Question

If $7^{th}$ term is $108$ and ${11}^{th}$ term is $212$ of an A.P then find its $n^{th}$ term.

Answer 1

We know

$t_n=a+(n-1)d$

$t_7=a+6d$

$\therefore a+6d=108$ -$\left(i\right)$

$t_{11}=a+10d$

$\therefore a+10d=212$ - $\left(ii\right)$

$\left(i\right)-\left(ii\right)$ $-4d=-104$

$\therefore d=26$

$a+6\times 26=108$

$a=108-156$

$\therefore a=-48$

$t_n=-48+(n-1)26$

$\therefore t_n=26n-74$