Proof: S12 = 3(S8 - S4)

Introduction: The given problem is related to the sum of the first n terms of an AP. In this solution, we will prove that S12 is equal to 3 times the difference between S8 and S4.

Step 1: Find the formula for the sum of n terms of an AP: The sum of the first n terms of an AP can be calculated using the following formula:

S_n = (n/2) [2a + (n-1)d]

where a is the first term of the AP, d is the common difference, and n is the number of terms in the series.

Step 2: Find S12: Using the above formula, we can find S12 as follows:

S12 = (12/2) [2a + (12-1)d]
= 6 [2a + 11d]

Step 3: Find S8 and S4: In order to find the difference between S8 and S4, we need to calculate S8 and S4. Using the same formula, we get:

S8 = (8/2) [2a + (8-1)d]
= 4 [2a + 7d]

S4 = (4/2) [2a + (4-1)d]
= 2 [2a + 3d]

Step 4: Find the difference between S8 and S4: We can now find the difference between S8 and S4 as follows:

S8 - S4 = [4(2a + 7d)] - [2(2a + 3d)]
= 8a + 28d - 4a - 12d
= 4a + 16d

Step 5: Prove that S12 = 3(S8 - S4): We can now substitute the values of S12 and S8 - S4 in the given equation and simplify:

S12 = 6 [2a + 11d]
= 3 [4a + 16d/2] + 3 [2a + 11d/2] - 3 [2a + 3d/2]
= 3 [S8 - S4]

Hence, we have proved that S12 is equal to 3 times the difference between S8 and S4.