Prove S12 = 3(S8 − S4) for an AP | Class 10 Maths

Quick answer: Using $S_n = \frac{n}{2}[2a + (n - 1)d]$: $S_4 = 4a + 6d$, $S_8 = 8a + 28d$ and $S_{12} = 12a + 66d$. Then $3(S_8 - S_4) = 3(4a + 22d) = 12a + 66d = S_{12}$. The same method proves the general result $S_{3n} = 3(S_{2n} - S_n)$ for any AP.

The sum formula

For an AP with first term $a$ and common difference $d$, the sum of the first $n$ terms is:

$S_n = \frac{n}{2}\left[2a + (n - 1)d\right]$

Step 1: write S₄, S₈ and S₁₂

Step 2: prove S₁₂ = 3(S₈ − S₄)

Start with the right-hand side.

Step 2a: Find S₈ − S₄.

$S_8 - S_4 = (8a + 28d) - (4a + 6d) = 4a + 22d$

Step 2b: Multiply by 3.

$3(S_8 - S_4) = 12a + 66d$

Step 2c: Compare with the left-hand side.

$S_{12} = 12a + 66d$

Since both sides equal $12a + 66d$:

$S_{12} = 3(S_8 - S_4)$

Hence proved.

Step 3: the general result S₃ₙ = 3(S₂ₙ − Sₙ)

Step 3a: Write the three sums.

$S_{3n} = \frac{3n}{2}\left[2a + (3n - 1)d\right]$

$S_{2n} = n\left[2a + (2n - 1)d\right]$

$S_n = \frac{n}{2}\left[2a + (n - 1)d\right]$

Step 3b: Find $S_{2n} - S_n$, taking $\frac{n}{2}$ common.

$S_{2n} - S_n = \frac{n}{2}\left[2\{2a + (2n - 1)d\} - \{2a + (n - 1)d\}\right]$

$= \frac{n}{2}\left[4a + (4n - 2)d - 2a - (n - 1)d\right]$

$= \frac{n}{2}\left[2a + (3n - 1)d\right]$

Step 3c: Multiply by 3.

$3(S_{2n} - S_n) = \frac{3n}{2}\left[2a + (3n - 1)d\right] = S_{3n}$

Hence proved. Putting $n = 4$ gives $S_{12} = 3(S_8 - S_4)$.

Check with numbers

Take the AP 2, 5, 8, 11, … ($a = 2$, $d = 3$).

Why it works

$S_8 - S_4$ is the sum of the 5th to 8th terms, and $S_{12} - S_8$ is the sum of the 9th to 12th terms. In an AP, each block of 4 terms is larger than the previous block by the same amount ($16d$). So the three blocks of four are themselves in AP, and their total, $S_{12}$, is 3 times the middle block, $S_8 - S_4$.

Summary table

Exam tips

• Write each sum in a separate line and number them; it makes the proof easy to follow.
• Start from the side that needs more working (here, the RHS).
• For the general version, take $\frac{n}{2}$ common early to avoid fractions inside the bracket.
• A quick numerical check with a simple AP confirms your algebra.

Common mistakes

• Thinking $S_8 - S_4 = S_4$. It is the sum of the 5th to 8th terms, not the first four.
• Forgetting to change the sign of $6d$ when subtracting.
• Writing $S_8 = \frac{8}{2}[2a + 8d]$ instead of $[2a + 7d]$.
• Ending the proof without a conclusion.

Frequently Asked Questions

How do you prove S₁₂ = 3(S₈ − S₄)?

Write S₄ = 4a + 6d, S₈ = 8a + 28d and S₁₂ = 12a + 66d. Then 3(S₈ − S₄) = 3(4a + 22d) = 12a + 66d, which equals S₁₂.

What is the formula for the sum of n terms of an AP?

Sₙ = n/2 [2a + (n − 1)d], where a is the first term and d is the common difference.

How do you prove S₃ₙ = 3(S₂ₙ − Sₙ)?

Find S₂ₙ − Sₙ, take n/2 common to get n/2 [2a + (3n − 1)d], and multiply by 3 to get 3n/2 [2a + (3n − 1)d], which is S₃ₙ.

Does this identity work for every AP?

Yes. The proof uses only the general sum formula, so it holds for any first term a and common difference d.

What does S₈ − S₄ represent?

It is the sum of the 5th, 6th, 7th and 8th terms of the AP.