Question

The sum of $4$th and $8$th terms of an A.P is $24$ and the sum of $6$th and$10$th terms is $44$. Find the A.P

Answer 1

we know that the $n^{th}$ term of the arithmetic progression is given by $a+(n-1)d$

Given that sum of $4^{th}$ and $8^{th}$ terms is $24$

Therefore, $[a+(4-1\left)d\right]+[a+(8-1\left)d\right]=24$

$⟹a+3d+a+7d=24$

$⟹2a+10d=24$

$⟹a+5d=12$-------(1)

Given that sum of $6^{th}$ and ${10}^{th}$ terms is $44$

Therefore, $[a+(6-1\left)d\right]+[a+(10-1\left)d\right]=44$

$⟹a+5d+a+9d=44$

$⟹2a+14d=44$

$⟹a+7d=22$-------(2)

subtracting (2) from (1) we get

$(a+7d)-(a+5d)=22-12$

$⟹2d=10$

$⟹d=5$

Substituting $d=5$ in (1) we get

$a+5\left(5\right)=12$

$⟹a=12-25=-13$

Therefore, the arithmetic progression with first term $a=-13$ and common difference $d=5$ is $-13,-13+5,-13+2\left(5\right),-13+3\left(5\right),...$

$-13,-8,-3,2,...$ is the required A.P.