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# The sum of n, 2n and 3n terms of an A.P are ${S_1},{\text{ }}{S_2},{\text{ }}{S_3}$ respectively.Prove that ${S_3}{\text{ }} = {\text{ }}3({S_2}{\text{ }} - {\text{ }}{S_1})$.  Answer Verified
Hint: First find the value of ${S_1},{\text{ }}{S_2}$ and ${S_3}$. And then subtract ${S_1}$ from ${S_2}$. After that multiply the equation by 3 to prove the result.

Let a and d be the first term and common difference of the given A.P.
As we know that sum of t terms of this A.P will be,
Let, ${{\text{S}}_t}$ be the sum of the first t terms of A.P. then,
$\Rightarrow {S_t}{\text{ }} = {\text{ }}\dfrac{t}{2}[2a{\text{ }} + {\text{ }}(t{\text{ }} - {\text{ }}1)d]$
As, we know that ${{\text{S}}_1}$ is the sum of n terms of A.P. So,
$\Rightarrow {S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(n{\text{ }} - {\text{ }}1)d]$ (1)
As, we know that ${{\text{S}}_2}$ is the sum of 2n terms of A.P. So,
$\Rightarrow {S_2}{\text{ }} = {\text{ }}\dfrac{{2n}}{2}[2a{\text{ }} + {\text{ }}(2n{\text{ }} - {\text{ }}1)d]$ (2)
As, we know that ${{\text{S}}_3}$ is the sum of 3n terms of A.P. So,
$\Rightarrow {S_3}{\text{ }} = {\text{ }}\dfrac{{3n}}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]$ (3)
Now, we know that we had to prove ${S_3}{\text{ }} = {\text{ }}3({S_2}{\text{ }} - {\text{ }}{S_1})$.
So, subtracting equation 1 from 2. We get,
$\Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}n[2a{\text{ }} + {\text{ }}(2n{\text{ }} - {\text{ }}1)d]{\text{ }} - {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(n{\text{ }} - {\text{ }}1)d]{\text{ }}$
On, solving above equation becomes,
$\Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}n\left\{ {\dfrac{{4a{\text{ }} + {\text{ }}4nd{\text{ }} - {\text{ }}2d{\text{ }} - {\text{ }}2a{\text{ }} - {\text{ }}nd{\text{ }} + {\text{ }}d}}{2}} \right\}$
On, solving above equation becomes,
$\Rightarrow {S_2} - {\text{ }}{S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}\left[ {2a{\text{ }} + {\text{ }}3nd{\text{ }} - {\text{ }}d} \right]$
Now, reducing the above equation to prove the result. We get,
$\Rightarrow {S_2}{\text{ }} - {\text{ }}{S_1}{\text{ }} = {\text{ }}\dfrac{n}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]$
Multiplying both sides of the above equation by 3. We get,
$\Rightarrow 3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}\dfrac{{3n}}{2}[2a{\text{ }} + {\text{ }}(3n{\text{ }} - {\text{ }}1)d]$
Now, using equation 3. We can write above equation as,
$\Rightarrow 3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}{S_3}$
Hence, $3({S_2}{\text{ }} - {\text{ }}{S_1}){\text{ }} = {\text{ }}{S_3}$

Note: Whenever we came up with this type of question then easiest and efficient way is to first, assume first term and common difference of the A.P. and then find sum of n, 2n and 3n terms of the A.P, using formula of sum of t terms of A.P and then put their values to get required result.
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