Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}$ are in AP with common difference $\ne 0$, then find the value of $\sum\limits_{i=1}^{5}{{{a}_{i}}}$ when ${{a}_{3}}=2$.

Answer 1

Hint: In order to find the solution of this question, we will consider the terms of the AP as a -2d, a-d, a, a+d, a+2d and then as per the given condition, that is ${{a}_{3}}=2$, we will try to find the value of a and then we will use that value to find the value of $\sum\limits_{i=1}^{5}{{{a}_{i}}}$.

Complete step-by-step solution -
In this question, we have been asked to find the value of $\sum\limits_{i=1}^{5}{{{a}_{i}}}$ for an AP ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}$ whose common difference $\ne 0$ and ${{a}_{3}}=2$. To solve this question, let us consider the common difference of the AP as $d\ne 0$ and the first term ${{a}_{1}}=a-2d$. So, we can say that,
$\begin{align}
  & {{a}_{1}}=a-2d \\
 & {{a}_{2}}=a-d \\
 & {{a}_{3}}=a \\
 & {{a}_{4}}=a+d \\
 & {{a}_{5}}=a+2d \\
\end{align}$
Now, we have been given that ${{a}_{3}}=2$. So, we can say, ${{a}_{3}}=a=2.........\left( i \right)$ from the above assumption.
Now, we have been asked to calculate the value of $\sum\limits_{i=1}^{5}{{{a}_{i}}}$. So, we can say that we have been asked to find the value of ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}=\sum\limits_{i=1}^{5}{{{a}_{i}}}$. Now, we will substitute the assumed values of ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}$ in the above equation. So, we get,
$\sum\limits_{i=1}^{5}{{{a}_{i}}}=\left( a-2d \right)+\left( a-d \right)+\left( a \right)+\left( a+d \right)+\left( a+2d \right)$
Now, we know that the like terms show algebraic summation. So, we can say,
$\begin{align}
  & \sum\limits_{i=1}^{5}{{{a}_{i}}}=a+a+a+a+a+\left( d+2d-d-2d \right) \\
 & \sum\limits_{i=1}^{5}{{{a}_{i}}}=5a \\
\end{align}$
Now, we will put the value of a from equation (i). So, we get,
$\sum\limits_{i=1}^{5}{{{a}_{i}}}=5\times 2$
Therefore, we get,
$\sum\limits_{i=1}^{5}{{{a}_{i}}}=10$
Hence, we can say that for an AP ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}$ whose common difference $\ne 0$ and ${{a}_{3}}=2$, the value of $\sum\limits_{i=1}^{5}{{{a}_{i}}}$ is 10.

Note: We can also solve this question by using the property of arithmetic mean, that is, if a, b and c are in AP, then $\dfrac{a+c}{2}=b$. So, we know that ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}$ are in AP. So, we can say that ${{a}_{1}},{{a}_{3}},{{a}_{5}}$ are in AP and ${{a}_{2}},{{a}_{3}},{{a}_{4}}$ are in AP. Therefore, we can say, $\dfrac{{{a}_{1}}+{{a}_{5}}}{2}={{a}_{3}}\Rightarrow {{a}_{1}}+{{a}_{5}}=2{{a}_{3}}$ and $\dfrac{{{a}_{2}}+{{a}_{4}}}{2}={{a}_{3}}\Rightarrow {{a}_{2}}+{{a}_{4}}=2{{a}_{3}}$. Hence, we can use this information to write, $\sum\limits_{i=1}^{5}{{{a}_{i}}}$ as ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}=2{{a}_{3}}+{{a}_{3}}+2{{a}_{3}}=5{{a}_{3}}$. And we have been given that ${{a}_{3}}=2$ so we get, $\sum\limits_{i=1}^{5}{{{a}_{i}}}=5{{a}_{3}}=5\times 2=10$, which is our answer.