The 6th and 17th terms of an A.P. are 19 and 41 respectively,

Solution:

Given:

$a_{6}=19$

$\Rightarrow a+(6-1) d=19 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+5 d=19 \quad \ldots(1)$

And, $a_{17}=41$

$\Rightarrow a+(17-1) d=41 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+16 d=41 \quad \ldots(2)$

Solving the two equations, we get,

$16 d-5 d=41-19$

$\Rightarrow 11 d=22$

$\Rightarrow d=2$

Putting $d=2$ in the eqn (1), we get:

$a+5 \times 2=19$

$\Rightarrow a=19-10$

$\Rightarrow a=9$

We know:

$a_{40}=a+(40-1) d \quad\left[a_{n}=a+(n-1) d\right]$

$=a+39 d$

$=9+39 \times 2$

$=9+78=87$