find two consecutive natural number the sum of whose reciprocal is 11/30
n = ist number
n+1 = second number
the sum: n + n + 1 or 2n + 1
reciprocal we would want 1/(2n+1)
and that should equal 11/30
Set the two fractions equal to each other
Cross multiply and then solve for n.
then find n+1
want the answer of Quadratic Eqution- Divide 20 into two parts so that their sum of their reciprocals is4/15
Let's assume the first natural number is x.
The second consecutive natural number is x + 1.
According to the given information, the sum of their reciprocals is 11/30, so we can write the equation:
1/x + 1/(x + 1) = 11/30
To solve this equation, we need to find a common denominator:
(1/x) * (x + 1)/(x + 1) + (1/(x + 1)) * (x/x) = 11/30
((x + 1) + x)/(x(x + 1)) = 11/30
(2x + 1)/(x(x + 1)) = 11/30
Cross-multiplying:
30(2x + 1) = 11x(x + 1)
60x + 30 = 11x^2 + 11x
Rearranging the equation:
11x^2 - 49x + 30 = 0
Factoring the equation:
(11x - 6)(x - 5) = 0
Setting each factor equal to zero:
11x - 6 = 0 or x - 5 = 0
Solving these equations:
11x = 6 or x = 5
x = 6/11 or x = 5
Since we are looking for natural numbers, x = 5 is the only valid solution.
Therefore, the first natural number is 5, and the second consecutive natural number is 6.
To find two consecutive natural numbers whose sum of reciprocals is equal to 11/30, we can set up the following equation:
1/n + 1/(n + 1) = 11/30
To solve the equation, we need to get rid of the denominators. We can do this by finding a common denominator for both fractions, which in this case is n(n + 1).
Multiplying both sides of the equation by n(n + 1), we get:
(n + 1) + n = (11/30)(n)(n + 1)
Simplifying the equation further, we have:
2n + 1 = (11/30)(n^2 + n)
To eliminate the fractional part, we can multiply both sides of the equation by 30, leading to:
60n + 30 = 11n^2 + 11n
Rearranging the equation, we get:
11n^2 - 49n - 30 = 0
To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula. In this case, factoring will give us the solution:
11n^2 - 49n - 30 = (11n + 2)(n - 15) = 0
Setting each factor equal to zero, we have two possible solutions:
11n + 2 = 0 or n - 15 = 0
Solving each equation separately, we get:
11n = -2 or n = 15
Dividing both sides of the first equation by 11, we find:
n = -2/11
However, since we are looking for natural numbers, -2/11 is not a valid solution. Hence, the only solution is n = 15.
Therefore, the two consecutive natural numbers whose sum of reciprocals is equal to 11/30 are 15 and 16.