Geometric progression – count terms up to a given term: How many terms are there in the GP 5, 20, 80, 320, …, 20480?

Introduction / Context:
Counting terms in a GP reduces to solving a_n = a * r^(n−1) for n, given first term, common ratio, and last term.

Given Data / Assumptions:

• a = 5
• Common ratio r = 20/5 = 4
• Last term L = 20480

Concept / Approach:
Set 5 * 4^(n−1) = 20480 ⇒ 4^(n−1) = 4096. Express 4096 as a power of 4 to find n − 1.

Step-by-Step Solution:
4^5 = 1024, 4^6 = 4096.Thus 4^(n−1) = 4^6 ⇒ n − 1 = 6 ⇒ n = 7.

Verification / Alternative check:
The sequence terms are 5, 20, 80, 320, 1280, 5120, 20480 — exactly 7 terms.

Why Other Options Are Wrong:
6 or 5 underestimate; 8 or 10 overestimate the power index required to reach 20480.

Common Pitfalls:
Dividing by r incorrectly or misidentifying r from adjacent terms.

Final Answer:
7