Sequence and Series
1 Board Maths previous year questions on Sequence and Series — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.
If \(8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}\left(3+p^2\right)+\ldots \infty\) then the value of \(p\) is (22 Jan, Shift I, Memory Based)
\(\frac{16}{5}\)
\(\begin{aligned}& 8=\left(3+\frac{3}{4}+\frac{3}{4^2}+\ldots+\infty\right)+\left(\frac{p}{4}+\frac{p^2}{4^2}+\ldots+\infty\right) \\& 8=3\left(1+\frac{1}{4}+\frac{1}{4^2}+\ldots+\infty\right)+\left(\frac{p}{4}+\frac{p^2}{4^2}+\ldots+\infty\right) \\& 8=3\left(\frac{1}{1-\frac{1}{4}}\right)+\frac{\frac{p}{4}}{1-\frac{p}{4}}\end{aligned}\)
\(\begin{aligned}& 8=3\left(\frac{4}{3}\right)+\frac{p}{4-p} \\& 4=\frac{p}{4-p} \\& \Rightarrow 16-4 p=p \\& \Rightarrow 5 p=16 \\& \Rightarrow p=\frac{16}{5}\end{aligned}\)
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