The sum of the 4^th and 8^th terms of an A.P is 24 and the sum of the 6^th and 10^th terms is 44, then the first three terms of the A.P is 

Given that in an A.P

$t_4+t_8=24and{t}_6+t_{10}=44$

$\Rightarrow a+3d+a+7d=24\text{ and }a+5d+a+9d=44$

$\Rightarrow 2a+10d=24\_\_\left(1\right)and2a+14d=44\_\_\_\left(2\right)$

$solving\left(1\right)\&\left(2\right)2a+10d=24$

                            $\underline{2a+14d=44}$

                               $-4d=-20$

                                     $d=5$

$2a+10\left(5\right)=24$

$\Rightarrow 2a=-26$

$\Rightarrow a=-13$

$\therefore t_1=a=-13$

$t_2=a+d=-13+5=-8$

$t_3=t_2+d=-8+5=-3$

The first three terms are $-13,-8,-3$