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the 4th term of a geometric progression is 8 and t
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.
CAT - 2013
CAT
Updated On:
Jul 31, 2025
256
512
1024
2048
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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: $a_4 = a r^3 = 8$, $a_7 = a r^6 = 64$.
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Step 2: Form ratio.
$\frac{a r^6}{a r^3} = \frac{64}{8} \implies r^3 = 8 \implies r = 2$.
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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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