Question

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

Answer 1

Answer:

Step-by-step explanation:

X=11.

See in attachment

Answer 2

Answer:

Required Numbers are (11,12)

$Or\\</p><p>(\frac{-12}{23},\frac{11}{23})$

Step-by-step explanation:

$Let\:x \:and \:(x+1) \:are\\</p><p>two \: consecutive\: numbers$

$Sum\:of\: the\: reciprocals=\frac{23}{132}$

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

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

$\implies \frac{2x+1}{x^{2}+x}=\frac{23}{132}$

$\implies 132(2x+1)=23(x^{2}+x)$

$\implies 264x+132=23x^{2}+23x$

$\implies 0=-264x-132+23x^{2}+23x$

$\implies 23x^{2}-241x-132=0$

/* Splitting the middle term, we get

$\implies 23x^{2}-263x+12x-132=0$

$\implies 23x(x-11)+12(x-11)=0$

$\implies (x-11)(23x+12)=0$

$\implies x-11=0\:Or\:23x+12=0$

$\implies x=11\:Or\:23x=12$

$\implies x=11\:Or\:x =\frac{-12}{23}$

Therefore,

Case 1:

If x = 11, x+1 = 11+1 = 12

case 2:

$If \:x =\frac{-12}{23}\\</p><p>x+1=\frac{-12}{23}+1\\=\frac{-12+23}{23}\\=\frac{11}{23}$

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