Question

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

Answer 1

$T_4+T_8=24$

$T_6+T_{10}=34$

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

$\Rightarrow 2a+10d=24$

$\Rightarrow a+5d=12$ …………$\left(1\right)$

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

$\Rightarrow 2a+14d=34$

$\Rightarrow a+7d=17$ …………$\left(2\right)$

Equation $\left(2\right)$ - Equation $\left(1\right)$

$\Rightarrow 2d=5$

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

$a=12-5d=12-\frac{25}{2}=\frac{-1}{2}$

AP$=\frac{-1}{2},2,\frac{9}{2},$….