How do you find #S_n# for the arithmetic series given #a_1=132#, d=-4, #a_n=52#?

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Answer 1 · 2016-10-28T02:00:42

The series has a sum of #1932#.

Let's start by finding the number of terms, #n#.

#t_n = a + (n - 1)d#

#52 = 132 + (n - 1)-4#

#52 = 132 - 4n + 4#

#4n = 84#

#n = 21#

We now apply the formula for sum of an arithmetic series using the formula #s_n = n/2(t_1+ t_n)#

#s_n = n/2(t_1 + t_n)#

#s_21= 21/2(132 + 52)#

#s_21 = 1932#

Hopefully this helps!