if SN denotes the sum of the first n term of an ap prove that S12= 3(S4-S4)
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Answered by
0
hello...
S12 is not =3 (S4-S4)
because S4-S4= 0 and 3 (0)=0 so ..
LHS is not equal to RHS ...
S12 is not =3 (S4-S4)
because S4-S4= 0 and 3 (0)=0 so ..
LHS is not equal to RHS ...
Answered by
3
it is (s8-s4)
On Equating the RHS we get ,
3(S8 - S4)
=> 3 { n/2 * [ 2a + (n-1)d ] - n/2 * [2a + (n-1)d ] }
=> Substituting 8 and 4 in n we get,
=> 3 { 8/2 * [ 2a + (8-1)d ] - 4/2 * [2a + (4-1)d ] }
=> 3 { 4 [2a + 7d] - 2[2a + 3d] }
=> 3 { 8a +28d - 4a - 6d }
=> 3 {4a +22d }
=> 6 { 2a + 11d } ( Taking 2 common}
=> 12/2 ( 2a + (12-1)d } (Rearranging the terms)
=> S12
On Equating the RHS we get ,
3(S8 - S4)
=> 3 { n/2 * [ 2a + (n-1)d ] - n/2 * [2a + (n-1)d ] }
=> Substituting 8 and 4 in n we get,
=> 3 { 8/2 * [ 2a + (8-1)d ] - 4/2 * [2a + (4-1)d ] }
=> 3 { 4 [2a + 7d] - 2[2a + 3d] }
=> 3 { 8a +28d - 4a - 6d }
=> 3 {4a +22d }
=> 6 { 2a + 11d } ( Taking 2 common}
=> 12/2 ( 2a + (12-1)d } (Rearranging the terms)
=> S12
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