S=a+(a+d)r+(a+2d)r2+… Then S=1−ra+(1−r)2dr,∣r∣<1 Since, r=71 and a=5,d=α $\begin{aligned}
& 7=\frac{5}{1-\frac{1}{7}}+\frac{\alpha \cdot \frac{1}{7}}{\left(1-\frac{1}{7}\right)^2} \
& \Rightarrow \alpha=6
\end{aligned}$
If 7=5+ 1 7(5+α)+ 1 7^2(5+2 α)+ 1 7^3(5+3 α)+… ∞, then the value of α is :
JEE Main 2025 — Mathematics Algebra
If 7=5+71(5+α)+721(5+2α)+731(5+3α)+…∞, then the value of α is :
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Options
- A
76
- B
6
- C
71
- D
1
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