Question:
The first and last term of a G.P. are 7 and 448 respectively. If the sum is 889, then the common ratio is
The first and last term of a G.P. are 7 and 448 respectively. If the sum is 889, then the common ratio is
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If $ar^{n-1}$ is a perfect power, try simple integer values of $r$ like 2 or 3.
Updated On: Apr 24, 2026
- $4$
- $2$
- $\frac{1}{2}$
- $\frac{1}{4}$
- $3$
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The Correct Option is B
Solution and Explanation
Concept:
• Last term: $l = ar^{n-1}$
• Sum of G.P.: $S_n = a \dfrac{r^n - 1}{r - 1}$
Step 1: Use last term formula
\[ 448 = 7r^{n-1} \Rightarrow r^{n-1} = 64 = 2^6 \]
Step 2: Assume $r=2$ and verify
\[ r^{n-1} = 2^6 \Rightarrow n-1 = 6 \Rightarrow n = 7 \]
Step 3: Check sum
\[ S_n = 7 \cdot \frac{2^7 - 1}{2 - 1} = 7(128 - 1) = 7 \cdot 127 = 889 \]
Step 4: Conclusion
Sum matches given value, so $r=2$ is correct. Final Conclusion:
Common ratio $r = 2$
• Last term: $l = ar^{n-1}$
• Sum of G.P.: $S_n = a \dfrac{r^n - 1}{r - 1}$
Step 1: Use last term formula
\[ 448 = 7r^{n-1} \Rightarrow r^{n-1} = 64 = 2^6 \]
Step 2: Assume $r=2$ and verify
\[ r^{n-1} = 2^6 \Rightarrow n-1 = 6 \Rightarrow n = 7 \]
Step 3: Check sum
\[ S_n = 7 \cdot \frac{2^7 - 1}{2 - 1} = 7(128 - 1) = 7 \cdot 127 = 889 \]
Step 4: Conclusion
Sum matches given value, so $r=2$ is correct. Final Conclusion:
Common ratio $r = 2$
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