Question

If the sum of first n, 2n, 3n terms of an A.P are ${{S}_{1}},{{S}_{2}}\text{ and }{{S}_{3}}$. Prove that ${{S}_{3}}=3\left( {{S}_{2}}-{{S}_{1}} \right)$.

Answer 1

Hint: In this question, we are given sum of first n, 2n, 3n terms of an A.P as ${{S}_{1}},{{S}_{2}}\text{ and }{{S}_{3}}$ and we have to prove that ${{S}_{3}}=3\left( {{S}_{2}}-{{S}_{1}} \right)$. For this, we will use general formula to find sum of n terms in A.P which is given as ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ where 'a' is the first term and 'd' is the common difference. In this formula, we will substitute n, 2n, 3n to find ${{S}_{1}},{{S}_{2}}\text{ and }{{S}_{3}}$ in terms of a and d. Using these we will prove the left hand side to be equal to the right hand side.

Complete step-by-step answer:
Here, we are given sum of first n terms, 2n terms and 3n terms of an A.P as ${{S}_{1}},{{S}_{2}}\text{ and }{{S}_{3}}$ respectively. Let us use basic formula of calculating sum of n terms of an AP given by
\[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\cdots \cdots \cdots \cdots \cdots \left( 1 \right)\]
Where a is the first term and d is a common difference.
For this AP also, let us suppose that 'a' is the first term of AP and d is the common difference. Sum of first n terms of an AP equal to ${{S}_{1}}$ becomes:
\[{{S}_{1}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\cdots \cdots \cdots \cdots \cdots \left( 2 \right)\]
For 2n terms, let us substitute 2n in place of n in (1) to get sum of first 2n terms equal to ${{S}_{2}}$ we get:
\[\begin{align}
  & {{S}_{2}}=\dfrac{2n}{2}\left( 2a+\left( 2n-1 \right)d \right) \\
 & {{S}_{2}}=n\left( 2a+\left( 2n-1 \right)d \right)\cdots \cdots \cdots \cdots \cdots \left( 3 \right) \\
\end{align}\]
For 3n terms, let us substitute 3n in place of n in (1) to get sum of first 3n terms equal to ${{S}_{3}}$ we get:
\[{{S}_{3}}=\dfrac{3n}{2}\left( 2a+\left( 3n-1 \right)d \right)\cdots \cdots \cdots \cdots \cdots \left( 4 \right)\]
Now, using (2), (3) and (4) we will prove ${{S}_{3}}=3\left( {{S}_{2}}-{{S}_{1}} \right)$.
Taking left hand side $3\left( {{S}_{2}}-{{S}_{1}} \right)$ from (2) and (3), let us subtract (2) from (3) and multiply difference by 3, we get:
\[3\left( {{S}_{2}}-{{S}_{1}} \right)=3\left( n\left( 2a+\left( 2n-1 \right)d \right)-\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \right)\]
Let us first open brackets we get:
\[\Rightarrow 3\left( 2an+n\left( 2n-1 \right)d-an-\dfrac{n}{2}\left( n-1 \right)d \right)\]
Let us simplify it more by opening bracket, we get:
\[\Rightarrow 3\left( 2an+2{{n}^{2}}d-nd-an-\dfrac{{{n}^{2}}d}{2}+\dfrac{nd}{2} \right)\]
Bringing like terms together, we get:
\[\Rightarrow 3\left( 2an-an+2{{n}^{2}}d-\dfrac{{{n}^{2}}d}{2}-nd+\dfrac{nd}{2} \right)\]
Solving like terms, we get:
\[\Rightarrow 3\left( an+\dfrac{3{{n}^{2}}d}{2}-\dfrac{nd}{2} \right)\]
Let us take $\dfrac{n}{2}$ common from all terms we get:
\[\Rightarrow \dfrac{3n}{2}\left( 2a+3nd-d \right)\]
Now, let us take d common from last two terms we get:
\[\Rightarrow \dfrac{3n}{2}\left( 2a+\left( 3n-1 \right)d \right)\]
As we can see this equation is equal to ${{S}_{3}}$ from (4) so we get:
$\Rightarrow {{S}_{3}}$
Which is equal to the right hand side of the equation that we had to prove. Hence,
${{S}_{3}}=3\left( {{S}_{2}}-{{S}_{1}} \right)$.

Note: In this question, students should not get confused with terms ${{S}_{1}},{{S}_{2}}\text{ and }{{S}_{3}}$. Here, they do not mean sum of 1 term, 2 terms and 3 terms respectively but they are just notations as given in question for sum of n, 2n and 3n terms. Since, we are dealing with same AP so a and d remains same for ${{S}_{1}},{{S}_{2}}\text{ and }{{S}_{3}}$. Student should do the calculations carefully and step by step to avoid mistakes.