 100.2 – 2 + 1, = 1 + 99.2100, , S = 100.2100 –, , Ans. [C], , Ex.16, , a, b, c are first three terms of a GP. If HM of a and, b is 12 and that of b and c is 36, then a equals(A) 24, (B) 8, (C) 72, (D) 1/3, , Sol., , Let given three terms be br, b, b/r, 12 =, , 2 ( br) b, 2 (br), =, br b, r 1, , ...(1), , and 36 =, , 2b ( b / r ), 2b, =, b (b / r ), r 1, , ...(2), , , , – S = – a + (a + d) – (a + 2d) + ...+, (a+ (2n – 1)d) – (a + 2nd), , 1 xn, – n xn, 1 x, , (1) ÷ (2) r = 1/3, PROGRESSIONS, , 46
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Then from (2) b = 24, , Ex.17, , Sol., , a = br = 8, , 1 2 1 1, = , b b a a, , Ans. [B], , Find three numbers a, b, c between 2 and 18 such, that - (i) Their sum is 25 (ii)The numbers 2, a,b, are consecutive terms of an A.P. (iii) The, , Also, , numbers b,c, 18 are three consecutive terms of a, G.P., (A) 4, 8, 16, (B) 3, 6, 12, (C) 4, 8, 13, (D) 5, 8, 12, a + b + c = 25, ...(1), , in second), , 2b, 2, a, b are in A.P. a =, 2, , (eliminating 1/a in first factor and, , 2 1, = , c b, , Sol., , If x, 1, z are in A.P. x, 2,z are in G.P. then, x, 4, z are in(A) AP, (B) GP, (C) HP, (D) None of these, Here 2 = x + z, ...(1), 4 = xz, ...(2), Now, , Ex.21, , Ans. [C], , (C) 2n, , 2, 1, , (A), bc b 2, , 2, (B) 2 –, ab, b, , 3, 2, (C), –, ac b 2, , (D) None of these, , Here a, b, c in H.P., , 3, , Ex.22, , (D) 1, , ab ( n 1), ab (n 1), =, bn a, b(n 1) (b a ), , ab (n 1), (interchange a and b), an b, , H1 a H n b, +, H1 a H n b, , =, , (2n 1) b a (2n 1) a b, +, ab, ba, , =, , 2nb b a – 2na – a b, = 2n, ba, , If a,b,c are in H.P. then, , Ans. [C], , b, c, a, ,, ,, will, bc ca ab, , (B) G.P., (D) None of these, , a, b, c are in HP, , , 1 1 1, , ,, are in AP, a b c, , , , abc abc abc, ,, ,, are in AP, a, b, c, , 1 +, , 1 1, 2, = +, a c, b, , 1 1 1 2 1, Now , b c a b b, , (B) n, , Here H1 =, , If a, b, c in H.P. then value of, 1 1 1 1 1 1, =, b c a c a b, , Sol., , (A) 0, , be in(A) A.P., (C) H.P., Sol., , Ex.19, , H1 a H n b, +, =, H1 a H n b, , Hence, , 8, 2xz, =, =4, 2, xz, , x, 4, z are H.P., , between a and b then, , Similarly Hn=, , c = 12, from (3) b = 8 and from (1) a = 5, Hence a, b, c = 5, 8,12, Ans. [D], , Ans [A, B], , If H1, H2, H3, ......,Hn be n harmonic means, , ...(2), , Sol., , 1 1, +, c a, , 2, 1, 1, – 2, =, b, bc, b, , , The third one is not an answer, Ex.20, , and b, c, 18 are in G.P. c2 = 18 b ...(3), Eliminating a and b from (1), (2) and (3), gives, the following equation c2 + 12c – 288 = 0, (c– 12) (c +24) = 0 c = 12, –24, [Leaving c = – 24 because this is not in, between 2 and 18], , Ex.18, , 2, 3, 1, = 2 –, b, ab, b, , ab, ca, bc, , 1+, , 1+, are in AP, b, c, a, , , , bc ca ab, ,, ,, are in AP, a, b, c, , , , a, b, c, ,, ,, are in HP.Ans.[C], bc ca ab, , If the (m+1)th, (n+1)th, (r+1)th terms of an A. P. are, in G. P. and m, n, r are in H.P. then the ratio of, PROGRESSIONS, , 47
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common difference to the first terms in the, A. P. isSol., , (A) n/2 (B) 2/n, (C) – n/2 (D) – 2/n, Let the first term of A.P. be a and common, difference be d., Given (a + md), (a + nd), (a + rd) in G.P., (a + nd)2 = (a + md) (a + rd), , Now a, b, c in AP, 1 – a, 1 – b, 1 – c in A. P., , , x, y, z in H. P, , a, x, y, z, b are in HP, then, , 2mr, mr, , 2mr, 2n , 2, d, n = 2 n mr = – 2, =, a, n mr n 2 , n, mr n 2, , Sol., , Sol., , Also, , ...(1), , 1 1 1 1 1, , , , , are in AP, so as above, a x y z b, , 1 1 1 3 1 1, + + = , x y z 2 a b, , Here e2 = df, Now dx2 + 2ex + f = 0 given, , , , dx2 + 2 df x + f = 0 x = –, , 3, (a + b), 2, , a + b = 10, , If d, e, f are in G.P. and two quadratic equations, ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a, common root then, d/a, e/b, f/c are in(A) H. P., (B) G. P., (C) A. P., (D) None of these, , 1 1 1 5, + + = ., x y z 3, , Numbers a, b are(A) 8, 2, (B) 11, 3, (C) 9, 1, (D) None of these, By property of A.P.x + z = a + b and y = 1/2 (a + b), x + y + z =, , Ans. [D], Ex.23, , Ans. [C], , Ex. 25 If a, x, y, z, b are in AP, then x+ y + z = 15 and if, , d 2n m r, =, a, mr n 2, , But m, n, r in H.P. n =, , 1, 1, 1, ,, ,, in H. P., 1 a 1 b 1 c, , f, d, , 10, 1 1, + =, a b, 9, , ab = 9, , ...(2), , From (1) and (2) a, b are 9, 1, , Ans. [C], , 2, , Putting in ax + 2bx + c = 0 we get, a, , Ex.26, , f, f, + c = 2b, d, d, , , , a, c 2b, + =, d f, e, , , , a b c, , ,, are in A.P., d e f, , Sol., , Series is :, , d e f, ,, ,, are in H.P., a b c, , Ans. [A], , , Ex.24, , If a, b, c in A.P. and x = a n , y =, n 0, , , , z=, , cn, , then x, y, z are in-, , n 0, , Sol., , (A) AP, (B) GP, (C) HP, (D) None of these, Here a, b, c in A.P. , given, Also x =, , If rth term of a series is (2r + 1)2–r, then sum of its, infinite terms is(A) 10, (B) 8, (C) 5, (D) 0, –r, Here Tr = (2r + 1)2, , 1, 1, 1, ,y=, ,z=, 1 a, 1 b, 1 c, , , , bn ,, , n 0, , 1, 2, , 5 7, , 3 2 2 .... , 2, , , , Obviously the series in the bracket is, Arithmetico-Geometrical Series. Therefore by the, formula, S=, , dr, a, +, 1 r (1 r ) 2, , We have, , 1 , 2 , , 1 3, 2 , , S=, =5, , 2, 1 1 2 , 1 , , 2 1 2 , , Ans. [C], PROGRESSIONS, , 48
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a1 – a2= d (a1a2), a2 – a3 = d (a2a3),....,, , Ex.27, , n 1, , (an–1 – an) = d (an–1 an ), , n 1, , b, is AM/GM/HM, between a and b if n, a n bn, is equal to respectivelya, , Sol., , a1–an = d(a1a2+ a2a3+ .....+an–1 an), , 1, 1, (B) 0,, ,–, 2, 2, , 1, (A) – 1, – , 0, 2, , (C) 0, –, , Adding these relations, we get, , 1, ,–1, 2, , Also, , Putting n = –, , , , a 01 b 01, a b, 0, , 1, ,, 2, , 0, , 1, 1, a 2, , a, , =, , , , 1, 2, , =, , ab, = A.M., 2, , 1, 1, b 2, , b, , 1, 1, 2, , a 0 b0, a 1 b 1, , (n –1) (a1 an) = a1a2 + a2a3 + ....+ an–1 an, Ans. [B], Ex.29, , Sol., , Alternately, a n 1 b n 1, , =, , n, , ab, 2, , a, , –a b=–b, n, , n+1, , (D) None of these, , Here the given series is not A.P., G.P., or HP, , 0 = 3 + 4 7 10 13 .... Tn, , , , , n, , + ab, , A.P., , Tn = 3 +, , n, , a, =1, b, , (n 1), [2 (4) + (n – 2) 3], 2, , = 1/2 (3n2 – n + 4), , , n = 0, similarly for GM and HM also., , Sol., , (C) 580, , after subtracting, , an (a – b) = + bn (a – b) , a b, , Ex.28, , (B) 570, , S = 3 + 7 + 14 + 24 + ...+ Tn, , 2an+1 + 2bn+1 = an+1 + bn+1+ anb + abn, n+1, , (A) 560, , Let S = 3 + 7 + 14 + 24 + 37 + ....+ Tn, , for AM, a b, , Sum of the series 3 + 7 + 14 + 24 + 37 +...10, terms, is –, , 2ab, =, ab, , H. M. Hence, option (C) is correct. Ans. [C], , n, , ...(2), , From (1) and (2), we have, , a b, 1, 1, , a, b, , =, , 1, 1, –, = (n –1) d, a n a1, , a1 – an = (n – 1) d (a1an), , ab = G.M., , n = –1, =, , 1, 1, = Tn =, + (n – 1) d, an, a1, , (D) None of these, , By trial,, putting n = 0,, , ...(1), , Sn = 1/2 [3n2 – n + 4n], , a1a2 + a2a3 + .... + an–1 an is equal to-, , n (n 1) (2n 1) n (n 1), , , 4n , = 1/2 3, 6, 2, , , , (A) na1 an, , (B) (n–1) a1an, , Putting n = 10, , (C) (n+1) a1an, , (D) None of these, , 10 11 21 10 11, , , 40 , S10 = 1/2 , 2, 2, , , , If, , a1,, , a2,, , a3 ....,, , an are, , in, , HP,, , then, , Let d be common difference of the corresponding, AP., So, , 1, 1, 1, 1, 1, 1, –, = d,, –, = d, ....,, –, =d, a 2 a1, a3 a2, a n a n 1, , = 1/2 [1155– 55 + 40] =, , 1140, = 570, 2, , Ans. [B], PROGRESSIONS, , 49
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Note :, , First apply the method of difference for n terms, , Ex.31, , and proceed., , Ex.30, , 21/4. 22/8. 23/16. 24/32...... is equal to(A) 1, , Sol., , (B) 2, , (C) 3/2, , 1 2 3 4, ..., 8 16 32, =, , Now S =, , , 3, 4, 1 2, + +, +, +...., 4 8 16 32, , ...(1), ...(2), , (1) – (2), , , 1, 1, 1 1, S= + +, + ...., 2, 4 8 16, , =, , 1/ 4, = 1/2, 1, 1, 2, , 8, [10n+1 – 9n – 10], 81, , (B), , 8, [10n – 9n – 10], 81, , (C), , 8, [10n+1 – 9n + 10], 81, , (D) None of these, , 2s (say), , 1 2, 3, 1, ..., S= , 8 16 32, 2, , (A), , (D) 5/2, , The given product, = 24, , Sum of n terms of the series, 8 + 88 + 888 + .... equals, , S = 1, , Sol., , Sum =, , 8, [9 + 99 + 999 + ...n terms], 9, , =, , 8, [(10–1) + (100–1) + (1000–1) + .... n terms], 9, , =, , 8, [ (10 + 102 +103 + ....+ 10n) – n], 9, , =, , 8, 9, , 10(10 n 1), 8, [10n+1 – 9n – 10], n =, , 10 1, 81, Ans. [A], , 1, , Product = 2 = 2 Ans. [B], , PROGRESSIONS, , 50
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LEVEL- 1, Question, based on, , Q.1, , Arithmetic Progression (A.P.), , Q.11, , In the following two A.P.’s how many terms are, identical?, 2, 5, 8, 11.... to 60 terms, 3, 5, 7, ..... 50 terms, (A) 15, (B) 16, (C) 17, (D) 18, , Q.12, , The first term of an A.P. is 2 and common, difference is 4. The sum of its 40 terms will be –, (A) 3200, (B) 1600, (C) 200, (D) 2800, , 10th term of the progression – 4 – 1+ 2 + 5 +.... is(A) – 23, (B) 23, (C) – 32, (D) 32, th, , th, , Q.2, , If 4 term of an AP is 64 and its 54 term is, – 61, then its common difference is –, (A) 5/2, (B) – 5/2 (C) 3/50, (D) – 3/50, , Q.3, , Which term of the series 3 + 8 + 13 + 18 + ... is, 498(A) 95th, (B) 100th (C) 102th, (D) 101th, , Q.13, , If nth term of an AP is 1/3 (2n + 1), then the sum, of its 19 terms is(A) 131, (B) 132 (C) 133, (D) 134, , Q.4, , The number of terms, 101 + 99 + 97 + .....+ 47 is(A) 25, (B) 28, (C) 30, , series, , Q.14, , The sum of numbers lying between 10 and 200, which are divisible by 7 will be(A) 2800, (B) 2835, (C) 2870, (D) 2849, , If (m + 2)th term of an A.P. is (m + 2)2 – m2, then, its common difference is(A) 4, (B) – 4 (C) 2, (D) – 2, , Q.15, , If the sum of n terms of an AP is 2n 2 + 5n, then its, nth term is(A) 4n-3, (B) 4n + 3, (C) 3n + 4, (D) 3n – 4, , Q.16, , If the ratio of sum of n terms of two A.P’s is, (3n + 8) : (7n + 15), then the ratio of 12th, terms is(A) 16 : 7, (B) 7 :16, (C) 7 : 12, (D) 12 : 5, , Q.17, , If the ratio of the sum of n terms of two AP’s is, 2n : (n + 1), then ratio of their 8th terms is(A) 15 : 8, (B) 8 : 13, (C) n : (n– 1), (D) 5 : 17, , Q.18, , The sum of three consecutive terms of an, increasing A.P. is 51. If the product of the first, and third of these terms be 273, then third term is(A) 13, (B) 17, (C) 21, (D) 9, , Q.19, , If we divide 20 into four parts which are in A.P., such that product of the first and the fourth is to, the product of the second and third is the same as, 2 : 3, then the smallest part is(A) 1, (B) 2, (C) 3, (D) 4, , Q.5, , Q.6, , in, , the, (D) 20, , If mth terms of the series 63 + 65 + 67 + 69 + ...., and 3 + 10 + 17 + 24 + ... be equal, then m =, (A) 11, (B) 12, (C) 13, (D) 15, , Q.7, , If the 9th term of an A.P. be zero, then the ratio of, its 29th and 19th term is(A) 1 : 2, (B) 2 : 1 (C) 1 : 3, (D) 3 : 1, , Q.8, , If fourth term of an A.P. is thrice its first term and, seventh term – 2 (third term) = 1, then its, common difference is(A) 1, (B) 2, (C) – 2, (D) 3, , Q.9, , If pth, qth and rth terms of an A.P. are a, b and c, respectively, then a(q – r) + b (r – p) + c (p – q) is, equal to (A) 0, (B) 1, (C) a + b + c, (D) p + q + r, , Q.10, , The 19th term from the end of the series, 2 + 6 + 10 + ....+ 86 is –, (A) 6, (B) 18, (C) 14, (D) 10, , PROGRESSIONS, , 51
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Q.20, , Three numbers are in A.P. The product of the, extremes is 5 times the mean, also the sum of the, two largest is 8 times the least, the numbers are(A) 3, 9, 15, (B) 6, 18, 30, (C) 3, 8, 13, (D) 6, 16, 26, , Q.21, , If the angles of a quadrilateral are in A.P. whose, common difference is 10º, then the angles of the, quadrilateral are(A) 65º, 85º, 95º, 105º, (B) 75º, 85º, 95º, 105º, (C) 65º, 75º, 85º, 95º, (D) 65º, 95º, 105º, 115º, , Q.22, , Q.23, , Q.24, , Q.25, , Q.26, , Three numbers are in A.P., If their sum is 33 and, their product is 792, then the smallest of these, numbers is –, (A) 14, (B) 11, (C) 8, (D) 4, The sum of first four terms of an A.P. is 56 and, the sum of its last four terms is 112. If its first, term is 11, then number of its terms is(A) 10, (B) 11, (C) 12, (D) None of these, If the numbers a, b, c, d, e form an A.P., then the, value of a – 4b + 6c – 4d + e is(A) 1, (B) 2, (C) 0, (D) None of these, If a2 (b + c), b2 (c + a), c2 (a + b) are in A.P., then, a, b, c, are in(A) A.P., (B) G.P., (C) H.P., (D) None of these, , If the roots of the equation (b – c) x2 + (c – a)x +, (a – b) = 0 are equal , then a, b, c will be in(A) A.P., (B) G.P., (C) H.P., (D) None of these, , Q.29, , If, , b c, (A) A.P., (C) H.P., , ,, , 1, c a, , ,, , Q.30, , Question, based on, , (B) G.P., (D) None of these, , Arithmetic Mean (A.M.), , If x, y, z are in A.P. and A.M. of x and y is a and, that to y and z is b, then A.M. of a and b is (A) x, (B) y, (C) z, (D) 1/2(x + y), , Q.32, , If A1, A2 be two arithmetic means between, 1/3 and 1/24, then their values are(A) 7/72, 5/36, (B) 17/72, 5/36, (C) 7/36, 5/72, (D) 5/72, 17/72, , Q.33, , The AM of 1, 3, 5, ...., (2n – 1) is –, (A) n + 1, (B) n + 2, 2, (C) n, (D) n, , Q.34, , Given two numbers a and b, let A denotes the, single A.M. and S denote the sum of n A.M.’s, between a and b, then S/A depends on(A) n, a, b, (B) n , b, (C) n, a, (D) n, , Question, based on, , Geometrical Progression (G.P.), , Q.35, , If the first term of a G.P. be 5 and common ratio, be – 5, then which term is 3125 –, (A) 6th, (B) 5th, (C) 7th, (D) 8th, , Q.36, , The fifth term of a GP is 81 and its 8th term is, 2187, then its third term is(A) 3, (B) 9, (C) 27, (D) None of these, , 1 1, 1 1, 1 1, If a , b , c are in A.P., b c, a b, c a, , then a, b, c are also(A) A.P., (C) H.P., , The middle term of the progression, 4, 9, 14,....104 is(A) 44, (B) 49, (C) 59, (D) 54, , Q.31, , 1, , are ina b, (B) G.P., (D) None of these, , 1, 1, 1, ,, ,, are in A.P. thenpq rp qr, , (A) p2, q2, r2 are in A.P., (B) q2, p2, r2 are in A.P., (C) q2, r2, p2 are in A.P., (D) p, q, r are in A.P., , If a, b, c are in A.P., then, 1, , Q.27, , Q.28, , PROGRESSIONS, , 52
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Q.37, , Q.38, , In any G.P. the first term is 2 and last term is 512, and common ratio is 2, then 5th term from end is(A) 16, (B) 32, (C) 64, (D) None of these, , (A) 9th, (C) 8th, , 512, ?, 729, (B) 10th, (D) None of these, , Q.39, , If third term of a G.P is 4, then product of first 5, term is(A) 43, (B) 44, (C) 45, (D) None of these, , Q.40, , If third and seventh terms of a GP are 15 and 135, respectively, then its fifth term will be(A) 5, (B) 9, (C) 45, (D) 90, For which values of x do the numbers 1, x2,, 6 – x2 taken in that order form a geometric, progression(A) x = ± 2, , (B) x = ± 2, , (C) x = ± 3, , (D) x = ± 3, , Q.42, , Three numbers a, b, 12 are in G.P. and a, b, 9 are, in A.P., then a and b are –, (A) 3, 6, (B) – 3, 6, (C) 3, – 6, (D) – 3, – 6, , Q.43, , The second; third and sixth terms of an A.P. are, consecutive terms of a G.P. The common ratio of, the G.P. is(A) 1, (B) 3, (C) – 1, (D) – 3, , Q.44, , Total number of terms in the progression, 96 + 48 + 24 + 12 + .....+ 3/16 is(A) 9, (B) 10, (C) 15, (D) 20, , Q.45, , Q.46, , The sum of the first 10 terms of a certain G.P. is, equal to 244 times the sum of the first 5 terms., Then the common ratio is(A) 3, (B) 4, (C) 5, (D) None, The sum of the infinite terms of, 1 – 1/3 + 1/32 – 1/33 + ... is(A) 3/4, (B) 4/3 (C) – 3/4, , (D) – 4/3, , The sum 1 +, is finite if –, (A) x < 2, (C) x < 1, , Which term of the progression, 18, –12, 8, .... is, , Q.41, , Q.47, , 8, 4, 2, + 2 + 3 + .... (upto ), x, x, x, , (B) x > 2, (D) x < 1/2, , Q.48, , If the sum to n terms of a series be 3(2n –1), then, it is(A) A.P., (B) G.P., (C) H.P., (D) None of these, , Q.49, , The value of 91/3. 91/9. 91/27... upto , is(A) 1, (B) 3, (C) 9, (D) None of these, , Q.50, , If 3 + 3+ 32 + ... =, equals(A) 15/23, (C) 7/15, , 45, (> 0); then, 8, , (B) 15/7, (D) 23/15, , Q.51, , If the sum of an infinite GP be 3 and the sum of, the squares of its term is also 3, then its first term, and common ratio are –, (A) 3/2, 1/2, (B) 1/2, 3/2, (C) 1, 1/2, (D) None of these, , Q.52, , Every term of an infinite GP is thrice the sum of, all the successive terms. If the sum of first two, terms is 15, then the sum of the GP is(A) 20, (B) 16, (C) 28, (D) 30, , Q.53, , A geometric progression consists of an even, number of terms. The sum of all the terms is three, times that of the odd terms, the common ratio of, the progression will be(A) 1/2, (B) 2, (C) 3, (D) 1/3, , Q.54, , If first term of a decreasing infinite G.P. is 1 and, sum is S, then sum of squares of its terms is(A) S2, (B) 1/S2, 2, (C) S / (2S – 1), (D) S2/(2S + 1), , Q.55, , If sum of three numbers of a G.P. is 19 and their, product is 216, then its c.r. is(A) 1/2, (B) 1/3 (C) 3/2, (D) 3/4, , Q.56, , If the product of three numbers in GP is 3375 and, their sum is 65, then the smallest of these, numbers is (A) 3, (B) 5, (C) 4, (D) 6, PROGRESSIONS, , 53
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Q.57, , Q.58, , Q.59, , Q.60, , Q.61, , If the product of three terms of G.P. is 512., If 8 added to first and 6 added to second term, so, that number may be in A.P., then the numbers, are(A) 2, 4, 8, (B) 4, 8, 16, (C) 3, 6, 12, (D) None of these, In the four numbers first three are in G.P. and last, three are in A.P. whose common difference is 6., If the first and last numbers are same, then first, will be(A) 2, (B) 4, (C) 6, (D) 8, Break the numbers 155 into three parts so that the, obtained numbers form a G.P., the first term, being less than the third one by 120(A) 5, 65, 125, (B) 10, 65, 120, (C) 5, 25, 125, (D) None of these, Find three numbers in G.P. such that their sum is, 14 and the sum of their squares is 84 (A) 3, 6, 12, (B) 2, 6, 18, (C) 1, 3, 9, (D) 2, 4, 8, , Q.65, , If a, b, c, d are in G.P. then a + b, b + c, c + d are, in(A) A.P., (B) G.P., (C) H.P., (D) None of these, , Q.66, , If a, b, c are in G.P. then, (A) A.P., (C) H.P., , Question, based on, , If three geometric means be inserted between, 2 and 32, then the third geometric mean will be(A) 8, (B) 4, (C) 16, (D) 12, , Q.68, , The product of three geometric means between, 4 and 1/4 will be (A) 4, (B) 2, (C) – 1, (D) 1, , Q.69, , The ratio between the GM’s of the roots of the, equations ax2 + bx + c = 0 and x2 + mx + n = 0, is-, , Determine the first term and the common ratio of, the geometric progression, the sum of whose first, and third terms is 40 and the second and fourth, term is 80 (A) 8, 3, (B) 8, 2, (C) 7, 3, (D) 7, 2, The sum of three positive numbers constituting an, arithmetic progression is 15. If we add 1,4,19 to, those numbers respectively. We get a geometric, progression, then the numbers are(A) 2, 5, 8, (B) 8, 5, 2, (C) 5, 8, 2, (D) All of these, , Q.64, , The fractional value of 0.1 2 5 is(A) 125/999, (B) 23/990, (C) 61/550, (D) None of these, If x, y, z are in G.P. then x2 +y2, xy + yz, y2 + z2, are in (A) A.P., (B) G.P., (C) H.P., (D) None of these, , (A), , b, an, , (B), , c, an, , (C), , an, c, , (D), , cn, a, , If G be the geometric mean of x and y, then, 1, G x, 2, , (A) G2, , 2, , +, , 1, G y2, 2, , (B), , 1, G, , 2, , =, (C), , 2, G2, , (D) 3G2, , Q.71, , The A.M. of two numbers is 34 and GM is 16, the, numbers are(A) 2 and 64, (B) 64 and 3, (C) 64 and 4, (D) None of these, , Q.72, , Two numbers are in the ratio 4 : 1. If their AM, exceeds their GM by 2, then the numbers are-(A), 4, 1, (B) 16, 4, (C) 12, 3, (D) None of these, , , , Q.63, , Geometrical Mean (G.M.), , Q.67, , Q.70, Q.62, , 1 1 1, ,, , are in a b c, (B) G.P., (D) None of these, , PROGRESSIONS, , 54
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Q.73, , Question, based on, , a, b, c are in A.P. If x is the GM between, a and b and y is the GM between b and c, then the, A.M. between x2 and y2 will be(A) a2, (B) b2, 2, (C) c, (D) None of these, , Q.79, , Sum to infinite of the series, 2, 4, 3, + 2 + 3 + .... is5 5, 5, (A) 5/4, (B) 6/5, (C) 25/16, (D) 16/9, , 1+, , Q.75, , The sum of infinite terms of the progression, 1+ 3x + 5x2 + 7x3 + ......(x < 1) is(A), (C), , Q.76, , 1 x, 1 x, 1 x, (1 x ) 2, , 1 x , (B) , , 1 x , , ab, (D) None of these, b (n 1) (a b), , 1, (C), 10, , 1, 5, , (D) 10, , (B), , mn, r 1, , (C), , mn, r, , (D), , mn, r 1, , b b, b, , , c – will be, 2 2, 2, , (B) G.P., (D) None of these, , If a, b, c are in A.P., then, , (A) A.P., (C) H.P., Question, based on, , 1, 1, The fifth term of the H.P. 2, 2 , 3 ,..... will be3, 2, , (B) 3, , r, mn, , bc, ab, ca, ,, ,, are inca ab bc ab bc ca, , Harmonic Progression (H.P.), , 1, 5, , (A), , in (A) A.P., (C) H.P., , 1+2(1+1/n) + 3(1+1/n)2 + ... terms, equals(A) n (1+1/n), (B) n2, (C) n(1+1/n)2, (D) None of these, , (A) 5, , If the mth term of a H.P. be n and n th term be m,, then the rth term will be-, , If a, b, c be in H.P. then a –, , (D) None of these, , If fourth term of an HP is 3/5 and its 8th term is, 1/3, then its first term is–, (A) 2/3, (B) 3/2, (C) 1/4, (D) None of these, , ab, b (n 1) (a b), , Q.82, , 2, , Q.84, Q.78, , (C), , (B), , If b + c, c + a, a + b are in H.P., then a 2, b2, c2 will, be in(A) A.P., (B) G.P., (C) H.P., (D) None of these, , Q.83, , Q.77, , ab, a (n 1) ab, , Q.81, , Not in AIEEE syllabus, Question, based on, , (A), , Arithmetic-Geometrical Progression, (A.G.P.), , Q.80, Q.74, , If first and second terms of a HP are a and b, then, its nth term will be-, , Harmonic Mean (H.M.), , The HM between 1/21 and – 1/5 is (A), , Q.85, , (B) G.P., (D) None of these, , 1, 8, , (B) –, , 1, 8, , (C), , 1, 4, , (D) –, , 1, 4, , If H is H.M. between two numbers a and b, then, 1, 1, +, equals Ha Hb, , (A) a – b, (C), , 1 1, –, a b, , (B) a + b, (D), , 1 1, +, a, b, , PROGRESSIONS, , 55
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Q.86, , The HM between, (A), (C), , Q.87, , 2ab, ab, 2ab, a 2 b2, , a, b, and isb, a, , (B), (D), , Q.93, , (A) 50, (C) 101, , 2a 2 b 2, a 2 b2, 2a 2 b 2, ab, , The number of terms in the sequence, 1, 3, 6, 10, 15, 21, .... , 5050 is-, , Q.94, , Sum of n terms of 1 + (1 + x) + (1 + x + x2), + (1 + x + x2 + x3) + .... is-, , If 4 HM’s be inserted between 2/3 and 2/13, then, the second HM is-, , (A), , 1 xn, 1 x, , (A) 2/5, (C) 2/11, , (B), , x (1 x n ), 1 x, , (B) 2/7, (D) 2/17, , (C), Question, based on, , (B) 100, (D) 105, , n (1 x ) x (1 x n ), (1 x ) 2, , Relation between A.M., G.M. & H.M., , (D) None of these, Q.88, , If A,G & 4 are A.M, G.M & H.M of two numbers, respectively and 2A + G2 = 27, then the numbers, are(A) 8, 2, (B) 8, 6 (C) 6, 3, (D) 6, 4, , n, , Q.95, , k 3 is equal to-, , k 1, , n, , (A) 2 k, , 2, , k 1, , Q.89, , Q.90, , If x, y, z are AM, GM and HM of two positive, numbers respectively, then correct statement is (A) x < y < z, (B) y < x < z, (C) z < y < x, (D) z < x < y, , n , (C) k , k 1 , , 3, , n , (B) k , k 1 , , 2, , n, , (D) 3 k 2, k 1, , If sum of A.M. and H.M. between two positive, numbers is 25 and their GM is 12, then sum of, numbers is(A) 9, (B) 18, (C) 32, (D) 18 or 32, , Q.91, , Question, based on, , Q.92, , The A.M. between two positive numbers exceeds, the GM by 5, and the GM exceeds the H.M. by 4., Then the numbers are(A) 10, 40, (B) 10, 20, (C) 20, 40, (D) 10, 50, Special Series, , Sum of the series 1+ 3+ 7 + 15 + 31+ .... to n, terms is(A) 2n – 2 – n, (B) 2n+1+2 + n, (C) 2n+1–2–n, , (D) None of these, PROGRESSIONS, , 56
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LEVEL- 2, Q.1, , Find the sum of all the even positive integers less, than 200 which are not divisible by 6(A) 6535, (B) 6539, (C) 6534, , Q.2, , Q.7, , 2, , The sum of n terms of the series, a2, a3, + log 2 + ... isb, b, , a, (A) n log , b, , (B) n log (ab), , Q.3, , (C), , a n, n2, log + log (ab), b 2, 2, , (D), , a n, n2, log, – log (ab), b 2, 2, , (C) 398 +, Q.4, , 219, , (B) a + c = b, , (C) ac –, , (D) None of these, , n terms of the series 1+, , n is(A) 11, (C) 10, , x11 1 , , + 20, x9 , , , , If 0 < x, y, a, b < 1, then the sum of the infinite, terms of the series, , (C), , Q.9, , Q.10, , (B) 9, (D) 8, , ax, 1 b, x, 1 b, , +, , +, , x, 1 y, x, 1 y, , (B), , (D), , x, 1 b, ax, 1 b, , +, , +, , x, 1 y, x, 1 y, , If sum of 3 terms of a G.P. is S. product is P, and, sum of reciprocal of its terms is R, then, (A) S, , (B) S3, , (C) 2S2, , (D) S2/R, , If A and G are respectively A.M. and G.M. of, roots of a quadratic equation, then it is(A) x2 + 2Ax + G2 = 0, (B) x2 – 2Ax + G2 = 0, , 1, 1 1, +, + + ....,, 2, 4 8, , 1, , then the least value of, 1000, , x ( ab + xy ), , P2 R3 equals to -, , If S denotes the sum to infinity and Sn the sum of, , such that S – Sn .<, , x18 1 , , (C) 2, x 1 , , , , (A), , Certain numbers appear in both the arithmetic, progressions 17, 21, 25.... and 16, 21, 26.... find, the sum of the first two hundred terms appearing, in both(A) 4022, (B) 402200, (C) 201100, (D) 398000, , Q.6, , x11 1 , , + 20, x9 , , , , terms are respectively, , (A) a = b = c, =0, , x18 1 , , (B) 2, x 1 , , , , + x (b a + y x ) + ... is-, , If first and (2n – 1)th terms of an A.P., G.P. and, , b2, , x 22 1 , , + 20, x 20 , , , , x( a+ x)+, , (D) None of these, , H.P. are equal and their, a, b, c, then -, , Q.5, , Q.8, , (B) 398 + 221, , nth, , x 20 1 , , (A) 2, x 1 , , , , (D) None of these, , The sum of 40 terms of the series, 1+ 2 + 3 + 4 + 5 + 8 + 7+ 16 + 9 + ... is(A) 398 + 220, , 2, , 2, , 1, 2 1 , 3 1 , , x + x 2 + x 3 + .... is x, x , x , , , , , (D) 6532, , log a + log, , The sum of 10 terms of the series, , (C) x2 – Ax + G = 0, (D) None of these, Q.11, , If tn be the nth term of an A.P. and if t7 = 9, then, the value of the c.d. that would make t1t2t7 least, is(A) 33/40, (C) 33/10, , (B) 33/20, (D) None of these, PROGRESSIONS, , 57
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Q.12, , Q.13, , If mth terms of the series 63 + 65 + 67 + 69 + ...., and 3 + 10 + 17 + 24 + ... be equal, then m =, (A) 11, (B) 12, (C) 13, (D) 15, A ball falls from a height of 100 mts. on a floor., If in each rebound it describes 4/5 height of the, previous falling height, then the total distance, travelled by the ball before coming to rest is(A) , (C) 1000 mts, , Q.14, , (B) 500 mts, (D) 900 mts, , Q.19, , In the following two A.P.’s how many terms are, identical ?, 2, 5, 8, 11.... to 60 terms, 3, 5, 7, ..... 50 terms, (A) 15, (B) 16, (C) 17, (D) 18, , Q.20, , If 1 + r + r2 + ....+ rn = (1+ r) (1+ r2) (1+ r4) (1+ r8) ,, then the value of n is(A) 13, (B) 14, (C) 15, (D) 16, , Q.21, , In an A.P. of which a is the first term, if the sum, of the first p terms is zero, then the sum of the, next q term is-, , If A,G and H are respectively A.M., G.M., and, H.M. of three positive numbers a, b and c, then, the equation whose roots are a, b and c is given by(A) x3 – 3Ax2 + 3G3 x + G3 = 0, , a ( p q )q, p 1, , (B) –, , (C), , a (p q)p, p 1, , (D) None of these, , (B) x3 – 3Ax2 + 3(G3/H) x – G3 = 0, (C) x3 + 3Ax2 + 3(G3/H) x – G3 = 0, (D), Q.15, , x3, , –, , 3Ax2, , –, , 3(G3/H), , =0, , If a and be the first and last term of an A.P. and, , Q.18, , 2 a 2, 2S a, , (B), , (C), , 2 a 2, 2S a, , (D) None of these, , ( x 2 yz), , (B), , (C), , (z 2 xy ), , (D) None of these, , If sum of infinite G.P. is x and sum of square of, its terms is y, then common ratio is(A), , x y, , 2, , x y, , 2, , 2, , (C), , (B), , x2 y, 2, , x2 y, , x y, , The sum of 10 terms of the series., 0.7 + .77 + .777 + ... is1 , 7 , 89 10 , 81 , 10 , , (A), , 1 , 7 , 89 10 , 9 , 10 , , (B), , (C), , 1 , 7 , 89 9 , 81 , 10 , , (D) None of these, , The, , value, , of, , x, alogb, , where, , a, , =, , 1 1 1, , b = 5 , x = ..... , is 4 8 16, , , (A) 1, Q.26, , (B) 2, , (C) 1/2, , (D) 4, , Find the sum of the series up to n term, 1.3.5 + 3.5.7 + 5.7.9 + .., (A) 8n3 + 12n2 – 2n– 3, (B) n (8n3 + 11n2 – n – 3), , x2 y, 2, , (D), , Q.25, , ( y 2 zx ), , (A), , x2 y, , Q.24, , 2 a 2, 2S a, , If x,y,z are in A.P. , then magnitude of its, common difference is-, , If a, b, c are in A.P. and x = 1 + a + a2 + ...,, y = 1 + b + b2 + ... and z = 1 + c + c2 + ....,, (where a, b, c < 1)), then x, y, z are in(A) A.P., (B) G.P., (C) H.P., (D) None of these, , (B) G.M., (D) None of these, , (A), , Find sum of the series, 1.32 + 2.52 + 3.72 + .... to 20 terms(A) 188090, (B) 94045, (C) 325178, (D) 812715, , Q.23, , S be the sum of its all terms; then its common, difference is-, , Q.17, , Q.22, , The G.M. of roots of the equation x2 – 2ax + b2 = 0 is, equal to which type of mean of roots of, x2 – 2bx + a2 = 0?, (A) A.M., (C) H.M., , Q.16, , x+, , G3, , a ( p q )q, p 1, , (A), , 2, , x 2 y2, , (C) n (2n3 + 8n2 + 7n – 2), (D) None of these, PROGRESSIONS, , 58, , 0.2,
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Q.27, , If A.M. between p and q (p q) is two times the, GM, then p : q is(A) 1 : 1, (B) 2 : 1, , 3 ) : (2 – 3 ), , (C) (2 +, (D) 3 :1, Q.28, , Q.33, , The sum of the infinite series, 12 + 22 x + 32 x2 + ..... is(A) (1+ x)/(1– x)3, , (B) (1+ x)/(1– x), , (C) x/ (1– x)3, , (D) 1/(1– x)3, , Q.34, , If the sum of four numbers in A.P. be 48 and that, the product of the extremes is to the product of, the means is 27 to 35 then the numbers are(A) 3, 9, 15, 21, (B) 9, 5, 7, 3, (C) 6, 10, 14, 18, (D) None of these, , Q.35, , The sum of infinite series 1–, , The sum of the first ten terms of the geometric, progression is S1 and the sum of the next ten, terms (11th through 20th) is S2. then the common, ratio will be(A) ± 10, , (C) ± 10, , Q.29, , S1, S2, , (B) ±, , S2, S1, , (D), , S2, S1, , (A), , S1, S2, , Q.36, , y = b – (b/r) + (b/r2) – .... and, z = c + (c/r2) + (c/r4) + ...., then (xy/z) isbc, (B), a, , ca, (C), b, , (B), , 2, 3, , (C) –, , 2, 9, , (D), , 9, 2, , If a, b, c are in G.P. and A.M. between a, b and b,, c are respectively p and q, then (a/p) + (c/q) is, equal to(A) 0, (B) 1, (C) 2, (D) 1/2, , If x = a + (a/r) + (a/r 2) + ............,, , ab, (A), c, , 2, 9, , 7, 3 5, + – + ... is8, 2 4, , Q.37, , The, , solution, , (8) (1|cos x||cos, , (D) abc, , 2, , of, , x||cos3 x |.....), , the, , equation, , = 43 in the interval, , (–, ) areQ.30, , The series of natural numbers is divided into, groups as follows ; (1), (2, 3), (4, 5, 6), (7, 8, 9, 10), , (A) ±, , (B) ±, , and so on. Find the sum of the numbers in the nth, group is-, , , , ,±, 3, 6, , (C) ±, , , 2, ,±, 3, 3, , (D) None of these, , (A), , 1, [n(n2+ 1)], 2, , 2n ( n 1), (C), 3, , Q.31, , n (n 1), 4, 2, , (B), , Q.38, , The sum to infinity of the following series, , (A) , , (B) 1, , (C) 0, , (D) None of these, , The number of terms in the sequence, 1, 3, 6, 10, 15, 21,...., 5050 is(A) 50, (B) 100 (C) 101, (D) 105, , If a, b, c, d are in G.P., then the value of, (a – c)2 + (b – c)2 + (b – d)2 – (a – d)2 is(A) 0, (B) 1, (C) a + d, (D) a – d, , n 2 (n 1), (D), 2, Q.39, , 1, 1, 1, +, +, + .... shall be1 . 2 2 .3 3 .4, , Q.32, , , ,±, 3, , The third term of an A.P. is 9 and the difference, of the seventh and the second term is 20. If the, number 2001 is the n th term of the sequence then, n is(A) equal to 499, (B) is equal to 500, (C) equal to 501, (D) can have no value, , Q.40, , Given the geometric progression 3, 6,12, 24,....., the term 12288 would occur as the(A) 11th term, (C) 13th term, , (B) 12th term, (D) 14th term, , PROGRESSIONS, , 59
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LEVEL- 3, Q.1, , The, , maximum, , sum, , of, , the, , series, , Q.7, , then x is equal to -, , 2, 1, 20 + 19 +18 +……. is 3, 3, , (A) 310, (C) 320, , (B) 300, (D) None of these, Q.8, , Q.2, , Let a, b be the roots of, , – 3x + p = 0 and let, , x2, , c, d be the roots of – 12 x + q = 0, where a, b,, c, d form an increasing G.P. Then the ratio of, q + p : q – p is equal to (A) 8 : 7, (B) 11 : 10, , Q.3, , (D) None of these, , a bx, =, a bx, , If, , b cx, =, b cx, , a, b, c, d are in (A) A.P., (C) H.P., Q.4, , 3, 12, , +, , 5, 12 2 2, , 6n, n 1, , (B), , 9n, n 1, , (C), , 12 n, n 1, , (D), , 15 n, n 1, , Q.6, , 3, (C), 4, , 3, (D), 8, , S2 2, , is equal to -, , (B) 3, , (C) 9, , (D) 10, , If Sn denotes the sum of n terms of an A.P., then, Sn+3 – 3Sn+2 + 3Sn+1 – Sn is equal to (A) 0, , Q.11, , (B) 1, , (C) 1/2, , (D) 2, , If a1, a2, a3, ......, a24 are in A.P. and, a1 + a5 + a10 + a15 + a20 + a24 = 225, then a1 + a2, , 3, 2, , (B), , S3 (1 8 S1 ), , Q.10, , + a3 + ...... + a23 + a24 is equal to-, , 1, If t r 2(3 1) n 1 , then lim , =, n, , , t, r 1, r 1 r, , (A) 3, , If S1, S2, S3 are the sums of first n natural, , The sum of three consecutive terms in a, geometric progression is 14. If 1 is added to the, first and the second terms and 1 is subtracted, from the third, the resulting new terms are in, arithmetic progression. Then the lowest of the, original terms is (A) 1, (B) 2, (C) 4, (D) 8, , n, , n, , (D) log3 0.25, , Q.9, , +……….. is -, , 12 2 2 32, , (A), , n, , Q.5, , +, , (C) 1 – log34, , (A) 1, , (B) G.P., (D) None of these, , 7, , (B) log34, , then, , c dx, (x 0), then, c dx, , The sum of the first n terms of the series, , (A) log43, , numbers, their squares, their cubes respectively,, , x2, , (C) 17 : 15, , If 1, log9 (31–x + 2) and log3 (4.3x – 1) are in A.P.,, , (A) 909, Q.12, , (C), , n, (a12 + a2n2), n 1, , (D) None of these, , 5, 1, 1, 1, +, +, is, if, 3, Z, X, Y, , a, X, Y, Z, b are in H.P., then a and b are(A) 1, 9, (B) 3, 7, (C) 7, 3, (D) 9, 1, / 4, , Q.13, , 2n, n, (a12 – a2n2) (B), (a2n2 –a12), n 1, 2n 1, , (D) 900, , The value of x + y + z is 15 if a, x, y, z, b are in, , then a12 – a22 + a32 – a42 + ...... + a2n–12 – a2n2 is, , (A), , (C) 750, , A.P. while the value of, , Let the sequence a1, a2, a3, ......, an form an A.P.,, equal to -, , (B) 75, , If In =, , tan, , n, , x sec2 x dx , then I1, I2, I3,... are in -, , 0, , (A) A. P., (C) H.P., , (B) G.P., (D) None of these, , PROGRESSIONS, , 60
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Q.14, , A G.P. consists of 2n terms. If the sum of the, terms occupying the odd places is S1 and that of, the terms at the even places is S2, then S2/S1 is -, , Then Sn–1 = t1 + t2 + t3 + ...... + tn–1, n > 1 (n N), , (B) Independent of r, (C) Independent of a and r, (D) Dependent on r, , subtracting, we get Sn – Sn–1 = tn, n > 1 Further if we put, n = 1 in the first sum then S1 = t1 .Thus we can write, , If x18 = y21 = z28, then 3, 3 logy x, 3 logz y,, 7 logx z are in (A) A.P., , Suppose a series of n terms is given by Sn = t1 + t2 + t3 +, ...... + tn,, , (A) Dependent on a, , Q.15, , Passage Based Questions (Q. 21 - 23), , (B) G.P. (C) H.P., , tn = Sn – Sn–1 and t1 = S1. The above result can be used to, find the terms of any kind of series, independent of its, nature, provided the sum to first n terms is given., , (D) None, Q.21, , Q.16, , Q.17, , The sum of integers from 1 to 100 that are, divisible by 2 or 3, is (A) 3300, (B) 3330, (C) 3000, (D) None of these, The sum of an infinitely decreasing G.P. is equal, to 4 and the sum of the cubes of its terms is equal, to, , Q.18, , (C) 9 a + 3 b, , 1, 4, , (B), , 1, 8, , (C), , 1, 16, , (D), , 1, 32, , If the sum of the first 2n terms of the A.P. 2, 5, 8,, ...... is equal to the sum of first n terms of the A.P., 57, 59, 61, ......, then n equals (A) 10, (B) 11, (C) 12, (D) 13, , (D) 16a + 4b, , Q.22, , The sum of n terms of a series is a.2n – b, where a, and b are constants then the series is –, (A) A.P., (B) G.P., (C) A.G.P., (D) G.P. from second term onwards, , Q.23, , If the sum to n terms of a series is given, , 64, . Then 5th term of the progression is 7, , (A), , If sum of n terms of a series is of the form, an2 + bn, where a and b are constants, then the, fourth term of the series is–, (A) a + b, (B) 7a + b, , by, , n (n 1) (n 2), then the nth term of the series, 6, , is –, (A) n2, , (B) (n)2, , (C)n, , (D)n + n, , Questions based on Statements (Q. 24-28), Q.19, , Let Sn =, , 1, 13, , +, , 1 2, 13 2 3, , + …..+, , 1 2 ....... n, 13 2 3 ..... n 3, , ;, , n = 1, 2, 3, ... Then Sn is not greater than(A) 1/2, (C) 2, Q.20, , (B) 1, (D) 4, , The number of common terms to the two, sequences 17, 21, 25, ......, 417 and 16, 21,, 26, ...... 466 is (A) 21, (B) 19, (C) 20, (D) 91, , Each of the questions given below consist of Statement – I, and Statement – II. Use the following Key to choose the, appropriate answer., (A) If both Statement- I and Statement- II are true,, and Statement - II is the correct explanation of, Statement– I., (B) If both Statement - I and Statement - II are true, but Statement - II is not the correct explanation of, Statement – I., (C) If Statement - I is true but Statement - II is false., (D) If Statement - I is false but Statement - II is true., PROGRESSIONS, , 61
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Q.24, , Statement I : If A and G be the A.M and G.M., between two positive real numbers a and b then a,, , Q.27, , and (a2 + b2) x2 – 2b (a + c)x + (b2 + c2) = 0 then, a, b, c are in G.P. and x is their common ratio., Statement II : If the ratio of the sum of m, , b are given by A ± (A G ) (A G ) ., Statement II : Using x2 – (a + b)x + ab = 0 ;, , terms and n terms of an A.P. is m 2 : n2 then the, , where a + b = 2A, ab = G2, we calculate x., Q.25, , Q.26, , ratio of its mth and nth terms will be, (2m – 1) : (2n–1), , Statement I : The sum of all numbers of the, form n3 which lie between 100 and 10,000 is, 53261., , Statement I : If a, b, c, x are all real numbers, , Q.28, , Statement I : 1 + 3 + 7 + 13 + ....... up to n, n ( n 2 2), ., 3, , a, ab, Statement II : If =, =, then a, b, c are, c, bc, , terms =, , in G.P., , Statement II :, , Statement I : The number of terms of the A.P., 3, 7, 11, 15, ..... to be taken so that the sum is 465, is 15., Statement II : The sum of the integers from, 1 to 100 which are not divisible by 3 or 5 is 2632., , of a and b if n = –, , a n 1 b n 1, a n bn, , is Harmonic mean, , 1, ., 2, , PROGRESSIONS, , 62
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ANSWER KEY, LEVEL- 1, Q.No., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , Ans., , B, , B, , B, , B, , A, , C, , B, , B, , A, , C, , C, , A, , C, , B, , B, , B, , A, , C, , B, , A, , Q.No. 21, , 22, , 23, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , 31, , 32, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , Ans., , B, , D, , B, , C, , A, , A, , A, , A, , A, , D, , B, , B, , D, , D, , B, , B, , B, , A, , C, , C, , Q.No. 41, , 42, , 43, , 44, , 45, , 46, , 47, , 48, , 49, , 50, , 51, , 52, , 53, , 54, , 55, , 56, , 57, , 58, , 59, , 60, , Ans., , B, , A, , B, , B, , A, , A, , B, , B, , B, , C, , A, , B, , B, , C, , C, , B, , B, , D, , C, , D, , Q.No. 61, , 62, , 63, , 64, , 65, , 66, , 67, , 68, , 69, , 70, , 71, , 72, , 73, , 74, , 75, , 76, , 77, , 78, , 79, , 80, , D, , B, , D, , C, , C, , Ans., , B, , A, , A, , B, , B, , B, , C, , D, , B, , B, , C, , B, , B, , C, , C, , Q.No. 81, , 82, , 83, , 84, , 85, , 86, , 87, , 88, , 89, , 90, , 91, , 92, , 93, , 94, , 95, , Ans., , B, , C, , A, , D, , C, , B, , C, , C, , C, , A, , C, , B, , C, , B, , A, , LEVEL- 2, Q.No., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , Ans., , C, , C, , B, , C, , B, , A, , A, , D, , B, , B, , B, , C, , D, , B, , A, , B, , B, , B, , C, , C, , Q.No. 21, , 22, , 23, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , 31, , 32, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , Ans., , A, , C, , B, , D, , C, , C, , C, , A, , A, , B, , B, , A, , C, , A, , C, , C, , A, , C, , C, , 11 12 13, D A, D C, , 14, D, , 15, A, , 16, D, , 17, B, , 18, B, , 19, C, , 20, C, , B, , LEVEL- 3, Q.No. 1, Ans. A, Q.No. 21, Ans. B, , 2, C, 22, D, , 3, B, 23, C, , 4, A, 24, A, , 5, D, 25, C, , 6, A, 26, B, , 7, C, 27, B, , 8, C, 28, C, , 9, B, , 10, A, , PROGRESSIONS, , 63
