Using S_n = \(a\frac{r^2 - 1}{r - 1}\)
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)
\((\frac{-6}{(r + r^2)})r^2 - 3r - \frac{-6}{(r + r^2)} = -3\)
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