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 ${a_3} = 2$.
A. $10$
B. $5$
C. $20$
D. None of these

Answer 1

Hint: In the given problem, we are given that ${a_1},{a_2},{a_3},{a_4},{a_5}$ are in AP with common difference $ \ne 0$, ${a_3} = 2$ and we need to find the value of $\sum\limits_{i = 1}^5 {{a_i}} $. Sequence ${a_1},{a_2},{a_3},{a_4},{a_5}$ is in AP i.e., the difference of a term and the previous term is always same. We will apply ${a_n} = a + \left( {n - 1} \right)d$ formula and obtain the obtain the value of ${a_1}$ and after this we will find the value of ${a_2}$, ${a_4}$ and ${a_5}$ by adding the common difference. At the end, we will substitute all the values in $\sum\limits_{i = 1}^5 {{a_i}} $ to get our required answer.

Formula Used:
Arithmetic progression formula for $n\,th$ term ${a_n} = a + \left( {n - 1} \right)d$

Complete step by step solution:
Given that, ${a_1},{a_2},{a_3},{a_4},{a_5}$ are in A.P. and ${a_3} = 2$
Let $a$ be the first term i.e., ${a_1}$ and $d$ be the common difference of an A.P.
We know that for $n\,th$ term, ${a_n} = a + \left( {n - 1} \right)d$. Therefore for ${a_3}$, we have
$ \Rightarrow {a_3} = a + \left( {3 - 1} \right)d$
$ \Rightarrow {a_3} = a + 2d$
We have, ${a_3} = 2$. Therefore, we get
$ \Rightarrow 2 = a + 2d$
$ \Rightarrow a = 2 - 2d$
Or,
$ \Rightarrow {a_1} = 2 - 2d$
Now, we have
$ \Rightarrow {a_2} = {a_1} + d = \left( {2 - 2d} \right) + d = 2 - d$
$ \Rightarrow {a_4} = {a_1} + 3d = \left( {2 - 2d} \right) + 3d = 2 + d$
$ \Rightarrow {a_5} = {a_1} + 4d = \left( {2 - 2d} \right) + 4d = 2 + 2d$
So, $\sum\limits_{i = 1}^5 {{a_i}} = {a_1} + {a_2} + {a_3} + {a_4} + {a_5}$
Now, substitute the values in the above written expression
$ \Rightarrow \sum\limits_{i = 1}^5 {{a_i}} = 2 - 2d + 2 - d + 2 + 2 + d + 2 + 2d$
$ \Rightarrow \sum\limits_{i = 1}^5 {{a_i}} = 10$
Hence, the value of $\sum\limits_{i = 1}^5 {{a_i}} $ is $10$.

Option ‘A’ is correct

Note: Here, we determined the terms of an A.P. using ${a_n} = a + \left( {n - 1} \right)d$ formula and found the value of $\sum\limits_{i = 1}^5 {{a_i}} $ by expansion. Remember that the sum of the terms of an A.P. can also be calculated by using the ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$ formula.
Alternate Method:
Let us determine $\sum\limits_{i = 1}^5 {{a_i}} $ by using ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$ where $n$ is total number of terms, $a$ is first term and $l$ is the last term.
We have, first term $ = 2 - 2d$
Last term $ = 2 + 2d$
Total number of terms $ = 5$
$ \Rightarrow {S_5} = \dfrac{5}{2}\left( {2 - 2d + 2 + 2d} \right)$
$ \Rightarrow {S_5} = \dfrac{5}{2} \times 4$
$ \Rightarrow {S_5} = 10$
Or,
$ \Rightarrow \sum\limits_{i = 1}^5 {{a_i}} = 10$