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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.
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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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