**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.**

**Ans: **

Let the Numerator be A,

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

Hence the fraction is:

And its reciprocal will be:

By given 2^{nd} condition:

∴

∴ 21 [ A^{2 } + (4 A^{2} + 4 A + 1) ] = 58 A (2 A + 1)

∴ 105 A^{2 } + 84 A + 21 = 116 A^{2} + 58 A

∴ 11 A^{2 } – 26 A – 21 = 0

∴ 11 A^{2 } – 33 A + 7 A – 21 = 0

∴ 11 A (A – 3) + 7 (A – 3) = 0

∴ (A – 3) (11 A + 7) = 0

∴ A = 3 and A =

Here, we reject A = because it is negative, and accept A = 3

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

**Therefore, the fraction is .**

*Check: If f**raction is: and its reciprocal is = *

*Hence, sum of the fraction and its reciprocal = *

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

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