Question

how many terms of the AP 18, 16, 14, .... be taken so that their sum is zero?​

Answer 1

Step-by-step explanation:

$s_{n} = 0 \\ \frac{n}{2} (2(18) - (n - 1) (- 2)) = 0\\ n(36 - 2n + 2) = 0 \times 2 \\ - 2 {n}^{2} + 38n = 0 \\ 38n - 2 {n}^{2} = 0 \\2 n(n - 19) = 0 \\ n = 19$

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