Answer :

Here, first term = a = 21

Common difference = d = 18 - 21 = - 3


Let first n terms of the AP sums to zero.


Sn = 0


To find: n


Now, Sn = (n/2) × [2a + (n - 1)d]


Since, Sn = 0


(n/2) × [2a + (n - 1)d] = 0


(n/2) × [2(21) + (n - 1)(-3)] = 0


(n/2) × [42 - 3n + 3)] = 0


(n/2) × [45 - 3n] = 0


[45 - 3n] = 0


45 = 3n


n = 15


15 terms of the given AP sums to zero.


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