Find the sums given below :34 + 32 + 30 + . . . + 10

Here, a = 34 and d = 32 – 34 = – 2.
Let, the nth term of the A.P. is 10.
Therefore, a_n = 10
⇒ a + (n -1)d = 10
⇒ 34 + (n – 1)(-2) = 10
⇒ (n -1)(- 2) = – 24
⇒ n -1 = 12
⇒ n = 13
The sum of n terms of an AP is given by
S_n = n/2[a + l]
⇒ S₁₃ = 13/2[34 + 10]

⇒ S₁₃ = 13/2[44]
= 13 × 22 = 286