how many terms of the sequence 18 16 14 should be taken so that their sum is zero
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We know that sum = n/2(2a + (n-1) d)
0 = n/2(2 * 18 + (n - 1) * (-2))
0 = n/2(36-2n+2)
0 = 38n - 2n^2
2n^2 - 38n = 0
2n(n-19) = 0
n - 19 = 0
n = 19.
Hope this helps!
0 = n/2(2 * 18 + (n - 1) * (-2))
0 = n/2(36-2n+2)
0 = 38n - 2n^2
2n^2 - 38n = 0
2n(n-19) = 0
n - 19 = 0
n = 19.
Hope this helps!
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