Find the number of terms of the A.P. 63, 60, 57, ... so that their sum is 693. Explain the double answer.

AP = 63, 60, 57,…

Here, a = 63, d = 60 – 63 = -3 and S_n = 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.

⇒ S₂₁ = 21 × 33

⇒ S₂₁ = 693

and

⇒ S₂₂ = 11 × 63

⇒ S₂₂ = 693

So, the terms may be either 21 or 22 both holds true.

We get the double answer because here the 22^nd term is zero and it does not affect the sum.