Math, asked by shafeeqshafeeq, 1 year ago

If Sn denotes the sum of first n terms of AP.Prove that
S12=3(S8-S4)

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Answered by tejadhith98
8
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Answered by Anonymous
1

 \Large \bf Solution :

Let the first term be a and common difference be d.

We have to prove :  \sf S_{12} = 3(S_{8} - S_{4})

 \underline {\bf In \: RHS, \: we \: have :} \sf 3(S_{8} - S_{4})

Now,

 \underline{\boxed{\sf S_{n} = \dfrac{n}{2} . [2a+(n-1)d]}}

 \sf : \implies 3 \Bigg[\dfrac{8}{2}(2a+(8-1)d) - \dfrac{4}{2}(2a+(4-1)d)\Bigg]

 \sf : \implies 3 \Bigg[\cancel{\dfrac{8}{2}}(2a+7d) - \cancel{\dfrac{4}{2}}(2a+3d)\Bigg]

 \sf : \implies 3 [4(2a+7d) - 2(2a+3d)]

 \sf : \implies 3 \times 2[2(2a+7d) - (2a+3d)]

 \sf : \implies 6(4a+14d - 2a+3d)

 \sf : \implies 6(2a+11d)

 \underline {\bf In \: LHS, \: we \: have :} \sf S_{12}

Now,

 \underline{\boxed{\sf S_{n} = \dfrac{n}{2} . [2a+(n-1)d]}}

 \sf : \implies \dfrac{12}{2} (2a + (12-1)d)

 \sf : \implies \cancel{\dfrac{12}{2}} (2a + 11d)

 \sf : \implies 6 (2a + 11d)

 \bf As, \: LHS = RHS,

 \underline{\underline{\bf Hence, \: Proved.}}

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