How do you find #S_n# for the arithmetic series given #a_1=40#, n=20, d=-3?

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Answer 1 · 2016-11-12T22:01:21

When # S_20 = 230 #

The sum of the first #n# terms of an AP whose terms are

# a, a+d, a+2d, ... , a+(n-1)d #

is given by

# S_n = n/2{2a+(n-1)d} #

So for the AP with #a=40# and #d=-1# we have:

# S_n = n/2{2(40)+(n-1)(-3)} #
# :. S_n = n/2( 80 -3n + 3 ) #
# :. S_n = n/2( 83 -3n ) #

When # n=20 => S_20 = 10( 83 -60 ) = 230 #