Question

If the terms \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}\] are in AP with common difference not equal to 0, then find the value of \[\sum\limits_{i=1}^{5}{{{a}_{i}}},\] when \[{{a}_{3}}=2.\]

Answer 1

Hint: We are given that \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}\] are in AP. We will then use the \[{{n}^{th}}\] term formula of AP given as \[{{a}_{n}}=a+\left( n-1 \right)d\] to unite each term in the term of a and d. Then we have \[{{a}_{3}}\] as 2. So, we will simplify and we get, a + 2d = 2. Now, at last, we will sum the first five terms of AP. We will get our summation in the form of a and d, then we will simplify and get the value of \[\sum\limits_{i=1}^{5}{{{a}_{i}}}.\]

Complete step-by-step solution:
We are given that \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}},{{a}_{5}}\] are in AP. We know that AP (Arithmetic Progression) is a sequence in which terms are increased by the same common difference. The general term of an AP is given by the form \[{{a}_{n}}=a+\left( n-1 \right)d\] where a is the first term and d is the common difference.
So, using this, we get, \[{{a}_{1}}\] mean n = 1 in \[{{a}_{n}}.\]
\[\Rightarrow {{a}_{1}}=a+\left( 1-1 \right)d\]
\[\Rightarrow {{a}_{1}}=a\]
So, we get,
\[{{a}_{1}}=a.....\left( i \right)\]
Now putting n as 2, we get,
\[{{a}_{2}}=a+\left( 2-1 \right)d\]
\[\Rightarrow {{a}_{2}}=a+d\]
So, we get,
\[{{a}_{2}}=a+d.....\left( ii \right)\]
Putting n = 3, we get,
\[{{a}_{3}}=a+2d.....\left( iii \right)\]
Now put n = 4, we will get,
\[{{a}_{4}}=a+\left( 4-1 \right)d\]
\[\Rightarrow {{a}_{4}}=a+3d\]
So, we get,
\[\Rightarrow {{a}_{4}}=a+3d......\left( iv \right)\]
Now, lastly putting n = 5, we will get,
\[\Rightarrow {{a}_{5}}=a+\left( 5-1 \right)d\]
\[\Rightarrow {{a}_{5}}=a+4d\]
So, we get,
\[\Rightarrow {{a}_{5}}=a+4d......\left( v \right)\]
Now, we also have, \[{{a}_{3}}=2.\]
\[\Rightarrow a+2d=2......\left( vi \right)\]
Now, we have to find \[\sum\limits_{i=1}^{5}{{{a}_{i}}}.\]
We know that \[\sum\limits_{i=1}^{5}{{{a}_{i}}}\] the mean sum of the first five terms.
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}\]
Now using (i), (ii), (iii), (iv) and (v), we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=a+a+d+a+2d+a+3d+a+4d\]
Further adding, we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=5a+10d\]
Taking 5 commons, we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=5\left( a+2d \right)\]
As we have $a + 2d$ as $24 using (vi). So, we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=5\left( a+2d \right)\]
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=5\times 2\]
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=10\]
So, we get, \[\sum\limits_{i=1}^{5}{{{a}_{i}}}\] as 10.

Note: The alternate method in an AP is the sum of the first and fifth term is the same as the sum of the second and fourth term, i.e.
\[\Rightarrow {{a}_{1}}+{{a}_{5}}={{a}_{2}}+{{a}_{4}}\]
And these terms are equal as 2 times the third term. So, we have, \[{{a}_{1}}+{{a}_{5}}={{a}_{2}}+{{a}_{4}}\] and these terms are equal as 2 times the third term. So, we have,
\[{{a}_{1}}+{{a}_{5}}={{a}_{2}}+{{a}_{4}}=2{{a}_{3}}......\left( vii \right)\]
\[\sum\limits_{i=1}^{5}{{{a}_{i}}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}\]
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}={{a}_{1}}+{{a}_{5}}+{{a}_{3}}+{{a}_{2}}+{{a}_{4}}\]
Now using (vii), we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=2{{a}_{3}}+{{a}_{3}}+2{{a}_{3}}\]
Simplifying further, we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=5{{a}_{3}}\]
As, \[{{a}_{3}}=2,\] so, we get,
\[\Rightarrow \sum\limits_{i=1}^{5}{{{a}_{i}}}=5\times 2=10\]