JAMB Mathematics 2023 Paper

Concept:Use $T_n = ar^{n-1}$ for the second term and $S_\infty = \frac{a}{1 - r}$ for the sum to infinity, remembering that $|r|$ must be less than 1 for convergence.Explanation:The second term is $T_2 = ar = \frac{-2}{3}$.The sum to infinity is $\frac{a}{1 - r} = \frac{3}{2}$.From the sum formula, $a = \frac{3}{2}(1 - r)$.Substitute this into $ar = \frac{-2}{3}$.$$\frac{3}{2}(1 - r)r = \frac{-2}{3}$$Multiply through by 6 to get $9r(1 - r) = -4$.Simplify to form the quadratic $9r^2 - 9r - 4 = 0$.Factorise as $(3r - 4)(3r + 1) = 0$.This gives $r = \frac{4}{3}$ or $r = \frac{-1}{3}$.But $\frac{4}{3}$ has an absolute value greater than 1, so the series would not have a finite sum to infinity.Therefore, the only valid common ratio is $r = \frac{-1}{3}$.Answer:$r = \frac{-1}{3}$, which is option A.