An A.P. consists of 21 terms. The sum of the three terms in the middle is 129 and of the last three is 237. Find the A.P.

Let 1^st term of A.P be a and common difference be d.

Now three middle terms of this A.P are $a_{10},a_{11}$  and $a_{12}$

According to the question $a_{10}+a_{11}+a_{12}=129$

$\Rightarrow \left(a+9d\right)+\left(a+10d\right)+\left(a+11d\right)=129$

$\Rightarrow 3a+30d=129$

$\Rightarrow a+10d=43$

$\Rightarrow a=43-10d\cdots \cdots \cdots \left(1\right)$

Also, last three terms are $a_{19},a_{20}$  and $a_{21}$

$\therefore a_{19}+a_{20}+a_{21}=237$

$\Rightarrow \left(a+18d\right)+\left(a+19d\right)+\left(a+20d\right)=237$

$\Rightarrow 3a+57d=237$

$\Rightarrow a+19d=79$

$\Rightarrow 43-10d+19d=79$

Using (1)

$9d=36\Rightarrow d=4$ and $a=3$

$\therefore$ A.P is 3,7,11,15,…