Question

If sn denotes the sum of first n terms of an ap then prove that s12 = 3 ( s8 - s4

Answer 1

Hi there !

 a = first term of the AP  
d = common difference 

sum of first n terms :-

Sn  = n/2 {2a+(n-1) d}
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S₁₂ = 12/2 {2a+(12-1) d}
    =12a+66d


S₈ = 8/2  {2a+7d} 
      = 8a+28d


S₄ = 4/2 {2a+3d}
     = 4a+6d


S₁₂ = 12a + 66d 


RHS ->   3 (S8-S4) 
            = 3 (8a+28d-4a-6d)
           =12a+66d
           = LHS


Therefore,
S₁₂ = 3 (S₈ - S₄)

Answer 2

sn= n/2(2a + (n-1) ).......( 1 )
s12= 12/2 (2a + 11d ) = 12a + 66d......( 2 )
s8= 8/2 (2a + 7d ) = 8a + 28d............( 3 )
s4= 4/2 (2a + 3d ) = 4a + 6d...............( 4 )
now, 3( s8-s4)
so... s8 - s4 = (8a + 28d - 4a - 6d)
= 4a + 22d........( 5 )
now multiply ( 5 ) by 3 and you get 12 a+ 66d which is equal to s12
hence proved