Mark the correct alternative in each of the following:

In the given problem, we are given 7th and 13th term of an A.P.

We need to find the 26th term

Here,

$a_{7}=34$

$a_{13}=64$

Now, we will find $a_{7}$ and $a_{13}$ using the formula $a_{n}=a+(n-1) d$

So,

$a_{7}=a+(7-1) d$

$34=a+6 d$ .......(1)

Also,

$a_{13}=a+(13-1) d$

$64=a+12 d$.......(2)

Further, to solve for a and d

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

$64-34=(a+12 d)-(a+6 d)$

$30=a+12 d-a-6 d$

$30=6 d$

$d=\frac{30}{6}$

$d=5$ ......$.(3)$

Substituting (3) in (1), we get

$34=a+6(5)$

$34=a+30$

$a=34-30$

$a=4$

Thus,

$a=4$

$d=5$

So, for $18^{\text {th }}$ term $(n=18)$,

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

$a_{18}=4+(18-1) 5$

$=4+17(5)$

$=4+85$

$=89$

Therefore, $a_{18}=89$

Hence, the correct option is (c).