The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34.

Solution:

Let a be the first term and d be the common difference. Then,

$a_{4}+a_{8}=24$

$\Rightarrow a+(4-1) d+a+(8-1) d=24$

$\Rightarrow a+3 d+a+7 d=24$

$\Rightarrow 2 a+10 d=24$

$\Rightarrow a+5 d=12 \quad \ldots(\mathrm{i})$

Also, $a_{6}+a_{10}=34$

$\Rightarrow a+(6-1) d+a+(10-1) d=34$

$\Rightarrow a+5 d+a+9 d=34$

$\Rightarrow 2 a+14 d=34$

$\Rightarrow a+7 d=17 \quad \ldots(\mathrm{ii})$

Solving $(\mathrm{i})$ and $(\mathrm{ii})$, we get :

$2 d=5$

$\Rightarrow d=\frac{5}{2}$

Substituing the value in $(\mathrm{i})$, we get :

$a+5\left(\frac{5}{2}\right)=12$

$\Rightarrow a+\frac{25}{2}=12$

$\Rightarrow a=\frac{-1}{2}$