Question

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

Answer 1

We know that the $n^{th}$ term of the arithmetic progression is given by $a+(n-1)d$

Given that ${10}^{th}$ term of an A.P. is $41$

Therefore, $a+(10-1)d=41$

$⟹a+9d=41$ ---------(1)

Given that ${18}^{th}$ term of an A.P. is $73$

Therefore, $a+(18-1)d=73$

$⟹a+17d=73$ ---------(2)

subtracting eqn (2) from eqn (1) gives $(a+17d)-(a+9d)=73-41$

$⟹8d=32$

$⟹d=4$

substituting $d=4$ in eqn (1) we get

$a+9\left(4\right)=41$

$⟹a=41-36=5$

Therefore, ${26}^{th}$ term is $a+(26-1)d=5+25\left(4\right)=105$