Question

The $6th$ and $17th$ term of an A.P are $19$ and $41$ respectively , find the $40th$ term ?

Answer 1

$T_n=a+(n-1)d$
Where $a$ is the first term, $d$ is the common difference and $T_n$ denotes the n-th term of an AP.
Given:
$19=a+5d$
$41=a+16d$
Solving the above equations gives: $d=2,a=9$
Therefore, $T_{40}=9+39\cdot 2=87$