If the sum of the first two terms of a G.P. is 3, and the sum of the second and the third terms is -6, find the sum of the first term and the common ratio

Answer Details

Using Sn = $a\frac{r^2-1}{r-1}$
we get S2 = 3 = $a\frac{r^2-1}{r-1}$
giving 3(r - 1) = a(r2 - 1)
3r - 3 = ar2 - a
ar2 - 3r - a = -3 ..... (1)
ar + ar2 = -6 ..... (2)
From (2), a = $\frac{-6}{(r+r^2)}$
Substitute $\frac{-6}{(r+r^2)}$ for a in (1)
$\left(\frac{-6}{(r+r^2)}\right)r^2-3r-\frac{-6}{(r+r^2)}=-3$
Multiply through by (r + r2) to get
-6r2 - 3r(r + r2) + 6 = -3(r + r2)
-6r2 - 3r2 - 3r3 + 6 = -3r - 3r2
Equating to zero, we have
3r3 - 3r2 + 3r2 + 6r2 - 3r - 6 = 0
This reduces to;
3r3 + 6r2 - 3r - 6 = 0
3(r3 + 2r2 - r - 2) = 0
By the factor theorem,
(r + 2): f(-2) = (-2)3 + 2(-2)2 - (-2) - 2
-8 + 8 + 2 - 2 = 0
giving r = -2 as the only valid value of r for the G.P.
From (3), = $\frac{-6}{-2+(-2{)}^2}=\frac{-6}{-2+4}$
a = -6/2 = -3
Hence (a + r) = (-3 + -2) = -5