Question

how many terms of AP 18, 16, 14 so on be taken so their sum is zero

Answer 1

Sn=n/2(2a+(n-1)d)=0=n/2(2*18+(n-1)(-2))
0=36-2n+2=>2n=38=>n=19

Answer 2

Heya User,

----> Term 1 = T1 = 18
----> Common d = c.d. = -2

---> Sn =
$\frac{n(2a + (n - 1)d)}{2}$
==> Sn =
$\frac{n(36 - 2n + 2)}{2} = 19n - {n}^{2}$
But given, Sn = 0
-->
$19n - {n}^{2} = 0$
==>n=19 or 0..

But n is a natural no. ....
Soooo. Sum of 19 terms of this AP equals 0