Given, 34 + 32 + 30 + ……….. + 10

For this A.P.,

first term, a = 34

common difference, d = a₂−a₁ = 32−34 = −2

n^th term, a_n= 10

Let 10 be the n^th term of this A.P., therefore,

a_n= a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

12 = n −1

n = 13

We know that, sum of n terms is;

S_n = n/2 (a +l) , l = 10

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

= 286