How many terms of the A.P 16, 14, 12 are needed to give the sum 60? Explain why we get two answers
Answers
Answer:
Step-by-step explanation:
16 , 14 , 12 ,...............
a=16 , d = -2 , Sn = 60 , n=?
Sn= n/2 [2a+(n-1)d]
60 = n/2[2×16 +(n-1)-2]
60×2=n [32-2n +2]
120= n (34 -2n)
120=34n -2n²
2n²-34n+120=0
÷2
n²-17n + 60=0
(60n²=-20n & +3n)
n²-20n+3n + 60
n(n-20)-3(n-20)
(n-20) or (n-3)
n-20=0 or n-3 =0
.:. n = 20 or n = 3
Its has 2 solutions because sum of 60 will comes till 20 terms or till 3 terms.
Given:
Sn = 60
a = 16
d = -2
To Find:
The value of n
Solution:
Using the formula -
S = n/2 [ 2a + ( n-1)d]
60 = n/2 [ 2 x 16 + ( n - 1) x -2]
60 = n/2 ( 34 - 2n)
60 = n( 17 - n)
n² - 17n + 60
Splitting the equation -
n² - 5n - 12n + 60
= n ( n - 5) - 12 (n - 5)
= ( n - 5) ( n - 12)
Thus, n = 5 or n = 12
THe common difference d of the A.P. is = − 2
Thus, terms are in descending order
Taking n = 5
First 5 terms = 16, 14, 12, 10 and 8 where sum is 60
Takin n = 12
Last 7 term = 6, 4, 2, 0, -2, -4, - 6 where sum is 0
Therefore,
The sum of first 12 terms is also 60.
Sum of first terms = sum of first 12 twelve terms
Thus, we get two answers