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.

∴ S_n = 693

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 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 22^nd term is 0. So, the sum doesn’t get affected.