Find the sums given below:

34 + 32 + 30 +….+10

The sum of the series given by

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

Where n is the last term in the series

d is the common difference

a is the first term.

a_n = 10, a = 34

d = n₂ – n₁ = 32 - 34 = 2

a_n = a + (n-1)d

10 = 34 + (n-1)(-2)

-24 = -2(n-1)

12 = n-1

n = 13

The sum of the series is S_n = [2a + (n-1) d]

= [ 2×34 + (13-1)(-2)]

= [ 44 ] ⇒ 286