lnm87 wrote:
eakabuah wrote:
We are given that the sum of n consecutive positive integers greater than 1 is 75. We are to determine the sum of all possible values of n.
We have an arithmetic progression with a common difference of 1, and the first term greater than 1
Generically, the sum, Sn = (a1+an)*n/2
Now Sn=75, and Let a be the first term, then the last term is a+n, where n is the number of terms
so 75=(a+a+n)n/2
150=2an+n^2
Hence n^2 + 2an - 150 = 0 ..........................(1)
We need sets of two pairs of positive integers whose product is 150
These pairs are: 2,75; 3,50; 5,30; 6,25; 10,15.
Substituting each of these pairs into (1) above, we are able to come out with the positive roots of the equation and these positive roots correspond to the smaller value of the integer pairs.
Hence the possible values of n=2, 3, 5, 6, and 10.
Summing the possible values of n=2+3+5+6+10=26
The answer is, therefore, option D.
eakabuahCan you shed more light on red part.
You would realize that by the use of the sum of n terms in an arithmetic progression, we came up with a quadratic equation: n^2 + 2an - 150 = 0
What is interesting about this equation is that it is expected to yield two roots, and you will notice that the nature of these roots depends on the value of 2a. Our focus is not to determine a, which is the first term in the series but rather to determine n. So, for simplicity, you can even replace 2a with b or any variable.
You are now left with the task of guessing the values whose sum equal b and whose product equals -150. The best way to take guesses is to find find the factors of 150.
Factors of 150 are: {1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150}.
The two numbers whose sum equals b and product equal -150 are definitely among the set of numbers above.
Note that n>1, hence 1, 150 pair is not part of these pairs. So, you are left with 2,75; 3, 50; 5,30; 6,25; and 10,15. So that's how the pairs came about.