
The sum of three numbers in H.P is $37$ and the sum of their reciprocal $\dfrac{1}{4}$. Find the smallest of those three numbers.
Answer
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Hint:Arithmetic Progression is a series whose consecutive elements have a constant difference. It is denoted as $a - d,a,a + d,a + 2d....,a + \left( {n - 1} \right)d$ where $a$= first term, $d$= difference between the consecutive terms and $n$= number of terms in A.P. Whereas, Harmonic Progression is a series containing the reciprocal of terms of an A.P. It is denoted as $\dfrac{1}{a},\dfrac{1}{{a + d}},....,\dfrac{1}{{a + \left( {n - 1} \right)d}}$ where $\dfrac{1}{a}$= first term and $n$=number of terms in an A.P. Here in this question, we have to find the smallest three numbers whose sum in H.P. is $37$and their reciprocal sums up to be $\dfrac{1}{4}$.
Complete step by step answer:
Given is that the sum of three numbers in harmonic progression is $37$ and the sum of their reciprocal is $\dfrac{1}{4}$. We have to find out those three smallest numbers. Let three numbers in H.P be $\dfrac{1}{{a - d}},\dfrac{1}{a},\dfrac{1}{{a + d}}$. Then, according to the question,
$\dfrac{1}{{a - d}} + \dfrac{1}{a} + \dfrac{1}{{a + d}} = 37$-----(1)
$ \Rightarrow a - d + a + a + d = \dfrac{1}{4}$-----(2)
Using equation (2) we find out the value of $a$,
$3a = \dfrac{1}{4} \\
\Rightarrow a = \dfrac{1}{{12}} \\ $
Using the value of $a$ in equation (1) we get,
$
\dfrac{{12}}{{1 - 12d}} + \dfrac{{12}}{{1 + 12d}} = 37 - 12 \\
\Rightarrow \dfrac{{12}}{{1 - 12d}} + \dfrac{{12}}{{1 + 12d}} = 25 \\
\Rightarrow \dfrac{{24}}{{1 - 144{d^2}}} = 25 \\
\Rightarrow 1 - 144{d^2} = \dfrac{{24}}{{25}} \\ $
$\Rightarrow {d^2} = \dfrac{1}{{25 \times 144}} \\
\Rightarrow d = \pm \dfrac{1}{{60}} \\ $
Hence, $d = \pm \dfrac{1}{{60}}$
$a - d = \dfrac{1}{{15}},\dfrac{1}{{10}} \\
\Rightarrow a = \dfrac{1}{{12}} \\
\therefore a + d = \dfrac{1}{{10}},\dfrac{1}{{15}} \\ $
Therefore, the three numbers in H.P are $15,12,10$ or $10,12,15$.
Note:As we can very clearly see that the question is asked in a very simple language so, the concepts of arithmetic progression and harmonic progression series should be very clear. Students should be aware of all the formulas related to arithmetic progression and harmonic progression as it would help them solve the questions easily. Do not forget to convert A.P terms into H.P terms by doing the reciprocal of A.P terms.
Complete step by step answer:
Given is that the sum of three numbers in harmonic progression is $37$ and the sum of their reciprocal is $\dfrac{1}{4}$. We have to find out those three smallest numbers. Let three numbers in H.P be $\dfrac{1}{{a - d}},\dfrac{1}{a},\dfrac{1}{{a + d}}$. Then, according to the question,
$\dfrac{1}{{a - d}} + \dfrac{1}{a} + \dfrac{1}{{a + d}} = 37$-----(1)
$ \Rightarrow a - d + a + a + d = \dfrac{1}{4}$-----(2)
Using equation (2) we find out the value of $a$,
$3a = \dfrac{1}{4} \\
\Rightarrow a = \dfrac{1}{{12}} \\ $
Using the value of $a$ in equation (1) we get,
$
\dfrac{{12}}{{1 - 12d}} + \dfrac{{12}}{{1 + 12d}} = 37 - 12 \\
\Rightarrow \dfrac{{12}}{{1 - 12d}} + \dfrac{{12}}{{1 + 12d}} = 25 \\
\Rightarrow \dfrac{{24}}{{1 - 144{d^2}}} = 25 \\
\Rightarrow 1 - 144{d^2} = \dfrac{{24}}{{25}} \\ $
$\Rightarrow {d^2} = \dfrac{1}{{25 \times 144}} \\
\Rightarrow d = \pm \dfrac{1}{{60}} \\ $
Hence, $d = \pm \dfrac{1}{{60}}$
$a - d = \dfrac{1}{{15}},\dfrac{1}{{10}} \\
\Rightarrow a = \dfrac{1}{{12}} \\
\therefore a + d = \dfrac{1}{{10}},\dfrac{1}{{15}} \\ $
Therefore, the three numbers in H.P are $15,12,10$ or $10,12,15$.
Note:As we can very clearly see that the question is asked in a very simple language so, the concepts of arithmetic progression and harmonic progression series should be very clear. Students should be aware of all the formulas related to arithmetic progression and harmonic progression as it would help them solve the questions easily. Do not forget to convert A.P terms into H.P terms by doing the reciprocal of A.P terms.
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