Question

- if the sum of first n, 2n and 3n terms of an A. P.be S_{1}, S_{2} and S_{3} respectively then prove that S_{3} = 3(S_{2} - S_{1})​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that a and d represents the first term and common difference of an AP series.

Given that, $S_{1}$, $S_{2}$, $S_{3}$ represents the sum of n, 2n and 3n terms of an AP respectively.

We know,

↝ Sum of n  terms of an arithmetic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term.

• d is the common difference.

• n is number of terms.

So, using this, we get

$\rm \: S_1\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)$

$\rm \: S_2\:=\dfrac{2n}{2} \bigg(2 \:a\:+\:(2n\:-\:1)\:d \bigg)$

$\rm \: S_3\:=\dfrac{3n}{2} \bigg(2 \:a\:+\:(3n\:-\:1)\:d \bigg)$

Now, Consider

$\rm \: 3(S_{2} - S_{1})$

$\rm \: = \: 3\bigg[\dfrac{2n}{2} \bigg(2 \:a\:+\:(2n\:-\:1)\:d \bigg) - \dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)\bigg]$

$\rm \: = \: \dfrac{3n}{2} \bigg[2\bigg(2 \:a\:+\:(2n\:-\:1)\:d \bigg) - \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)\bigg]$

$\rm \: = \: \dfrac{3n}{2} \bigg[4 \:a\:+\:(4n\:-\:2)\:d - 2 \:a\: - \:(n\:-\:1)\:d \bigg]$

$\rm \: = \: \dfrac{3n}{2} \bigg[2 \:a\:+\:(4n\:-\:2 \: - n \: + \: 1)\:d \bigg]$

$\rm \: = \: \dfrac{3n}{2} \bigg[2 \:a\:+\:(3n\:-\:1)\:d \bigg]$

$\rm \: = \: S_{3}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:S_{3} = 3(S_{2} - S_{1}) \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

↝ nᵗʰ term of an arithmetic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the progression.

• n is the no. of terms.

• d is the common difference.