Question

If \[{S_1}\], \[{S_2}\], \[{S_3}\] are the sums of the terms of \[n\] three series in A.P., the first term of each being 1 and the respectively common difference being 1, 2, 3: Prove that \[{S_1} + {S_3} = 2{S_2}\].

Answer 1

Hint: Here, we will find the common difference and then use the formula sum of \[n\]th term of the arithmetic progression \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference. Apply this formula, and then substitute the value of \[a\],\[d\] and \[n\] in the obtained equation to find the required A.P.

Complete step-by-step answer:
We are given that the first term of each being 1 and the respectively common difference being 1, 2, 3.
Let us assume that \[{S_1}\], \[{S_2}\], \[{S_3}\] are the sums of the terms of \[n\] three series in A.P.
We know that the arithmetic progression is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
Considering \[{S_1}\], we get
Using the formula of sum of \[n\]th term of the arithmetic progression A.P., that is,\[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference, we get
\[
   \Rightarrow {S_1} = \dfrac{{n\left( {2 + \left( {n - 1} \right)} \right)}}{2} \\
   \Rightarrow {S_1} = \dfrac{{n\left( {2 + n - 1} \right)}}{2} \\
   \Rightarrow {S_1} = \dfrac{{n\left( {n + 1} \right)}}{2} \\
   \Rightarrow {S_1} = \dfrac{{{n^2} + n}}{2} \\
 \]
Considering \[{S_2}\], we get
\[d\]
Using the formula of sum of \[n\]th term of the arithmetic progression A.P., that is,\[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference, we get
\[
   \Rightarrow {S_2} = \dfrac{{n\left( {2 + \left( {n - 1} \right)2} \right)}}{2} \\
   \Rightarrow {S_2} = \dfrac{{n\left( {2 + 2n - 2} \right)}}{2} \\
   \Rightarrow {S_2} = \dfrac{{n\left( {2n} \right)}}{2} \\
   \Rightarrow {S_2} = {n^2} \\
 \]

Considering \[{S_3}\], we get

Using the formula of sum of \[n\]th term of the arithmetic progression A.P., that is,\[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and is the common difference, we get
\[
   \Rightarrow {S_3} = \dfrac{{n\left( {2 + 3\left( {n - 1} \right)} \right)}}{2} \\
   \Rightarrow {S_3} = \dfrac{{n\left( {2 + 3n - 3} \right)}}{2} \\
   \Rightarrow {S_3} = \dfrac{{n\left( {3n - 1} \right)}}{2} \\
   \Rightarrow {S_3} = \dfrac{{3{n^2} - n}}{2} \\
 \]

Substituting the value of \[{S_1}\] and \[{S_2}\] in the left hand side of the equation \[{S_1} + {S_3} = 2{S_2}\], we get
\[
   \Rightarrow \dfrac{{n + {n^2}}}{2} + \dfrac{{3{n^2} - n}}{2} \\
   \Rightarrow \dfrac{{{n^2} + n + 3{n^2} - n}}{2} \\
   \Rightarrow \dfrac{{4{n^2}}}{2} \\
   \Rightarrow 2{n^2}{\text{ }} \\
 \]
Substituting the value of \[{S_3}\] in the right hand side of the equation \[{S_1} + {S_3} = 2{S_2}\], we get
\[
   \Rightarrow 2 \cdot {n^2} \\
   \Rightarrow 2{n^2} \\
 \]
Therefore, the LHS is equal to RHS
Hence, proved.

Note: In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and their sums. Some students use the formula of sum, \[S = \dfrac{n}{2}\left( {a + l} \right)\], where \[l\] is the last term, but have the to find the value of , so it will be wrong. We can also find the value of \[n\]th term by find the value of \[{S_n} - {S_{n - 1}}\], where \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference. But this is a longer method, which takes time, so we will use the above method. One should know the \[{a_n}\] is the \[n\]th term in the arithmetic progression.