If the sum of the first two terms of a G P is 3... - JAMB Mathematics 2013 Question

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