JEE Mains · Maths · STD 11 - 8. sequence and series
If \(8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2 p)+\frac{1}{4^3}(3+3 p)+\ldots \infty,\) then the value of \(p\) is
- A \(9\)
- B \(5\)
- C \(6\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(9\)
Step-by-step Solution
Detailed explanation
\(8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}\) (sum of infinite terms of A.G.P \(=\frac{\mathrm{a}}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1-\mathrm{r})^2}\) ) \(\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9\)
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