Question

Find the ${40}^{th}$ term of an A.P where $t_8=56$ and $t_{15}=91$

Answer 1

$t_8=a+7d=56$

$t_{15}=a+14d=91$

$\Rightarrow a+14d-a-7d=91-56$

$\Rightarrow 7d=35$

$\therefore d=\frac{35}{7}=5$

$\left(1\right)\Rightarrow a+7\times 5=56$

$\Rightarrow a=56-35=21$

$\therefore t_{40}=a+39d=21+39\times 5=21+195=216$

$\therefore {40}^{th}$ term is $216$