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 :

Correct answer: B. 6 S=a+(a+d) r+(a+2 d) r^2+ Then S=(a)/(1-r)+(d r)/((1-r)^2),|r| < 1 Since, r=(1)/(7) and a=5, d=α aligned & 7=(5)/(1-(1)/(7))+(α · (1)/(7))/((1-(1)/(7))^2) & α=6 aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Sequences and Series

If 7=5+(1)/(7)(5+α)+(1)/(7^2)(5+2 α)+(1)/(7^3)(5+3 α)+ ∞ , then the...

7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\frac{1}{7^3}(5+3 \alpha)+\ldots \infty

, then the value of

\alpha

is :

Options

A $\frac{6}{7}$
B6
C $\frac{1}{7}$
D1

Correct answer

B. 6

Step-by-step solution

S=a+(a+d) r+(a+2 d) r^2+\ldots

Then

S=\frac{a}{1-r}+\frac{d r}{(1-r)^2},|r| < 1

Since,

r=\frac{1}{7}

and

a=5, d=\alpha

\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}

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