Question

The ${11}^{\text{th}}$ term and the ${21}^{\text{th}}$ term of an A.P. are $16$ and $29$ respectively, the then find ${41}^{\text{th}}$ term of that A.P.

Answer 1

Let $a$ and $d$ be the first term and the common difference of the A.P.

Then according to the problem,

$a+10d=16$......(1) and $a+20d=29$......(2).

Now, (2)$-$(1) we get,

$10d=13$

or, $d=1.3$.

Then $a=16-13=3$.

Now, ${41}^{th}$ term will be $a+40d=3+4\times 13=52+3=55$.