Question

What is the sum of the arithmetic sequence 140, 134, 128…, if there are 33 terms?

Answer 1

difference is -6

Answer 2

d/2 (a1 + an)

Answer 3

a1=140

an=

140 -6(32)=

Answer 4

\[\S_{n} = \ \frac{n(a_{1} + a_{n})}{2} \]

Answer 5

So you can find a(33) = 140 + (-6)(33 - 1) = 140 - 192 = -52
33(140 - 52)/2 = 1452