waecmaths question:
If 6, P and 14 are consecutive terms in an Arithmetic Progression (A.P), find the value of P.
Option A:
9
Option B:
8
Option C:
10
Option D:
6
waecmaths solution:
$\begin{align} & a=\text{ }1st\text{ }term,\text{ }d=common\text{ }difference \\ & {{T}_{n}}=a+(n-1)d \\ & {{T}_{1}}=a=6 \\ & {{T}_{2}}=a+d=P \\ & {{T}_{3}}=a+2d=14 \\ & d={{T}_{2}}-{{T}_{1}}=P-6--(i) \\ & d={{T}_{3}}-{{T}_{2}}=14-P--(ii) \\ & \text{Equating }(i)\text{ and }(ii) \\ & P-6=14-P \\ & 2P=20 \\ & P=10 \\\end{align}$
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