AP = 63, 60, 57,…
Here, a = 63, d = 60 – 63 = -3 and Sn = 693
We know that,
⇒ 1386 = n[129 – 3n]
⇒ 3n2 – 129n + 1386 = 0
⇒ n2 – 43n + 462 = 0
⇒ n2 – 22n – 21n + 462 = 0
⇒ n(n – 22) - 21(n – 22) = 0
⇒ (n – 21)(n – 22) = 0
⇒ n – 21 = 0 or n – 22 = 0
⇒ n = 21 or n = 22
So, n = 21 and 22
If n = 21, a = 63 and d = -3
a21 = 63 + (21 – 1)(-3)
a21 = 63 + 20 × -3
a21 = 63 – 60
a21 = 3
and If n = 22, a = 63 and d = -3
a22 = 63 + (22 – 1)(-3)
a22 = 63 + 21 × -3
a22 = 63 – 63
a22 = 0
Now, we will check at which term the sum of the AP is 693.
So, the terms may be either 21 or 22 both holds true.
We get the double answer because here the 22nd term is zero and it does not affect the sum.