First term is 5 and the common difference is 3. Starting with the first term, how do you find the number of terms that have a sum of 3925?

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Answer 1 · 2016-04-01T20:05:05

Use the method shown to find number of terms is #50#

The general term of an arithmetic sequence can be written:

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

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

Then:

#sum_(n=1)^N a_n = N * (a_1 + a_N)/2#

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

#=aN +d/2*N(N-1)#

#=d/2 N^2 + (a-d/2) N#

In our example, #a = 5#, #d = 3# and we want to solve

#3925 = sum_(n=1)^N a_n#

#= d/2 N^2 + (a-d/2) N#
#= 3/2 N^2 + (5-3/2) N#

#= 3/2 N^2 + 7/2 N#

Multiply both ends by #2# to get:

#3N^2 + 7N = 7850#

Subtract #7850# from both sides to get:

#3N^2 + 7N - 7850 = 0#

#N = (-7 +-sqrt(7^2 + 4*3*7850))/6#

#= (-7+-sqrt(49+94200))/6#

#= (-7+-sqrt(94249))/6#

#= (-7+-307)/6#

That is #N=-314/6 = -157/3# or #N=300/6 = 50#

Discard the negative solution to find #N=50#