1. If 6, 24, 96 are in G. P. then insert two more numbers in this progression so that it again forms a G. P.

A. 12 and 48

B. 16 and 48

C. 12 and 40

D. 20 and 32

Sol : Option A G.M. between 6 & 24 = GM between 24 & 96 = = 48 If we insert 12 between 6 and 24 and 48 between 24 & 96, then 6, 12, 24, 48, 96 from a G.P. So the required numbers are 12 & 48.

2. Find the value of ‘a’ given that the geometric mean between x and y is

A. -2/3

B. -1/4

C. -3/2

D. 7/6

Sol : Option C The G.M between x & y = Now = x ^{1/2} y ^{1/2} ⇒ x^{a+2} + y^{a+2} = x^{1/2} y^{1/2} (x^{a+1} + y^{a+1}) ⇒ x^{a+2} + y^{a+2} = x^{a+3/2} y^{1/2}+ x^{1/2} y^{a+3/2} ⇒ x^{a+2} - x^{a+3/2} y^{1/2} = x^{1/2} y^{a+3/2} – y^{a+2} ⇒ x^{a+3/2}(x^{1/2} - y^{1/2}) = y^{a+3/2}(x^{1/2}– y^{1/2}) ⇒ x^{a+3/2} = y^{a+3/2} ⇒ = 1 = ⇒ a + = 0 ⇒ a = -

3. Find the sum of the geometric series 3, 12, 48, 192, .....up to 8 terms.

A. 65355

B. 65535

C. 63555

D. 63535

Sol : Option B The given GP is 3, 12, 48, 192, ---- upto 8 terms. Here a = 3, r = 4 ∴ Sum = = 4^{8} – 1 = 65536 – 1 = 65535.

4. Find the sum of the geometric series 6, -3, -3/2, -3/4, …. Up to 10 terms

A. 512/255

B. 256/1021

C. 1024/515

D. 1023/256

Sol : Option D The given GP is 6, -3, ---- upto 10 terms Here a = 6, r = -1/2.

=

5. How many minimum terms in the G.P.: 1, 1.2, 1.44, 1.728, . . . are needed so that the sum of the terms is greater than 25? (Take log 2 = 0.301, log 3 = 0.4771

A. 8

B. 9

C. 10

D. 13

Sol : Option C The given GP is 1, 1.2, 1.44, 1.728, ----- Here a = 1, r = 1.2 Now Sn = > 25 ⇒ > 25 ⇒ 1.2^{n} - 1 > 5 ⇒ 1.2^{n}> 6 ⇒ n log (1.2) > log 6 ⇒ n[log12 – log10] > log 6 ⇒ n [2 log 2 + log 3 – log 10) > log 2 + log 3 ⇒ n [2 × 0.301 + 0.4771 – 1) > 0.301 + 0.4771 ⇒ 0.07491 n > 0.7781 ⇒ = 9.83 ⇒ n > 9.83 So the least value of n is 10

6. What is the 8^{th} term of the sequence 1, 1/8, 1/27 …?

A. 1/81

B. 1/512

C. 1/256

D. 1/244

Sol : Option B The given sequence is or Hence the 8^{th} term is

7. Sum of three numbers in GP with common ratio greater than 1 is 105. If the first two numbers are multiplied by 4 and the 3^{rd} number is multiplied by 3, then the resulting terms are in AP. What is the highest of the three numbers given?

A. 60

B. 50

C. 30

D. 45

Sol : Option A Let the numbers are a/r, a, ar Given that a/r + a + ar = 105 ⇒ a + ar + ar^{2} = 105r ---(1) Also , 4a and 3ar in Ap ⇒ 2 × 4a = + 3ar ⇒ 8 = + 3r ⇒ 3r^{2} – 8r + 4= 0 ⇒ 3r^{2} – 6r – 2r + 4= 0 ⇒ 3r (r – 2) – 2 (r – 2) = 0 (r – 2) (3r – 2) = 0 r = 2, 2/3 Since r > 1, put r = 2, in (1) we get a + 2a + 4a = 210 7a = 210 a = 30 ∴ the numbers are 15, 30, 60

8. There are three terms x, y, z between 4&40 such that (i) their sum is 37, (ii) 4, x, y are consecutive terms of an A.P. and (iii) y, z, 40 are the consecutive terms of a G.P. Find value of z.

A. 20

B. 10

C. 12

D. 15

Sol : Option A Here 4, x, y are in AP ⇒ 2x = 4 + y ---(1) Again y, z and 40 are in GP ⇒ z2 = 40y ⇒ y = ---(2) Also x + y + z = 37 ⇒ x = 37 – y – z ---(3) Put the value of (3) in (1) we get 2 (37 – y – z) = 4 + y ⇒ 74 – 2y – 2z = 4 +y ⇒ 3y = 70 – 2z Putting the value of y from (2), we get = 70 – 2z ⇒ 3z2 = 2800 – 80z ⇒ 3z2 + 80z – 2800 = 0 Solving we get z = 20, - 140/3 Rejecting z = we get z = 20 ∴ y = 10 & x = 7

9. There is a set of four numbers p, q, r and s respectively in such a manner that first three are in G.P. and the last three are in A.P. with a difference of 3. If the first and the fourth numbers are the same, find the value of p.

A. 8

B. 2

C. 6

D. 4

Sol : Option D Let the first three numbers are a/r, a, ar So the 4th number is ar + 3. Given that a/r = ar + 3 --(1) Also ar = a + 3 ⇒ar – a = 3 ⇒ a (r – 1) = 3 ⇒ a = --(2) Put in (1) we get = + 3 ⇒ = + 1 ⇒ r (r – 1) + r^{2} = 1 ⇒ 2r^{2} – r – 1 = 0 ⇒ 2r^{2} – 2r + r – 1 = 0 ⇒ 2r (r – 1) + 1(r – 1) = 0 ⇒ (2r + 1) (r – 1) = 0 ⇒ r = 1, -1/2 Now r ≠ 1 ⇒ r = -1/2. Put in (2) we get a = So the terms in G.P are or 4, -2, 1. Since the last term and the first terms are same so, the terms are 4, -2, 1, 4

10. The sum of three numbers in GP is 21/2 and their product is 27. What are the numbers?

A. ¼, 1, 5/4

B. 4, 1, 1/4

C. 6, 2, 2/3

D. 6, 3, 3/2

Sol : Option D Let the numbers are a/r, a, ar. Given that a/r × a × ar = 27 ⇒ a^{3} = 27 ⇒ a = 3. Also + a + ar = + 3 + 3r = ⇒ + 1 + r = + r = ⇒ 2r^{2} – 5r + 2 =0 ⇒ 2r^{2>/sup> – 4r – r + 2 = 0⇒ 2r (r – 2) – 1 (r – 2) = 0⇒ (2r – 1) (r – 2) = 0 ⇒ r = 2, 1/2.So the numbers are 3/2, 3, 6 or 6, 3, 3/2.}