Question

If sum of infinite gp p+1+1/p+1/p2+........... (p>2) is 49/6 then sum of 3rd term and 4th of the gp is

Answer 1

Given:

Infinite GP has sum $\frac{49}{6}$.

The GP is:

$p+1+\dfrac{1}{p}+\dfrac{1}{p^2} ......$

To find:

3rd and 4th term of GP = ?

Solution:

First term of GP = p

Common Ratio of GP =

$\frac{2nd\ term}{1st\ term}\\\Rightarrow \dfrac{1}{p}$

Given that p > 2, so common ratio $\frac{1}{p} <1$

Formula for infinite sum of a GP where r < 1 :

$S_{\infty} = \dfrac{a}{1-r}$

Putting the values:

$\dfrac{49}{6} = \dfrac{p}{1-\dfrac{1}{p}}\\\Rightarrow \dfrac{49}{6} = \dfrac{p\times p}{p-1}\\\Rightarrow p = 7, \dfrac{14}{12}$

Given that p > 2, so p = 7 is the only solution.

Now, sum of 3rd and 4th term is:

$\dfrac{1}{7}+\dfrac{1}{7^2}\\\Rightarrow \dfrac{1}{7}+\dfrac{1}{49}\\\Rightarrow \dfrac{7+1}{49}\\\Rightarrow \dfrac{8}{49}$