Given that AP is 27, 24, 21 …

a1 = 30 ;a2=27; a3 = 24 and d =a2-a1= 27- 30 = (-3) ; sn =0 (given )

We know that

Sn=n/2 [2a+(n−1)d]

Since the sum is 0,

n/2[2×30+(n−1)(−3)]=0

n(60−3n+3)=0

n(60−3n)=0

n=0 and 60−3n=0

60=3n

n = 60/3

n = 20

∴n=0 and n=20

Since the number of terms cannot be equal to 0 so number of terms (n) = 20

• This reply was modified 3 months ago by  Rajesh Saini. Reason: Type mistake