Three times the reciprocal of the sum of two consecutive odd integers is 3/20. Find the two consecutive off integers.
Let 2n-1 be the smaller odd integer, then
2n+1 is the larger.
Given
3/(2n-1 + 2n+1)=3/20
3/(4n)=3/20
4n=20
n=5
The integers are 2*5-1=9 and 2*5+1=11.
To solve this problem, we need to first set up an equation based on the given information.
Let's assume that the first odd integer is represented by "x", and the second consecutive odd integer is "x + 2" (since consecutive odd integers have a difference of 2).
Now, we can set up the equation using the information given in the problem:
3 * (1 / (x + (x + 2))) = 3/20
To simplify the equation, we can find the common denominator on the right side and multiply both sides by that denominator:
3 * (1 / (2x + 2)) = 3/20
Multiplying both sides by (2x + 2):
3 = (3/20) * (2x + 2)
Now, we can simplify the equation further:
3 = (3/20) * (2x + 2)
3 = (3/20) * 2(x + 1)
3 = (6/20) * (x + 1)
3 = (3/10) * (x + 1)
Multiplying both sides by 10 to eliminate the fraction:
30 = 3(x + 1)
Now, we can solve for "x":
30 = 3x + 3
27 = 3x
x = 9
Therefore, the first odd integer is 9, and the second consecutive odd integer is 9 + 2 = 11.