Question

the sum of the reciprocals of two consecutive numbers is 23/132 find the numbers

Answer 1

4) Sum of Reciprocals is..
Let no. 1 is 'x'. Therefore it's Reciprocal is 1/x. Since we are given sum of consecutive nos. So let no. 2 be 'x + 1' and it's Reciprocal is 1/x+1.

So, there sum is...
1/x + 1/x+1 = 23/132
==> (2x + 1)/x^2 +x = 23/132 (taking L.C.M.)

==> 132(2x + 1) = 23(x^2 + x)
==> 264x + 132 = 23x^2 + 23x
==> 23x^2 +23x - 264x -132 = 0
==> 23x^2 -241x - 132 = 0
==> 23x^2 - 253x + 12x - 132 = 0
==> 23x ( x - 11) + 12 ( x - 11) = 0
==> (23x + 12)(x - 11) = 0
So we get x = 11 because x = -12/23 can not be our answer as it is negative.

So, the required numbers are 11 and 12. You can answer by the following way :--

Reciprocal of 11 = 1/11
Reciprocal of 12 = 1/12

So,
1/11 + 1/12 = 23/132
Therefore x = 11 is correct and numbers are 11 and 12.
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Answer 2

Let the two consecutive numbers be x and x + 1.

Their reciprocals are (1/x) and (1/x + 1).

Given that sum of the reciprocals of two consecutive numbers is 23/132.

$= \ \textgreater \ \frac{1}{x} + \frac{1}{x + 1} = \frac{23}{132}$

$= \ \textgreater \ \frac{(x + 1) + x}{x(x + 1)} = \frac{23}{132}$

= > 132x + 132 + 132x = 23x^2 + 23x

= > 132 + 264x = 23x^2 + 23x

= > 23x^2 + 23x - 264x - 132 = 0

= > 23x^2 - 241x - 132 = 0

= > 23x^2 - 253x + 12x - 132 = 0

= > 23x(x - 11) + 12(x - 11) = 0

= > (23x + 12)(x - 11) = 0

= > x = -12/23, x = 11.

Neglect the -ve value.

Hence, the value of x = 11. and x + 1 = 12.

Therefore the two consecutive numbers are 11 and 12.


Hope this helps!