In AP, if s6+s7=167 And s10 …

S_n = {tex}\frac{n}{2}{/tex}[2a+(n-1)d]

Now, S₅+S₇=167
{tex}\Rightarrow{/tex} {tex}\frac{5}{2}{/tex}[2a+4d]+{tex}\frac{7}{2}{/tex}[2a+6d]=167
{tex}\Rightarrow{/tex} {tex}\frac{{5 \times 2}}{2}{/tex}[a+2d]+|{tex}\frac{{7 \times 2}}{2}{/tex}[a+3d]=167
{tex}\Rightarrow{/tex} 5a+10d+7a+21d=167
{tex}\Rightarrow{/tex} 12a+31d=167...(i)
Also, S₁₀=235
{tex}\Rightarrow{/tex} {tex}\frac{{10}}{2}{/tex}[2a+9d]=235
{tex}\Rightarrow{/tex} 5[2a+9d]=235
{tex}\Rightarrow{/tex} 10a+45d=235
{tex}\Rightarrow{/tex} 2a+9d=47...(ii)
Multiplying equation (ii) by 6, we get
12a+54d=282...(iii)
subtracting (i) from (iii), we get 
23d = 115
{tex}\Rightarrow{/tex} d = 5
{tex}\Rightarrow{/tex} 2a+9(5)=47
{tex}\Rightarrow{/tex} 2a = 2
{tex}\Rightarrow{/tex} a = 1
First term = a = 1
Second term = a+d=1+5=6
Third term = a+2d=1+2(5)=11
Thus, the Ap is 1,6,11,...