If Sₙ denotes the sum of first n terms of an AP, prove that S₁₂ = 3(S₈ - S₄)
Solution:
Given, Sₙ denotes the sum of first n terms of an AP.
We have to prove that S₁₂ = 3(S₈ - S₄).
The sum of the first n terms of an AP is given by
Sₙ = (n/2)[2a + (n-1)d]
LHS: S₁₂
When n = 12, S₁₂ = (12/2)[2a + (12-1)d]
S₁₂ = 6[2a + 11d]
S₁₂ = 12a + 66d
RHS: 3(S₈ - S₄)
When n = 8, S₈ = (8/2)[2a + (8-1)d]
S₈ = 4[2a + 7d]
S₈ = 8a + 28d
When n = 4, S₄ = (4/2)[2a + (4-1)d]
S₄ = 2[2a + 3d]
S₄ = 4a + 6d
Now, S₈ - S₄ = (8a + 28d) - (4a + 6d)
= 8a + 28d - 4a - 6d
= 8a - 4a + 28d - 6d
= 4a + 22d
3(S₈ - S₄) = 3(4a + 22d)
= 12a + 66d
= S₁₂
LHS = RHS
Therefore, it is proved that S₁₂ = 3(S₈ - S₄).
✦ Try This: The sum of n, 2n and 3n terms of an AP are S₁, S₂ and S₃ respectively. Prove that S₃ = 3(S₂ - S₁)
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 26
If Sₙ denotes the sum of first n terms of an AP, prove that S₁₂ = 3(S₈ - S₄)
Summary:
An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same. If Sₙ denotes the sum of first n terms of an AP, it is proved that S₁₂ = 3(S₈ - S₄)
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