Question:
If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :
If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :
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Terms 6 to 8 are \(r^4\) times terms 2 to 4, so \(r^4=81\); then find the first term from the first condition.
Updated On: Nov 5, 2026
- \( \frac{2}{13}\left(3^{50}-1\right) \)
- \( \frac{1}{13}\left(3^{50}-1\right) \)
- \( \frac{1}{26}\left(3^{49}-1\right) \)
- \( \frac{1}{26}\left(3^{50}-1\right) \)
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The Correct Option is D
Solution and Explanation
Let the first term be \(a\) and the common ratio \(r>0\).
\(ar+ar^2+ar^3=3\) and \(ar^5+ar^6+ar^7=243\).
Dividing: \(r^4=81\Rightarrow r=3\).
\(ar(1+r+r^2)=3\Rightarrow a\cdot3\cdot13=3\Rightarrow a=\frac{1}{13}\).
\(S_{50}=a\,\frac{r^{50}-1}{r-1}=\frac{1}{13}\cdot\frac{3^{50}-1}{2}=\frac{1}{26}\left(3^{50}-1\right)\).
Answer: (4) \( \frac{1}{26}\left(3^{50}-1\right) \)
\(ar+ar^2+ar^3=3\) and \(ar^5+ar^6+ar^7=243\).
Dividing: \(r^4=81\Rightarrow r=3\).
\(ar(1+r+r^2)=3\Rightarrow a\cdot3\cdot13=3\Rightarrow a=\frac{1}{13}\).
\(S_{50}=a\,\frac{r^{50}-1}{r-1}=\frac{1}{13}\cdot\frac{3^{50}-1}{2}=\frac{1}{26}\left(3^{50}-1\right)\).
Answer: (4) \( \frac{1}{26}\left(3^{50}-1\right) \)
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