Question

If

$$\begin{gathered} 3+3\alpha +3{\alpha }^2+\dots .\infty =45/8, \\ \alpha >0 \end{gathered}$$

then the value of α will be

• A. $15/23$
• B. $7/15$
• C. $7/8$
• D. $15/7$

Answer 1

The correct option is B

$7/15$

The explanation for the correct option:

Here $\alpha =3$

Common ratio, $r=\alpha$

$\begin{array}{rcl}SumofinfinitetermsofGP& =& a/(1-r)\\ & \Rightarrow & 3+3\alpha +3{\alpha }^2+\dots .\infty =3/(1-\alpha )\\ & \Rightarrow & 45/8=3/(1-\alpha )\left(given\right)\\ & \Rightarrow & 45–45\alpha =24\\ & \Rightarrow & 45\alpha =45–24\\ & \Rightarrow & 45\alpha =21\\ & \Rightarrow & \alpha =21/45\\ & \Rightarrow & =7/15\end{array}$

Hence, option (B) is the correct option.