Question

If sum of the $3^{rd}$ and the $8^{th}$ terms of an A.P. is 7 and the sum of the $7^{th}$ and the ${14}^{th}$ terms is -3, find the ${10}^{th}$ term.

Answer 1

It is given that,

$a_3+a_8=7$

$a_7+a_1=-3$

Now,

$a_n=a+(n-1)4d$

$a_3+a_8=7$

$\Rightarrow (a+(3-1\left)d\right)+(a+(8-1\left)d\right)=7$

$\Rightarrow a+2d+a+7d=7$

$\Rightarrow 2a+9d=7$

$\Rightarrow 2a=7-9d...\left(1\right)$

$a_7+a_{14}=-3$

$\Rightarrow (a+(7-1\left)d\right)+(a+(14-1\left)d\right)=-3$

$\Rightarrow a+6d+a+13d=-3$

$\Rightarrow 2a+19d=-3$

$\Rightarrow 7-9d+19d=-3\left(from\right(1\left)\right)$

$\Rightarrow 7+10d=-3$

$\Rightarrow 10d=-3-7$

$\Rightarrow 10d=-10$

$\Rightarrow d=1$

$\Rightarrow 2a=7-9(-1)\left(from\right(1\left)\right)$

$\Rightarrow 2a=7+9$

$\Rightarrow 2a=16$

$\Rightarrow a=8$

Now,

$a_{10}=a+(10-1)d$

$=8+9(-1)$

$=8-9$

$=-1$

Hence, the ${10}^{th}$ term is $-1$