Question

The sum of the reciprocals of three numbers in H.P. is 12 and the product of the numbers is 1÷48
Find the numbers.​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ three \: requried \: numbers \: in \: HP \\ \\ so \: let \: the \: three \: numbers \: in \: HP \: be \\ \frac{1}{a - d \: } \: , \frac{1}{a} \: , \: \frac{1}{a + d} \\ \\ since \: they \: are \: in \: HP \\ a - d \: , \: a, \: a + d \: are \: in \: AP \\ \\ according \: to \: first \: condition \\ a - d + a + a + d = 12 \\ 3a =1 2 \\ a = 4 \\ \\ according \: to \: second \: condition \\ \frac{1}{a} \times \frac{1}{a - d} \times \frac{1}{a + d} = \frac{1}{48} \\ \\ \frac{1}{4} \times \frac{1}{a {}^{2} - d {}^{2} } = \frac{1}{48} \\ \\ \frac{1}{a {}^{2} - d {}^{2} } = \frac{1}{12} \\ \\ applying \: invertendo \\ we \: get \\ \\ a {}^{2} - d {}^{2} = 12 \\ \\ d {}^{2} = a {}^{2} - 12 \\ \\ = (4) {}^{2} - 12 \\ \\ = 16 - 12 \\ \\d {}^{2} = 4 \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \\ \\ d = 2 \: \: or \: \: d = - 2$

$so \: if \: we \: take \: d = 2 \\ then \: required \: numbers \: are \\ \\ \frac{1}{a - d} = \frac{1}{4 - 2} = \frac{1}{2} \\ \\ \frac{1}{a} = \frac{1}{4} \\ \\ \frac{1}{a + d} = \frac{1}{4 + 2} = \frac{1}{6}$

$if \: we \: take \: d = - 2 \\ then \: required \: numbers \: are \\ \\ \frac{1}{a - d} = \frac{1}{4 - ( - 2)} = \frac{1}{4 + 2} = \frac{1}{6} \\ \\ \frac{1}{a} = \frac{1}{4} \\ \\ \frac{1}{a + d} = \frac{1}{4 + ( - 2)} = \frac{1}{4 - 2} = \frac{1}{2}$