WAEC 2014 · Paper 2 · Q11

1. The sum to infinity: $\dfrac{a}{1 - r} = 25$, so $a = 25(1 - r)$.
2. The first two terms: $a + ar = 16$, so $a(1 + r) = 16$.
3. Substitute: $25(1 - r)(1 + r) = 16$, so $1 - r^2 = \frac{16}{25}$ and $r^2 = \frac{9}{25}$.
4. The ratio is positive, so $r = \frac35$.
5. Then $a = 25 \times \frac25 = 10$.

(a)

1. $T_5 = ar^4$
   $= 10 \times \dfrac{81}{625}$
   $= \dfrac{162}{125}$
   $= 1\dfrac{37}{125}$.

(b)

1. $S_4 = \dfrac{a(1 - r^4)}{1 - r}$
   $= 25\left(1 - \dfrac{81}{625}\right)$
   $= 25 \times \dfrac{544}{625}$
   $= \dfrac{544}{25}$
   $= 21\dfrac{19}{25}$.