Question

The sum of $4th$ and $8th$ terms of an A.P. is $24$ and the sum of the $6th$ and $10th$ terms is $34.$ Find the first term and the common difference of the A.P.

Answer 1

We know that the $n^{th}$ term of an A.P is given by $T_n=a+(n-1)d$

$T_4+T_8=24$ [given]
$\Rightarrow a+3d+a+7d=24$
$\Rightarrow 2a+10d=24.........Eq\left(1\right)$

And, $T_6+T_{10}=34$ [given]
$\Rightarrow a+5d+a+9d=34$
$\Rightarrow 2a+14d=\mathrm{34............}Eq\left(2\right)$
Subtracting eq.(1) from eq.(2), we have
$2a+14d-(2a+10d)=34-24$
$\Rightarrow 4d=10$
$\Rightarrow d=\frac{10}{4}$$=\frac{5}{2}$
Put $d=\frac{5}{2}$ in $eq.\left(1\right)$, we have
$2a+10\times \frac{5}{2}=24$
$\Rightarrow 2a+25=24$
$\Rightarrow 2a=24-25$
$\Rightarrow 2a=-1$
$\Rightarrow a=\frac{-1}{2}$