The 6th and 17th terms of an A.P.

In the given problem, we are given $6^{\text {th }}$ and $17^{\text {th }}$ term of an A.P.

We need to find the $40^{\text {th }}$ term

Here,

$a_{6}=19$

$a_{17}=41$

Now, we will find $a_{6}$ and $a_{17}$ using the formula $a_{n}=a+(n-1) d$

So,

$a_{6}=a+(6-1) d$

$19=a+5 d$  .......(1)

Also,

$a_{17}=a+(17-1) d$

$41=a+16 d$.......(2)

So, to solve for a and d

On subtracting (1) from (2), we get

$a+16 d-a-5 d=41-19$

$11 d=22$

$d=\frac{22}{11}$

$d=2$.......$(3)$

So, to solve for a and d

Substituting (3) in (1), we get

$19=a+5(2)$

$19-10=a$

$a=9$

Thus,

$a=9$

$d=2$

$n=40$

Substituting the above values in the formula $a_{n}=a+(n-1) d$

$a_{40}=9+(40-1) 2$

$a_{40}=9+80-2$

$a_{40}=87$

Therefore, $a_{40}=87$