# Geometric Progression-Exercise 2

The sum of the numbers of an infinite decreasing geometric progression is 8.

Find y1 dhe q, considering that the sum of their cubes is $\frac{512}{7}$

Solution:

$\frac{y_{1}}{1-q}=8$

$\frac{y_{1}^{3}}{1-q^{3}}=\frac{512}{7}$

Make fraction in each side.

$\frac{y_{1}^{3}}{1-q^{3}}\cdot&space;\frac{1-q}{y_{1}}=\frac{512}{7}\cdot&space;\frac{1}{8}=\frac{64}{7}$

$\frac{y_{1}^{2}}{1+q+q^{2}}=\frac{64}{7}$

$7y_{1}^{2}=64(1+q+q^{2})$         $y_{1}=8(1-q)$

$7\cdot&space;64(1-q)^{2}=64\cdot&space;(1+q+q^{2})$

$7\cdot&space;(1-2q+q^{2})=&space;1+q+q^{2}$

$7q^{2}-14q+7-1-q-q^{2}=0$

$6q^{2}-15q+6=0$

$2q^{2}-5q+2=0$         $D=25-16=9$

$_{2}q_{1}=\frac{5\pm&space;3}{4}$                                                  $q_{1}=2$        $q_{2}=\frac{1}{2}\rightarrow$the quotinent of decreasing geometriv progression(we don’t take q2, because then, it wont be an infinite decreasing geometric progression anymore.

$q_{2}=\frac{1}{2}\Rightarrow&space;y_{1}=8\cdot&space;\frac{1}{2}=4$