Question:

The 4th term of a geometric progression is 8, and the 7th term is 64. Find the 10th term.

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For GP, use the ratio of terms to find $r$, then solve for $a$ and compute the required term.
Updated On: Jul 31, 2025
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The Correct Option is B

Solution and Explanation


- Step 1: GP formula. $n$-th term: $a_n = a r^{n-1}$. Given: $a_4 = a r^3 = 8$, $a_7 = a r^6 = 64$.
- Step 2: Form ratio. $\frac{a r^6}{a r^3} = \frac{64}{8} \implies r^3 = 8 \implies r = 2$.
- Step 3: Find $a$. $a r^3 = 8 \implies a \cdot 8 = 8 \implies a = 1$.
- Step 4: 10th term. $a_{10} = a r^9 = 1 \cdot 2^9 = 512$.
- Step 5: Verify. $a_4 = 1 \cdot 2^3 = 8$, $a_7 = 1 \cdot 2^6 = 64$. Matches.
- Step 6: Conclusion. Option (2) 512 is correct.
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