If \(S_n\) is the sum of first \(n\) terms of an AP, prove \(S_{12}=3(S_8-S_4)\).
Proved.
Step 1: Recall the formula for the sum of first \(n\) terms of an arithmetic progression (AP):
\[ S_n = \dfrac{n}{2} \big( 2a + (n-1)d \big) \]
where:
Step 2: Find \(S_{12}\).
\[ S_{12} = \dfrac{12}{2} \big( 2a + (12-1)d \big) \]
\[ S_{12} = 6(2a + 11d) \]
\[ S_{12} = 12a + 66d \]
Step 3: Find \(S_8\).
\[ S_8 = \dfrac{8}{2} \big( 2a + (8-1)d \big) \]
\[ S_8 = 4(2a + 7d) \]
Step 4: Find \(S_4\).
\[ S_4 = \dfrac{4}{2} \big( 2a + (4-1)d \big) \]
\[ S_4 = 2(2a + 3d) \]
Step 5: Calculate \(S_8 - S_4\).
\[ S_8 - S_4 = 4(2a + 7d) - 2(2a + 3d) \]
\[ = (8a + 28d) - (4a + 6d) \]
\[ = 4a + 22d \]
Step 6: Multiply the result by 3.
\[ 3(S_8 - S_4) = 3(4a + 22d) \]
\[ = 12a + 66d \]
Step 7: Compare with \(S_{12}\).
We found that:
Since both are equal, the given relation is proved.