In an AP, it is given that

Solution:

$S_{5}+S_{7}=167$

$\Rightarrow \frac{5}{2}(2 a+4 d)+\frac{7}{2}(2 a+6 d)=167 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

$\Rightarrow 5 a+10 d+7 a+21 d=167$

$\Rightarrow 12 a+31 d=167$                $\ldots \ldots(1)$

Also,

$S_{10}=235$

$\Rightarrow \frac{10}{2}(2 a+9 d)=235$

$\Rightarrow 5(2 a+9 d)=235$

$\Rightarrow 2 a+9 d=47$

Multiplying both sides by 6, we get

$12 a+54 d=282$             $\cdots \cdot(2)$

Subtracting (1) from (2), we get

$12 a+54 d-12 a-31 d=282-167$

$\Rightarrow 23 d=115$

$\Rightarrow d=5$

Putting d = 5 in (1), we get

$12 a+31 \times 5=167$

$\Rightarrow 12 a+155=167$

$\Rightarrow 12 a=167-155=12$

$\Rightarrow a=1$

Hence, the AP is $1,6,11,16, \ldots$.