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.

Answer 1

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 S_n = 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.