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
⇒ xa+2 + ya+2 = x1/2 y1/2 (xa+1 + ya+1)
⇒ xa+2 + ya+2 = xa+3/2 y1/2+ x1/2 ya+3/2
⇒ xa+2 - xa+3/2 y1/2 = x1/2 ya+3/2 – ya+2
⇒ xa+3/2(x1/2 - y1/2) = ya+3/2(x1/2– y1/2)
⇒ xa+3/2 = ya+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 = = 48 – 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 ⇒ 1.2n - 1 > 5
⇒ 1.2n> 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 8th 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 8th 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 3rd 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 + ar2 = 105r ---(1)
Also , 4a and 3ar in Ap
⇒ 2 × 4a = + 3ar
⇒ 8 = + 3r ⇒ 3r2 – 8r + 4= 0
⇒ 3r2 – 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) + r2 = 1
⇒ 2r2 – r – 1 = 0
⇒ 2r2 – 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 ⇒ a3 = 27 ⇒ a = 3.
Also + a + ar = + 3 + 3r =
⇒ + 1 + r = + r = ⇒ 2r2 – 5r + 2 =0
⇒ 2r2>/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.