The 10th and 18th terms of an A.P. are 41 and 73 respectively.

Given:

$a_{10}=41$

$\Rightarrow a+(10-1) d=41 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+9 d=41$

And, $a_{18}=73$

$\Rightarrow a+(18-1) d=73 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+17 d=73$

Solving the two equations, we get:

$\Rightarrow 17 d-9 d=73-41$

$\Rightarrow 8 d=32$

$\Rightarrow d=4$              ...(i)

Putting the value in first equation, we get:

$a+9 \times 4=41$

$\Rightarrow a+36=41$

$\Rightarrow a=5$         ...(ii)

$a_{26}=a+(26-1) d \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a_{26}=a+25 d$

$\Rightarrow a_{26}=5+25 \times 4$     (From (i) and (ii))

$\Rightarrow a_{26}=5+100=105$