Find the sum :

32 + 30 + 28 +……..+ 10.

Let total number of terms in the A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

⇒ 10 = 32 + (n - 1).(-2)

⇒ 10 - 32 = (n - 1).(-2)

⇒ -22 = (n - 1).(-2)

⇒ $\dfrac{-22}{-2}$ = (n - 1)

⇒ 11 = n - 1

⇒ n = 11 + 1

⇒ n = 12.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₁₂ = $\dfrac{12}{2}$ (32 + 10)

= 6 × 42

= 252.

Hence, S₁₂ = 252.