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The sum of first $n, \space 2n$, and $3n$ terms of an AP is $S_1, \space S_2$, and $S_3$ respectively. Prove that $S_3 = 3(S_2 - S_1).$

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Answer:

Step by Step Explanation:

1. We are told that
   
   $S_1$ = Sum of first $n$ terms
   $S_2$ = Sum of first $2n$ terms
   $S_3$ = Sum of first $3n$ terms
   
2. We know that the sum of first $n$ terms of an AP is given by $S_n = \dfrac { n } { 2 } (2a + (n-1)d),$ where $a$ is the first term and $n$ is the number of terms in the AP.
   
   Therefore, we have $$\begin{aligned} & S_1 = \dfrac { n } { 2 } (2a + (n-1)d) \\ & S_2 = \dfrac { 2n } { 2 } (2a + (2n-1)d) \\ & S_3 = \dfrac { 3n } { 2 } (2a + (3n-1)d) \end{aligned}$$
3. Now, $$\begin{aligned} 3(S_2 - S_1) & = 3 \bigg(\dfrac { 2n } { 2 } (2a + (2n-1)d) - \dfrac { n } { 2 } (2a + (n-1)d) \bigg) \\ & = \dfrac { 3n } { 2 } ( 2(2a + (2n-1)d) - (2a + (n-1)d) ) \\ & = \dfrac { 3n } { 2 } ( 4a + 4nd - 2d) - (2a + nd - d) ) \\ & = \dfrac { 3n } { 2 } ( 4a + 4nd - 2d - 2a - nd + d ) \\ & = \dfrac { 3n } { 2 } ( 2a + 3nd - d ) \\ & = \dfrac { 3n } { 2 } ( 2a + (3n - 1)d ) \\ & = S_3 \end{aligned}$$
4. Thus, $S_3 = 3(S_2 - S_1).$

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