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 

`⇒ 3n^2 - 129n + 1386 = 0`

`⇒  3n^2 - 66n -63 n + 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. 

`a_22 = 63+ (22-1) xx (-3) = 63-63 =0`

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