# math

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Evaluate the sum of the first 120 terms of the series whose nth term is equal to (4 n - 14)

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You can make use of:

Sum from n = 0 to N of n = 1/2 N (N+1)

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n.....1.....2.....3.....4.....5.....6
N....-10...-6....-2.....2.....6.....10

The 120th term derives from L = a + (n - 1)d where a = the first term, n = the number of terms and d = the common difference.

Then, the sum is S = n(a + L)/2.

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