First slide

Geometric progression

Question

If sum of the infinite G.P. $p + 1 + \frac{1}{p} + \frac{1}{p^{2}} + \dots, (p > 2)$ is 49/6, then sum of the 3rd term and the 4th term of the G.P. is

Moderate

A

$\frac{8}{49}$
B

$\frac{1}{7}$
C

$\frac{6}{7}$
D

$\frac{23}{49}$

Solution

For $p > 1,$ $\frac{49}{6} = p + 1 + \frac{1}{p} + \frac{1}{p^{2}} + \dots = \frac{p}{1 - 1 / p}$
$\begin{array}{ll} \Rightarrow \frac{49}{6} = \frac{p^{2}}{p - 1} \\ \Rightarrow 6 p^{2} - 49 p + 49 = 0 \\ \Rightarrow 6 p^{2} - 42 p - 7 p + 49 = 0 \\ \Rightarrow (6 p - 7) (p - 7) = 0 \end{array}$
As $p > 2, p = 7$
Thus, $\begin{array}{l} a_{3} + a_{4} = p {(\frac{1}{p})}^{2} + p {(\frac{1}{p})}^{3} \\ = \frac{1}{7} + \frac{1}{49} = \frac{8}{49} \end{array}$

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