Geometric Progression Questions with answers are provided here. Class 10 and 11 students can practise the questions based on geometric progression to prepare for the exams. These geometric progression problems are prepared by our subject experts, as per the NCERT curriculum and latest CBSE syllabus (2022-2023). Learn more: Geometric Progression.
Geometric Progression Questions and Solutions
Q.1: If the first term of a G.P. is 20 and the common ratio is 4. Find the 5th term.
Solution: Given,
First term, a=20
Common ratio, r=4
We know,
Nth term of G.P.,
an = arn-1
⇒ a5 = 20×44
= 20×256
= 5120
Q.2: The sum of the first three terms of a G.P. is 21/2 and their product is 27. Find the common ratio.
Solution: Let three terms of G.P. be a/r, a, ar.
Given,
Product of first three terms = 27
⇒ (a/r) (a) (ar) = 27
⇒ a3 = 27
⇒ a = 3.
Sum of first three terms = 21/2
⇒ (a / r + a + ar) = 21/2
⇒ a (1 / r + 1 + 1r) = 21/2
⇒ (1 / r + 1 + 1r) = (21/2)/3 = 7/2
⇒ (r2 + r + 1) = (7/2) r
⇒ r2 – (5/2) r + 1 = 0
⇒ r = 2 and ½
Q.3: Find a Geometric Progress for which the sum of first two terms is -4 and the fifth term is 4 times the third term.
Solution: Let the first term of the geometric series be a and the common ratio is r.
Sum of the first two terms = -4
a + ar = -4 ………………(i)
Fifth term is 4 times the third term.
ar4 = 4ar2
r2 = 4
r = ±2
If we consider r = 2, then putting value of r in eq.(i)
a(1+2) = -4
a = -4/3
ar = -8/3
ar2 = -16/3
Thus, the G.P. is -4/3, -8/3, -16/3, …..
Q.4: The number 2048 is which term in the following Geometric sequence 2, 8, 32, 128, . . . . . . . . .
Solution: Here a = 2 and r = 4
nth term G.P is an = arn-1
⇒ 2048 = 2 x ( 4) n-1
⇒ 1024 =( 4) n-1
⇒ ( 4) 5 = ( 4) n-1
⇒ n = 6
Q.5: In a G.P, the 6th term is 24 and the 13th term is 3/16 then find the 20th term of the sequence.
Solution: Let first term be ‘a’ and common ratio is ‘r’
Given,
a6 = 24 ————- ( i)
a13 = 3/16 ————– ( ii)
⇒ a6 = a r6-1
a13 = a r13-1
⇒ 24 = a r5
3/16 = a r12
⇒ r7 = 3 / 24 x 16 = 1 / (2)7
⇒ r = 1/2 ———– (iii)
Thus,
⇒ a6 = 24 = a (1/2)5
⇒ a = 3 x 28
Now a20 = a r20-1
a20 = 3 x 28 x ( 1/2 )19 = 3 / 211
Q.6: Find the sum of the geometric series:
4 – 12 + 36 – 108 + ………….. to 10 terms
Solution: The first term of the given Geometric Progression = a = 4 and its common ratio = r = −12/4 = -3.
Sum of the first 10 terms of geometric series:
S10 = a. (rn – 1/r-1)
= 4. ((-3)10 – 1)/(-3-1)
= – (-3)10 – 1
= – 59048
Q.7: ‘x’ and ‘y’ are two numbers whose AM is 25 and GM is 7. Find the numbers.
Solution: Here x’ and ‘y’ are two numbers then
Arithmetic mean = AM = (x+y)/2
25 = (x+y)/2
x+y = 50 ………(i)
Geometric mean, GM = √(xy)
7 = √(xy
72 = xy
xy = 49 ………..(ii)
Solving equation (i) and (ii), we get;
x = 1 and y = 49.
Q.8: Determine the common ratio r of a geometric progression with the first term is 5 and fourth term is 40.
Solution: Given,
First term, a1 = 5
Fourth term, a4 = 40
a4/a1 = 40/5
a1r3/a1 = 40/5
r3 = 8
r = 2
Q.9: If the nth term of a GP is 128 and both the first term a and the common ratio r
are 2. Find the number of terms in the GP.
Solution: nth term of a GP, an = 128
First term of GP, a = 2
Common ratio, r = 2
Nth term of G.P., an = a.rn-1
128 = 2.2n-1
64 = 2n-1
26 = 2n-1
n- 1 = 6
n = 7
Therefore, there are 7 terms in GP.
Q.10: What is the sum of infinite geometric series with first term equal to 1 and common ratio is ½?
Solution: By the formula of sum of infinite geometric series, we have;
S = a1 1/(1 – r)
S = 1. 1/(1-½)
S = 1/(½)
S = 2
Hence, the required sum is 2.
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