arithmetic series

Question

find the sum of the first 10 terms:

40, 37, 34, 31

Answer 1

x = starting number = 40

n = number of terms = 10

We can write the series (40, 37, 34, 31...) as x-0*3, x-1*3, x-2*3, x-3*3, ...

Sum of the first 10 terms is

sum_i(x-3*i), where i is from 0 to 9, x is 40

sum_i(x-3*i) = sum_i(x) -sum_i(3*i) = sum_i(x) - 3*sum_i(i)

sum_i(x) where i is from 0 to 9, means we sum x 10 times

sum_i(x) where i is from 0 to 9 = 10*x = 400

sum_i(i) , where i is from 0 to 9 = sum_i(i) , where i is from 1 to 9 = n*n(+1)/2 = 9*10/2 = 45

sum_i(x-3*i) = sum_i(x) -sum_i(3*i) = sum_i(x) - 3*sum_i(i) = 400 - 3*45 = 400 - 135 = 265

Answer 2

S_n = sum of first n terms = (n/2)[a₁ + a_n]

For the arithmetic series 40, 37, 34, 31, ..., a_n = a₁ + (n-1)d = 40 -3(n-1) = -3n + 43

So, a₁₀ = -30 + 43 = 13

Therefore, S₁₀ = (10/2)[a₁ + a₁₀] = 5(40 + 13) = 265