Question

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Answer 1

Answer :

Let the two consecutive odd terms be x and x + 2.

As per your question,

(1/x) + ( 1/x + 2 ) = 12/35

35 (x + 2) + 35x = 12x( x + 2 )

35x + 70 + 35x = 12x^2 + 24x

70x +70 = 12x^2 + 24x

12x^2 - 70x + 24x - 70 = 0

12x^2 - 46x - 70 = 0

6x^2 - 23x - 35 = 0

Hence, we get

x = 5 or x = -7/6

But -7/6 isn't acceptable.

So, x = 5

x + 2 = 5 + 2

So, the two odd consecutive terms are 5 and 7.

Verification :

1/5 + 1/7 = 12/35

(7 + 5) / 7 × 5 = 12 / 35

12/35 = 12/35

Hence, the two odd consecutive terms are 5 and 7.

Answer 2

$\mathfrak{\huge{Answer:}}$

Let the two numbers be = x and x + 2

According to the question :-

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

Solve this formed equation further

=》 $\tt{\frac{x + 2 + x}{x^{2} + 2x} = \frac{12}{35}}$

Solve it further

=》 $\tt{70x + 70 = 12x^{2} + 24x}$

Some more steps to go

=》 $\tt{12x^{2} - 46x - 70}$

Last one step

=》 $\tt{6x^{2} - 23x - 35}$

Using the completing the square method, find the value of x :-

Make the coefficient of x = 1

=》 $\bf{x^{2} - \frac{23}{6} - \frac{35}{6}= 0}$

=》 $\bf{x^{2} - \frac{23}{6} + (\frac{23}{12})^{2} - (\frac{23}{12})^{2} = \frac{35}{6}}$

Make the set up and do the squares

=》 $\bf{( x - \frac{23}{12})^{2} = \frac{529}{144} + \frac{35}{6}}$

Take out the square roots

=》 $\bf{x - \frac{23}{12} = \pm \sqrt{ \frac{1369}{144}}}$

Don't forget to put both the (+) and the (-) signs as we are unaware that the square root will be of positive or negative value.

=》 $\bf{x - \frac{23}{12} = \pm \frac{37}{12}}$

Do the required operations

=》 $\bf{x = \pm \frac{37 + 23}{12}}$

Values come out to be

=》 $\bf{x = 5, \frac{-7}{6}}$

Now, given is that the numbers are natural numbers, thus, they'll be positive.

Therefore, only 5 is acceptable as the value of x.

x = 5

Other number will be = x + 2 = 5 + 2 = 7

Finally, the numbers will be = $\sf{\huge{5\:and\:7}}$