Question

the sum of the reciprocal of two consective odd natural and numbers is 12/35 find those number?​

Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

the sum of the reciprocal of two consective odd natural and numbers is 12/35 find those number?

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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Let the first odd natural number be 'x'

Second odd natural number be 'x+2'

It's reciprocal becomes 1/x & 1/x+2 respectively.

Now , According to the Question:

$= > \frac{1}{x} + \frac{1}{x + 2} = \frac{12}{35}$

$= > 35(2x + 2) = 12( {x}^{2} + 2x)$

$= > 70x + 70 = 12 {x}^{2} + 24x$

$= > 12 {x}^{2} + 24x - 70x - 70 = 0$

$= > 12 {x}^{2} - 46x - 70 = 0$

Taking 2 as common:

$= > 2(6 {x}^{2} - 23x - 35) = 0$

$= > 6 {x}^{2} - 30x + 7x - 35 = 0$

$= > 6x(x - 5) + 7(x - 5) = 0$

$= > (6x + 7)(x - 5) = 0$

$= > x = - \frac{7}{6}$

$= > x = 5$

=>x=-7/6 can't be a number because it is negative and according to Question sum is given is positive.

First odd natural number is 1/x=$\bold{\pink{1/5}}$

second odd natural number is 1/x+2=$\bold{\red{1/7}}$

check :

$= > \frac{1}{5} + \frac{1}{7} = \frac{7 + 5}{35} = \frac{12}{35}$

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