JEE Mains · Maths · STD 11 - 8. sequence and series
If \(7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\frac{1}{7^3}(5+3 \alpha)+\ldots \infty\), then the value of \(\alpha\) is :
- A \(\frac{6}{7}\)
- B 6
- C \(\frac{1}{7}\)
- D 1
Answer & Solution
Correct Answer
(B) 6
Step-by-step Solution
Detailed explanation
\(S=a+(a+d) r+(a+2 d) r^2+\ldots\) Then \(S=\frac{a}{1-r}+\frac{d r}{(1-r)^2},|r| < 1\) Since, \(r=\frac{1}{7}\) and \(a=5, d=\alpha\)…
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