Question

The sum of the second and third terms of a geometric sequence is 96. The sum to infinity of this sequence is 500. Find the possible values for the common ratio, r.

Answer 1

Let, first term be a and common ratio is r.

So, $\dfrac{a}{r}+\dfrac{a}{r^2}=96$

Sum of series, $S = \dfrac{a}{1-r}=500$

a = 500( 1 - r )

$a( \dfrac{1}{r}+\dfrac{1}{r^2})=96\\\\r=0.916$

Therefore, the value of common ratio is 0.916 .

Hence, this is the required solution.