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

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.