# 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 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.

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 22nd term is zero and it does not affect the sum.

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