This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6
In this question we have been given that Sn denotes the sum of first n terms of an A.P.,
Now we have to prove that S12 = 3(S8 – S4)
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Class:- 10th
Solutions of CBSE Mathematics
Question 57
By using the formula of the sum of n terms of an A.P.
Sn = n[2a +(n – 1)d] / 2
So,
Therefore, L.H.S. = S12 = 12[2a + (12 – 1)d] / 2
= 6[12a + 11d]
= 12a + 66d
R.H.S. = 3(S8 – S4)
= 3[8(2a + (8 – 1)d) / 2 – 4(2a + (4 – 1)d) / 2]
= 3[4(2a + 7d) – 2(2a + 3d)]
= 3[8a + 28d – 4a – 6d]
= 3[4a + 22d]
= 12a + 66d
= L.H.S.
Hence proved.