Question

How many terms of ap 18,16,14....be sum 0

Answer 1

Sum = n/2(2a+(n-1)d)
0=n/2(2(18)+(n-1)(-2))
0=n/2(36-2n+2)
0=38-2n
-2n=-38
n=19.

Answer 2

Let the first term = a.

Let the common difference = d.

Given first term a = 18.

Given Last term sn = 0

Common difference d = 16 - 18

                                      = -2.


We know that sum of n terms sn = n/2(2a + (n - 1) * d)

                                                   0 = n/2(2(18) + (n - 1) * (-2))

                                                   0 * 2 = n(36 - 2n + 2)

                                                    0 = n(38 - 2n)

                                                    2n^2 - 38n = 0

                                                    2n(n - 19) = 0

                                                      n = 19.



Therefore the sum of 19 terms is 0.


Hope this helps!