If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Given, S₆ = 36 and S₁₆ = 256.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₆ = $\dfrac{6}{2}[2a + (6 - 1)d]$
⇒ 36 = 3[2a + 5d]
⇒ 2a + 5d = 12
⇒ 2a = 12 - 5d     (Eq 1)

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₁₆ = $\dfrac{16}{2}[2a + (16 - 1)d]$
⇒ 256 = 8[2a + 15d]
⇒ 2a + 15d = 32
⇒ 2a = 32 - 15d

Putting value of 2a from Eq 1 in above equation,

⇒ 12 - 5d = 32 - 15d
⇒ -5d + 15d = 32 - 12
⇒ 10d = 20
⇒ d = 2.

∴ From Eq 1,
⇒ 2a = 12 - 5d
⇒ 2a = 12 - 5(2)
⇒ 2a = 2
⇒ a = 1.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₁₀ = $\dfrac{10}{2}[2 \times 1 + (10 - 1)2]$
⇒ S₁₀ = 5[2 + 18]
⇒ S₁₀ = 5 × 20
⇒ S₁₀ = 100.

Hence, the sum of first 10 terms of the A.P. is 100.