Geometric Progressions: Solved Examples

Q.1. Find 9thterm of the following series: 5, 10, 20, 40 ……
A. 1024
B. 980
C. 1280
D. 320
Answer: Option C
Explanation: T9 = arn-1 ⇒ r = 10/5 = 2
T9= 5(2)9-1
Therefore, T9= 5 × 256 = 1280.
Q.2. Ram gives his son Rs. 100 on one day, Rs. 50 on the second day, Rs. 25 on third day and so on. What will be total amount given by Ram to his son starting from the first day, if he lives forever?
A. 200
B. 120
C. 250
D. 100
Answer: Option A
Explanation: In this question, everyday he is giving half the amount, he has given the previous day. And he has to pay forever. This makes it an infinite GP series.
Q.3. The 3rd and the 8th term of a G. P. are 4 and 128 respectively. Find the G. P.
A. 2,3, 4,5
B. 1,2,4,8
C. -12,144,-1728,
D. -1,2,-4,8
Answer: Option B
Explanation: Let a be the first term and r be the common ratio of the G. P., then ar3 = 8 …(1)
ar7=128 …(2)
Dividing 2 by 1:
r4 = 16
r = 2
putting r=2 in equation(1)
we get a=1.
So the required GP= 1,2,4,8,16…
Q.4. Which term of the G. P.: 6, –12, 24, – 48, ... is 384?
A. 5
B. 7
C. 8
D. 10
Answer: Option B
Explanation: Let the term be nth.
(ar)n-1 = 384
a=6, and r= (-2)
Solving for n:
6 (-2)n-1 = 384
(-2) n-1= 64
n-1 = 6
n=7
Q.5. The 1st term of a GP is 64 and the 5th term is 4. If the sum of all terms is 128, what is the common ratio?
A. 1/8
B. 2/5
C. 1/2
D. 1/4
Answer: Option C
Explanation:
Since the 1st term is 64 and the 5th term is 4, it is obvious that successive terms decrease in value. Therefore, r < 1. So, the sum of all terms is a/(1 – r) = 128. Solving, we get r = ½.
Q.6. Find the sum of GP.: 1, 2,4,8, ... up to the 10th term.
A. 210-1
B. 210
C. 216
D. 210/2
Answer: Option A
Explanation: Using the formula for summation of n terms of a GP:
Sn = a(rn -1)/(r-1)
Putting the value of a=1, r =2 , n=10
We get the solution :
210 - 1
Q.7. Find the sum of an infinite GP 3,1,1/3,…. ?
A. 9
B. 9/2
C. 12
D. 24
Ans. Option B
Explanation: The sum of an infinite GP is given by,
Sn = a / 1-r
a=3, r=1/3.
Putting the values in the equation :
3 / (1-1/3)= 9/2
Q.8. How many terms of the following G. P.: 64, 32, 16, … has the sum 127 ½ ?
A. 8
B. 5
C. 12
D. 24
Answer: Option A
Explanation: : Here a = 64, r =
Using the formula , we get

......(given)
or
or
or
∴ n = 8
Q.9. Find the sum up to n terms of the sequence: 0.7, 0.77, 0.777, …
A. 7n-1
B.
C. 7n/81
D. 7n+1/10n
Answer: Option B
Explanation: Let S denote the sum, then
S = 0.7 + 0.77 + 0.777 +..... to n term
= 7(0.1 + 0.11 + 0.111 +..... to n term )
= (0.9 + 0.99 + 0.999 + .... to n terms)
= {(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + .... to n terms}
= {(1+1+1+....n terms) - (0.1 + 0.01 + 0.001 + ... to n terms)}
=
= (Since r< 1)
=
=
=
Q.10. Express the following recurring decimal as an infinite G. P. and find its value in rational form:
A. 1/3
B. 1/5
C. 1/7
D. 1/9
Answer: Option A
Explanation: = 0.333333.....
=0.3+0.03+0.003+0.0003+......
=
The above is an infinite G.P. with the first term a = and
Hence, by using the formula we get

Hence, the recurring decimal