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Question
How many terms of the AP `20, 19 1/3 , 18 2/3, ...` must be taken so that their sum is 300? Explain the double answer.
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Solution
The given AP is: `20, 19 1/3 , 18 2/3, ...`
First term (a): a = 20
Common difference (d): d = `19 1/3 - 20 = -2/3`
The sum of the first n terms of an AP is given by:
`S_n = 300, a = 20, and d= -2/3`, we get
`S_n = n/2 [2a+(n-1)d]`
Substituting Sn = 300, a = 20, and d = `−2/3`, we get:
`300 = n/2 [2(20)+(n-1)(-2/3)]`
`300=n/2 [40-2/3(n-1)]`
Multiply through by 2 to eliminate the fraction:
`600 = n [40 - 2/3 (n-1)]`
`600=n[40-2n/3 + 2/3]`
`600 = n [120/3 - (2n)/3 + 2/3]`
`600 = n [(122-2n)/3]`
1800 = n(122 − 2n)
1800 = 122n − 2n2
2n2 − 122n + 1800 = 0
Divide through by 2 to simplify: n2 − 61n + 900 = 0
`n = (-b +- sqrtb^2-4ac)/(2a)`
`n = (-(-61) +- sqrt((-61)^2 - 4(1)(900)))/2`
`n = (61 +- sqrt(3721 - 3600))/2`
`n = (61 +- sqrt121)/2`
`n = (61 +- 11)/2`
`n = (61+11)/2 = 72/2 = 36`
`n = (61-11)/2 = 50/2 = 25`
The two solutions, n = 25 and n = 36, occur because the AP has a negative common difference (d = `−2/3`), meaning the terms decrease as n increases.
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