waecmaths question:
Find the 6th term of the sequence: \[\frac{2}{3},\frac{7}{15},\frac{4}{15},\cdot \cdot \cdot \]
Option A:
$-\frac{1}{3}$
Option B:
$-\frac{1}{5}$
Option C:
$\frac{1}{15}$
Option D:
$\frac{1}{5}$
waecmaths solution:
$\begin{align} & a=\frac{2}{3} \\ & common\text{ }difference=\frac{7}{15}-\frac{2}{3}=\frac{7-10}{15}=-\frac{1}{5} \\ & {{T}_{n}}=a+(n-1)d \\ & {{T}_{6}}=\tfrac{2}{3}+(6-1)(-\tfrac{1}{5})=\tfrac{2}{3}+(5)(-\tfrac{1}{5})=\tfrac{2}{3}-1 \\ & {{T}_{6}}=-\tfrac{1}{3} \\ & \text{The }{{6}^{th}}-term\text{ is }-\tfrac{1}{3} \\\end{align}$
maths year:
maths topics: