The 10th and 18th terms of an A.P.

In the given problem, we are given 10th and 18th term of an A.P.

We need to find the 26th term

Here,

$a_{10}=41$

$a_{18}=73$

Now, we will find $a_{10}$ and $a_{18}$ using the formula $a_{n}=a+(n-1) d$

So,

$a_{10}=a+(10-1) d$

$41=a+9 d$........(1)

Also,

$a_{18}=a+(18-1) d$

$73=a+17 d$.......(2)

So, to solve for a and d

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

$8 d=32$

$d=\frac{32}{8}$

$d=4$

Substituting d=4 in (1), we get

$41=a+9(4)$

$41-36=a$

$a=5$

Thus,

$a=5$

$d=4$

$n=26$

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

$a_{26}=5+(26-1) 4$

$a_{26}=5+100$

$a_{26}=105$

Therefore, $a_{26}=105$