Italy
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).^@
Answer:
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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
- 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} @^ - 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} @^
- Thus, ^@ S_3 = 3(S_2 - S_1). ^@
