# Q : 18     The sum of the  $\small 4$th and  $\small 8$th terms of an AP is  $\small 24$ and the sum of the $\small 6$th and $\small 10$th terms is $\small 44$  Find the first three terms of the AP.

It is given that
sum of the  $\small 4$th and  $\small 8$th terms of an AP is  $\small 24$ and the sum of the $\small 6$th and $\small 10$th terms is $\small 44$
i.e.
$a_4+a_8=24$
$\Rightarrow a+3d+a+7d=24$
$\Rightarrow 2a+10d=24$
$\Rightarrow a+5d=12 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_6+a_{10}=44$
$\Rightarrow a+5d+a+9d=44$
$\Rightarrow 2a+14d=44$
$\Rightarrow a+7d=22 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$a= -13 \ and \ d= 5$
Therefore,first three of AP with a = -13 and d = 5 is
-13,-8,-3

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