Question

how many terms of Ap 18,16,14,12...... should be taken so that their sum is zero​

Answer 1

Step-by-step explanation:

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.

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