With a = 2 and r = 4, explicit formula is tn = 2 x 4ⁿ⁻¹ and recursive formula is t1 = 2, tn = 4tn-1 for n >= 2. Solving 2 x 4ⁿ⁻¹ = 131072 gives 4ⁿ⁻¹ = 65536 = 4⁸, so n = 9.
Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.
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For the GP 2, 8, 32, …:
First term a = 2, common ratio r = 8 / 2 = 4.
Explicit formula:
tn = a x rⁿ⁻¹ = 2 x 4ⁿ⁻¹
Recursive formula:
t1 = 2 and tn = 4 x tn-1 for n >= 2.
To find which term is 131072:
2 x 4ⁿ⁻¹ = 131072
4ⁿ⁻¹ = 65536
Since 65536 = 4⁸ (as 4⁸ = 2¹⁶ = 65536):
n – 1 = 8
n = 9.
Hence, 131072 is the 9th term of the geometric progression.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/