Question

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

Answer 1

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

& \Rightarrow 836=715+{{n}^{2}} \\\ & \Rightarrow n=\sqrt{121} \\\ & \Rightarrow n=11 \\\ \end{aligned}$$ So, the number of terms in the sequence is 11. We know that if $${{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{aligned} & \Rightarrow 715=\dfrac{11}{2}\left( {{a}_{1}}+{{a}_{n}} \right) \\\ & \Rightarrow \left( {{a}_{1}}+{{a}_{n}} \right)=65\times 2=130 \\\ \end{aligned}$$ Now by dividing the both sides with 8 we get $$\begin{aligned} & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=\dfrac{130}{8} \\\ & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=16.25 \\\ \end{aligned}$$ Therefore, the value of $$\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{aligned} & {{a}_{1}}^{\prime }={{a}_{1}}+1 \\\ & {{a}_{n}}^{\prime }={{a}_{n}}+\left( 2n+1 \right) \\\ \end{aligned}$$ By using the sum of terms in arithmetic progression we can write $$\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{aligned} & \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{aligned}$$ Now by dividing the both sides with 8 we get $$\begin{aligned} & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=\dfrac{130}{8} \\\ & \Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)=16.25 \\\ \end{aligned}$$ Therefore, the value of $$\left( \dfrac{{{a}_{1}}+{{a}_{n}}}{8} \right)$$is 16.25.