The terms of an arithmetic sequence add up to 715. The first term of sequence is increased by 1, second term is increased by 3, and soon, in general, the \[{{k}^{th}}\]term is increased by \[{{k}^{th}}\]odd positive integer. The terms of new sequence add up to 836. If \[{{a}_{1}}\]and \[{{a}_{n}}\]denotes the first and last term of the original sequence, then the value of \[\left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)\]is

Hint: We solve this problem by taking the sum of original sequence as \[{{S}_{n}}\]and then we add sum of first \['n'\]odd positive numbers to \[{{S}_{n}}\]we get the sum of changed sequence as \[{{S}_{n}}^{\prime }\]because we added \[{{k}^{th}}\]odd positive number to \[{{k}^{th}}\]term. We use the result of sum of first \['n'\]odd positive numbers as \[\sum\limits_{k=1}^{n}{\left( 2k-1 \right)}={{n}^{2}}\]to find number of terms. Then, finally we use sum of \['n'\]in arithmetic progression as
 \[{{S}_{n}}=\dfrac{n}{2}\left[ {{a}_{1}}+{{a}_{n}} \right]\]to get the required result.

Complete step-by-step answer:
Let us assume that the sum of original sequence as
 \[{{S}_{n}}=715\]
Let us assume that after adding \[{{k}^{th}}\]odd positive number to \[{{k}^{th}}\]term we get
 \[{{S}_{n}}^{\prime }=836\]
We are given that in each term of original sequence \[{{k}^{th}}\]odd positive number to \[{{k}^{th}}\]term.
So, by converting this statement to sum of terms we get
 \[\Rightarrow {{S}_{n}}^{\prime }={{S}_{n}}+\left( 1+3+5+.......+\left( 2n-1 \right) \right)\]
We know that the sum of first \['n'\]odd positive numbers as \[\sum\limits_{k=1}^{n}{\left( 2k-1 \right)}={{n}^{2}}\]
By using this result in above equation we get
 \[\Rightarrow {{S}_{n}}^{\prime }={{S}_{n}}+{{n}^{2}}\]
By substituting the required values in above equation we get
 \[\begin{align}
  & \Rightarrow 836=715+{{n}^{2}} \\
 & \Rightarrow n=\sqrt{121} \\
 & \Rightarrow n=11 \\
\end{align}\]
So, the number of terms in the sequence is 11.
We know that if \[{{a}_{1}}\]and \[{{a}_{n}}\]denotes the first and last term of the sequence then the sum of sequence is given as
 \[{{S}_{n}}=\dfrac{n}{2}\left[ {{a}_{1}}+{{a}_{n}} \right]\]
By using this formula and substituting the required values we get
 \[\begin{align}
  & \Rightarrow 715=\dfrac{11}{2}\left( {{a}_{1}}+{{a}_{n}} \right) \\
 & \Rightarrow \left( {{a}_{1}}+{{a}_{n}} \right)=65\times 2=130 \\
\end{align}\]
Now by dividing the both sides with 8 we get
 \[\begin{align}
  & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=\dfrac{130}{8} \\
 & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=16.25 \\
\end{align}\]
Therefore, the value of \[\left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)\]is 16.25.

Note: We can find the value of \[\left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)\]from the second sequence.
In the second sequence the first and last terms are given as
 \[\begin{align}
  & {{a}_{1}}^{\prime }={{a}_{1}}+1 \\
 & {{a}_{n}}^{\prime }={{a}_{n}}+\left( 2n+1 \right) \\
\end{align}\]
By using the sum of terms in arithmetic progression we can write
 \[\Rightarrow {{S}_{n}}^{\prime }=\dfrac{n}{2}\left( {{a}_{1}}^{\prime }+{{a}_{n}}^{\prime } \right)\]
By substituting the required values in above equation we get
 \[\begin{align}
  & \Rightarrow 836=\dfrac{11}{2}\left( {{a}_{1}}+1+{{a}_{n}}+\left( 2\times 11-1 \right) \right) \\
 & \Rightarrow 76\times 2={{a}_{1}}+{{a}_{n}}+22 \\
 & \Rightarrow {{a}_{1}}+{{a}_{n}}=130 \\
\end{align}\]
Now by dividing the both sides with 8 we get
 \[\begin{align}
  & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=\dfrac{130}{8} \\
 & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=16.25 \\
\end{align}\]
Therefore, the value of \[\left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)\]is 16.25.