Question

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

Answer 1

We know that the $n^{th}$ term of the arithmetic progression is given by $a+(n-1)d$

Given that $6^{th}$ term of an A.P. is $19$

Therefore, $a+(6-1)d=19$

$⟹a+5d=19$ ---------(1)

Given that ${17}^{th}$ term of an A.P. is $41$

Therefore, $a+(17-1)d=41$

$⟹a+16d=41$ ---------(2)

subtracting eqn (2) from eqn (1) gives $(a+16d)-(a+5d)=41-19$

$⟹11d=22$

$⟹d=2$

substituting $d=2$ in eqn (1) we get

$a+5\left(2\right)=19$

$⟹a=19-10=9$

Therefore, ${40}^{th}$ term is $a+(40-1)d=9+39\left(2\right)=87$