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The sum of $n,2n,3n$ terms of an AP are${S_1},{S_2},{S_3}$ respectively. Then ${S_3} = 3\left( {{S_2} - {S_1}} \right)$ .
$A.$ True
$B.$ False

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Last updated date: 22nd Jun 2024
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Answer
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Hint- In order to solve this question we will use the formula of sum of series in an AP as it is given that the sum of $n,2n,3n$ terms of an AP are${S_1},{S_2},{S_3}$ respectively.

Complete step-by-step answer:
In order to prove that ${S_3} = 3\left( {{S_2} - {S_1}} \right)$ .
We have given that
$n,2n,3n$ terms of an AP are${S_1},{S_2},{S_3}$ respectively
Let the first term of an AP be $a$ and the common difference be $d$ .
Now, according to the question
$
  {S_1} = \dfrac{n}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\} - - - - - - - - - \left( i \right) \\
  {S_2} = \dfrac{{2n}}{2}\left\{ {2a + \left( {2n - 1} \right)d} \right\} - - - - - - - - - \left( {ii} \right) \\
  {S_3} = \dfrac{{3n}}{2}\left\{ {2a + \left( {3n - 1} \right)d} \right\} - - - - - - - - - \left( {iii} \right) \\
 $
And we have to prove that
${S_3} = 3\left( {{S_2} - {S_1}} \right)$
Now $R.H.S = 3\left( {{S_2} - {S_1}} \right)$
Putting the values of ${S_2},{S_1}$ from $\left( {ii} \right)$ and $\left( i \right)$ we get
$
   = 3\left[ {\dfrac{{2n}}{2}\left\{ {2a + \left( {2n - 1} \right)d} \right\} - \dfrac{n}{2}\left\{ {2a + \left( {n - 1} \right)d} \right\}} \right] \\
   = 3 \times \dfrac{n}{2}\left\{ {4a + \left( {4n - 2} \right)d - 2a - \left( {n - 1} \right)d} \right\} \\
   = \dfrac{{3n}}{2}\left\{ {2a + \left( {4n - 2 - n + 1} \right)d} \right\} \\
   = \dfrac{{3n}}{2}\left\{ {2a + \left( {3n - 1} \right)d} \right\} \\
   = {S_3} \\
   = L.H.S \\
 $
Thus it is true.
Therefore the correct option is $\left( A \right)$ .

Note- Whenever we face such types of questions the key concept is that we should write what is given to us. Then write the formula of sum of series in an AP and then put the formula in what we have asked to prove and thus we get the answer.