If sum of the 3rd and the 8th terms

Solution:

$a_{3}+a_{8}=7$ and $a_{7}+a_{14}=-3$

$\Rightarrow \quad a+(3-1) d+a+(8-1) d=7 \quad\left[\because a_{n}=a+(n-1) d\right]$

and $\quad a+(7-1) d+a+(14-1) d=-3$

$a+2 d+a+7 d=7$

and $\quad a+6 d+a+13 d=-3$

$2 a+9 d=7$ ...(i)

and $2 a+19 d=-3$ $..... (ii)

On subtracting Eq. (i) from Eq. (ii), we get

$10 d=-10 \Rightarrow d=-1 \quad$ [from Eq. (i)]

$2 a+9(-1)=7$

$\Rightarrow \quad 2 a-9=7$

$\Rightarrow \quad 2 a=16 \Rightarrow a=8$

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

$=8+9(-1)$

$=8-9=-1$