Geometric progression question

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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 ?

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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