Question

1 no.
solve it plz........​

Answer 1

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

Sum of n terms Sn = (n/2)[2a + (n - 1) * d]

LHS :

Sum of 12 terms S₁₂ = (12/2)[2a + (12 - 1) * d]

                                 = 6[2a + 11d]

RHS:

S₈ = (8/2)[2a + (8 - 1) * d]

   = 4[2a + 7d]

  = 8a + 28d

S₄ = (4/2)[2a + (4 - 1) * d]

   = 2[2a + 3d]

   = 4a + 6d

Then:

3(S₈ - S₄) = 3[(8a + 28d) - (4a + 6d)]

               = 3[4a + 22d]

              = 12a + 66d

              = 6[2a + 11d]

              = S₁₂

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