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कृपया अपनी पसंदीदा भाषा चुनें
यदि
$8=3+\frac{1}{4}(3+p)+\frac{1}{4^{2}}(3+2 p)+\frac{1}{4^{3}}(3+3 p)+\ldots \infty$,
है, तो $p$ का मान है
$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^{2}}$
(ए.जी.पी. के अनंत पदों का योग $=\frac{a}{1-r}+\frac{d r}{(1-r)^{2}}$ )
$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$
बीटा
