The denominator of a fraction is one more than twice the numerator. If the sum of the

Question

Q) The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2 16/21, find the fraction.

Sapling Academy · Accepted Answer

Let the Numerator be A,

then by given 1st condition, denominator = 2 A + 1

Hence the fraction is: $\frac{A}{2 A + 1}$

And its reciprocal will be: $\frac{2 A + 1}{A}$

By given 2nd condition: $\frac{A}{2 A + 1} + \frac{2 A + 1}{A} = 2 \frac{16}{21}$

∴ $\frac{A^2 + (2 A + 1)^2}{A (2 A + 1)} = \frac{58}{21}$

∴ 21 [ A2  + (4 A2 + 4 A + 1) ] = 58 A (2 A + 1)

∴ 105 A2  + 84 A + 21 = 116 A2 + 58 A

∴ 11 A2  - 26 A - 21 = 0

∴ 11 A2  - 33 A + 7 A - 21 = 0

∴ 11 A (A - 3) + 7 (A - 3) = 0

∴ (A - 3) (11 A + 7) = 0

∴ A = 3 and A = $\frac{- 7}{11}$

Here, we reject A = $\frac{- 7}{11}$ because it is negative, and accept A = 3

Hence, the fraction's Numerator, A = 3 and Denominator, 2A + 1 = 7

Therefore, the fraction is $\frac{3}{7}$.

Check: If fraction is: $\frac{3}{7}$ and its reciprocal is = $\frac{7}{3}$

Hence, sum of the fraction and its reciprocal =  $\frac{3}{7} + \frac{7}{3} = \frac{9 + 49}{21} = \frac{58}{21} = 2 \frac{16}{21}$

Since it matches with the given value, hence our answer is correct.

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