# find two consecutive natural numbers the sum of whose reciprocals is 11/30

2
by RUTVI1

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2016-03-06T22:48:41+05:30

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Two consecutive integers: x, (x+1)
Reciprocals = 1/x , 1/(x+1)

Sum of reciprocals = 11/30
1/x  + 1/(x+1)       = 11/30

Multiply equation by 30x(x+1) and you have:
= 30x + 30(x+1) = 11x(x+1)
= 30x + 30x+30 = 11x^2 + 11x
= 60x+30           = 11x^2+11x
0 = 11x^2+11x-60x-30

So, this is a quadratic equation. We can simplify the expression further.
11x^2+11x-60x-30 = 0
11x^2 - 49x - 30     = 0

Now, the factors of the expression are :
=(11x + 6)((x - 5) = 0

In (x-5), we can say 5 is the integral solution and therefore,
x = 5

So, the first number is 5
The second number = x+1   = 5+1
= 6

The two numbers = 5,6
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2016-03-06T22:50:43+05:30

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Let The Number be x and x + 1
According To Given Statement
1  +     1       = 11
---     ------       -----
x       x + 1       30
x + 1 + x   = 11
------------    ------
x² + x         30
60x + 30 = 11x² + 11x
11x² + 11x - 60x - 30  = 0
11x² - 49x- 30  = 0
11x² - 55x + 6x - 30 = 0
11x(x - 5) +6(x - 5) = 0
(11x + 6)(x - 5) = 0
x = -6/11 OR x = 5

Therefore the first No . is 5 as -6/11 is Not a Natural No.
So ,
Second Natural No = 6

So the Numbers are 5 and 6

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