Geometric Progressions Practice Problems: Level 02

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.

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

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.

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

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

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 8^th term of the sequence 1, 1/8, 1/27 …?

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?

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.

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.

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?

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
⇒ 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.}