Given A.P. is 27, 24, 21. . .
We know that,
Sn = \(\frac{n}{2}\)[2a + (n − 1)d]
Here we have, the first term (a) = 27
The sum of n terms (Sn) = 0
Common difference of the A.P. (d) = a2 – a1 = 24 – 27 = -3
On substituting the values in Sn, we get
⟹ 0 = \(\frac{n}{2}\)[2(27) + (n − 1)( − 3)]
⟹ 0 = (n)[54 + (n – 1)(-3)]
⟹ 0 = (n)[54 – 3n + 3]
⟹ 0 = n [57 – 3n] Further we have, n = 0 Or, 57 – 3n = 0
⟹ 3n = 57
⟹ n = 19
The number of terms cannot be zero,
Hence, the numbers of terms (n) is 19.
