Given

Arithmetic Progression, AP= 21, 18, 15, ……

First term, a = 21

Second term = 18

Required Sum = 0

Solution

Common difference, d = Second term - first term =18-21=-3

The sum for n terms, S_{n} = \frac{n}{2}[2a+(n-1)d]

Where n = number of terms

a = first term

d = common difference
As the required sum is 0.

So, substituting the given values to find the value of n, i,e., number of terms.

S_{n} = \frac{n}{2}[2a+(n-1)d]

\Rightarrow 0 = \frac{n}{2}[2.21+(n-1)(-3)]

\Rightarrow 0 = n (42 - 3n +3)

\Rightarrow 0 = n (45 - 3n)

\Rightarrow 0 = 45n - 3n^{2}

\Rightarrow 0 = 3n (15 - n)

\Rightarrow n = 0 or n = 15

For n = 0, this means 0 number term, which is impossible.

So, n = 15

Hence, 15 terms of the AP 21, 18, 15,.......... must be added to get the sum 0.

Conclusion

For the given A.P. 21,18,15…; sum of 15 terms adds to get zero.