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Question
How many terms of the AP 63, 60, 57, 54, ….. must be taken so that their sum is 693? Explain the double answer.
Solution
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,
sn = 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 22nd 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.
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