Question

How many terms of the A.P. 16,14,12……….. are needed to given the sum `60` ? Explain why do we get two answers.

Answer 1

Here `S_(n)=60, a=16, d=-2, n=?`
`S_(hn)=n/2[(2a+(n-1)d]`………..(Formula)
`:.60=n/2[2xx16+(n-1)xx(-2)]` ….[Substitutig the values]
`:.60=n/2(32-2n+2)`
`:.60=n/2(34-2n)`
`:.60=n/2xx2(17-n)`
`:.60=n(17-n)`
`:.60=17n-n^(2)`
`:.n^(2)-17n+60=0`
`:.n^(2)-5n-12n+60=0`
`:.n(n-5)-12(n-5)=0`
`:.(n-5)(n-12)=0`
`:.n-5=0` or `n-12=0`
`:.n=5` or `n=12`
The required terms are 5 or 12.
Explanation:
THe common difference d of the A.P. is `-2`
`:.` The terms of the A.P. are in descending order.
Taking `n=5` the first 5 terms are 16,14,12,10,8. The sum is 60.
Taking `n=12`, the last 7 terms `(12-5)` are `6,4,2,0,-2,-4,-5`
The sum of these seven terms is 0.
`:.` The sum of first 12 terms is also 60.
The sum of the first terms `=` the sum of the first twelve terms.
`:.` we get two answers.
Ans. 5 terms or 12 terms.

Answer 2

Given series 16, 14, 12....

So, a = 16 and d = 14 - 16 = -2

Sum = n/2 [2a + (n−1)d]

60 = n/2 ​[2 × 16 + (n−1)(−2)]

60 × 2 = n[32 + 2 − 2n]

120 = 34n − 2n²

2n² − 34n + 120 = 0

n² − 17n + 60 = 0

n² − 12n − 5n + 60 = 0

n(n − 12) − 5(n − 12) = 0

(n − 12)(n − 5) = 0

n = 5 or 12

Hence, the number of terms are 5 or 12.