# How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.

Here, first term = a =63

Common difference = d = 60 - 63 = - 3

Let first n terms of the AP sums to 693.

Sn = 693

To find: n

Now, Sn = (n/2) × [2a + (n - 1)d]

Since, Sn = 693

(n/2) × [2a + (n - 1)d] = 693

(n/2) × [2(63) + (n - 1)(-3)] = 693

(n/2) × [126 - 3n + 3)] = 693

(n/2) × [129 - 3n] = 693

n[129 - 3n] = 1386

129n - 3n2 = 1386

3n2 - 129n + 1386 = 0

(n - 22)( n - 21)= 0

n = 22 or n = 21

n= 22 or n = 21

Since, a22 = a + 21d

= 63 + 21(-3)

= 0

Both the first 21 terms and 22 terms give the sum 693 because the 22nd term is 0. So, the sum doesn’t get affected.

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