If $$3+14(3+d)+142(3+2d)+$$…∞$$=8$$ then the value of ' d ' is

$$3+34+342+$$…∞$$+d41+24+$$…∞⇒$$31$$−$$1/4+d.1/41$$−$$142=8$$⇒$$4+4d9=8$$→$$d=9$$

Arithmetico-Geometric Progression

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Question

$If 3 + \frac{1}{4} (3 + d) + \frac{1}{4^{2}} (3 + 2 d) + \dots \infty = 8 then the value of ' d' is$

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Solution

$(3 + \frac{3}{4} + \frac{3}{42} + \dots \infty) + \frac{d}{4} (1 + \frac{2}{4} + \dots \infty)$
$\begin{aligned} \Rightarrow \frac{3}{1 - 1 / 4} + \frac{d .1 / 4}{{(1 - \frac{1}{4})}^{2}} = 8 \\ \Rightarrow 4 + \frac{4 d}{9} = 8 \to d = 9 \end{aligned}$

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