question_answer

In an A.P., if \[~{{S}_{5}}+{{S}_{7}}=167\] and \[{{S}_{10}}=235\], find the A.P. where \[{{S}_{n}}\] denotes the sum of its first n terms.

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....