WAEC Mathematics 2017 Paper

Concept:The terms form an arithmetic progression (AP), so the 6th term can be found using the AP formula after determining the common difference.Explanation:First term: $a = \frac{2}{3} = \frac{10}{15}$.Common difference: $d = \frac{7}{15} - \frac{2}{3} = \frac{7 - 10}{15} = \frac{-3}{15} = -\frac{1}{5}$.Check the next term: $\frac{4}{15} - \frac{7}{15} = -\frac{3}{15} = -\frac{1}{5}$, so the sequence is an AP.Use the formula for the $n$th term: $T_n = a + (n-1)d$.For the 6th term, $T_6 = a + 5d$.Substitute the values: $T_6 = \frac{2}{3} + 5\left(-\frac{1}{5}\right)$.Simplify: $T_6 = \frac{2}{3} - 1 = \frac{2 - 3}{3} = -\frac{1}{3}$.Answer:The 6th term is $-\frac{1}{3}$, which is Option A.