Answer:
Given, \[{{S}_{5}}+{{S}_{7}}=167\] \[\Rightarrow \frac{5}{2}(2a+4d)+\frac{7}{2}\times (2a+6d)=167\] \[\Rightarrow \frac{5}{2}\times 2(a+2d)+\frac{7}{2}\times 2(a+3d)=167\] \[\Rightarrow 5a+10d+7a+21d=167\] \[\Rightarrow 12a+31d=167\] ?(i) \[\Rightarrow \frac{10}{2}(2a+9d)=235\] \[\Rightarrow 10a+45d=235\] \[\Rightarrow 2a+9d=47\] ?(ii) On multiplying equation (ii) by 6, we- get: \[12a+54d=282\] ?(iii) On subtracting equation (i) from (iii), we get: \[_{\begin{smallmatrix} 12a\,\,+\,\,31d\,\,=\,\,167 \\ -\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,- \\ \overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,23d\,\,=\,\,115\,\,} \end{smallmatrix}}^{12a\,\,+\,\,54d\,\,=\,\,282}\] \[\Rightarrow d=5\] Substituting value of d in equation (i), we get \[12a+31\times 5=167\] \[12a+155=167\] \[\Rightarrow 12a=12\] \[\Rightarrow a=1\] Hence A.P., is 1, 6, 11....
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