What is the sum of the arithmetic sequence 152, 138, 124 if there are 24 terms?

1 Answer
George C.
Mar 19, 2018

#-216#

Explanation:

The general term of an arithmetic sequence can be written in the form:

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

where #a# is the initial term and #d# the common difference.

Then:

#2 sum_(n=1)^N a_n = sum_(n=1)^N (a_n + a_(N+1-n))#

#color(white)(2 sum_(n=1)^N a_n) = sum_(n=1)^N (a + d(n-1) + a + d(N-n))#

#color(white)(2 sum_(n=1)^N a_n) = sum_(n=1)^N (2a + dN - d)#

#color(white)(2 sum_(n=1)^N a_n) = N(2a + dN - d)#

So:

#s_N = sum_(n=1)^N a_n = 1/2 N(2a+dN-d)#

Equivalently, the sum of the first #N# terms of an arithmetic sequence is the number of terms multiplied by the average term.

So:

#s_N = N * (a_1 + a_N)/2#

#color(white)(s_N) = 1/2 N (a+d(1-1) + a+d(N-1))#

#color(white)(s_N) = 1/2 N(2a+dN-d)#

For our example:

#a = 152#

#d = 138-152 = -14#

#N = 24#

So:

#s_24 = 1/2 (color(blue)(24))(2(color(blue)(152))+(color(blue)(-14))(color(blue)(24))-color(blue)(-14))#

#color(white)(s_24) = 12(304-336+14)#

#color(white)(s_24) = 12(-18)#

#color(white)(s_24) = -216#

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