JAMB Mathematics 2023 Paper

$T_2 = \frac{-2}{3}; S_\infty \frac{3}{2}$

$T_n = ar^n - 1$

∴ $T_2 = ar = \frac{-2}{3}$---eqn.(i)

$S_\infty = \frac{a}{1 - r} = \frac{3}{2}$---eqn.(ii)

= 2a = 3(1 - r)

= 2a = 3 - 3r

∴ a = $\frac{3 - 3r}{2}$

Substitute $\frac{3 - 3r}{2}$ for a in eqn.(i)

= $\frac{3 - 3r}{2} \times r = \frac{-2}{3}$

= $\frac{3r - 3r^2}{2} = \frac{-2}{3}$

= 3(3r - 3r$^2$) = -4

= 9r - 9r$^2$ = -4

∴ r = $\frac{4}{3} \text{ or } -\frac{1}{3}$

For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.

∴ r = -$\frac{1}{3}$ (since |-$\frac{1}{3}$| < 1)