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  • If the sum of 3 numbers in HP is  and sum of their reciporcals is 9. Then the numbers maybe:  <di

If the sum of 3 numbers in HP is \frac{23}{15} and sum of their reciporcals is 9. Then the numbers maybe:

 

Option: 1

1,\frac{1}{2},\frac{1}{3}


Option: 2

1,\frac{1}{3},\frac{1}{5}


Option: 3

\frac{1}{2},\frac{1}{3},\frac{1}{4}


Option: 4

None of these


Answers (1)

best_answer

Let numbers be,  \frac{1}{a-d},\frac{1}{a},\frac{1}{a+d}

So,  \frac{1}{a-d},\frac{1}{a},\frac{1}{a+d}=\frac{23}{15}                .....(1)

and

(a-d)+a+(a+d)=9

\Rightarrow 3a=9\Rightarrow a=3

Putting a = 3 in (1)

\Rightarrow \frac{1}{3-d}+\frac{1}{3}+\frac{1}{3+d}=\frac{23}{15}

\Rightarrow \frac{1}{3-d}+\frac{1}{3+d}=\frac{18}{15}

\Rightarrow \frac{3+d+3-d}{9-d^2}=\frac{18}{15}

\Rightarrow \frac{6}{9-d^2}=\frac{6}{5}

\Rightarrow 9-d^2=5

\Rightarrow d^2=4

\Rightarrow d=\pm 2

So the numbers are

1,\frac{1}{3},\frac{1}{5}      (for d = 2)

or

\frac{1}{5},\frac{1}{3},1             (for d = -2)

Posted by

vishal kumar

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