Question

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Answer 1

Given:
$a_{10}=41$
$\Rightarrow a+(10-1)d=41$ $[a_n=a+(n-1\left)d\right]$
$\Rightarrow a+9d=41$
And, $a_{18}=73$
$\Rightarrow a+(18-1)d=73$ $[a_n=a+(n-1\left)d\right]$
$\Rightarrow a+17d=73$
Solving the two equations, we get:
$\Rightarrow 17d-9d=73-41\Rightarrow 8d=32\Rightarrow d=4\dots \dots \left(i\right)$
Putting the value in first equation, we get:
$a+9\times 4=41$
$\Rightarrow a+36=41$
$\Rightarrow a=5\dots \dots \left(ii\right)$
$a_{26}=a+(26-1)d$ $[a_n=a+(n-1\left)d\right]$
$\Rightarrow a_{26}=a+25d$
$\Rightarrow a_{26}=5+25\times 4$ [From (i) and (ii)]
$\Rightarrow a_{26}=5+100=105$