# Geometric progression question

a GP with positive common ratio is such that the sum of first 2 terms is 17.5 and third term is $4\frac{2}{3}$ compute first term and the common difference

$17.5 = \frac{a(1-r^n)}{1-r}$

$ar^2 = 4\frac{2}{3}$

how do I solve these 2 to get a and r ?

assume that the terms are $a, ar, ar^2$
the sum of the first two terms is $a+ar=17.5$
the third term is $ar^2=14/3$
thus you can write $a$ as $\frac{14}{3r^2}$ and put in the first equation
$$\frac{14}{3r^2}+\frac{14}{3r}=17.5$$ $$14+14r=17.5*3r^2$$ now can you solve it