Question

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

Answer 1

The given AP is 63, 60, 57, 54, ...

Here, a = 63 and d = 60 – 63 = –3

Let the required number of terms be n. Then,

S_n = 693 

⇒ `n/2 [2 xx 63 + (n - 1) xx (-3)] = 693`   ...`{S_n = n/2 [2a + (n - 1)d]}`

⇒ `n/2 (126 - 3n + 3) = 693`

⇒ n(129 – 3n) = 1386 

⇒ 3n² – 129n + 1386 = 0

⇒ 3n² – 66n – 63n + 1386 = 0

⇒ 3n(n – 22) – 63(n – 22) = 0

⇒ (n – 22) (3n – 63) = 0

⇒ n – 22 = 0 or 3n – 63 = 0

⇒ n = 22 or n = 21 

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22^nd term of the AP is 0. 

a₂₂ = 63 + (22 – 1) × (–3) 

= 63 – 63 

= 0

Hence, the required number of terms is 21 or 22.