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Question: The sum of three numbers in H.P is \(37\) and the sum of their reciprocal \(\dfrac{1}{4}\). Find the...

The sum of three numbers in H.P is 3737 and the sum of their reciprocal 14\dfrac{1}{4}. Find the smallest of those three numbers.

Explanation

Solution

Arithmetic Progression is a series whose consecutive elements have a constant difference. It is denoted as a−d,a,a+d,a+2d....,a+(n−1)da - d,a,a + d,a + 2d....,a + \left( {n - 1} \right)d where aa= first term, dd= difference between the consecutive terms and nn= 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 1a,1a+d,....,1a+(n−1)d\dfrac{1}{a},\dfrac{1}{{a + d}},....,\dfrac{1}{{a + \left( {n - 1} \right)d}} where 1a\dfrac{1}{a}= first term and nn=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 3737and their reciprocal sums up to be 14\dfrac{1}{4}.

Complete step by step answer:
Given is that the sum of three numbers in harmonic progression is 3737 and the sum of their reciprocal is 14\dfrac{1}{4}. We have to find out those three smallest numbers. Let three numbers in H.P be 1a−d,1a,1a+d\dfrac{1}{{a - d}},\dfrac{1}{a},\dfrac{1}{{a + d}}. Then, according to the question,
1a−d+1a+1a+d=37\dfrac{1}{{a - d}} + \dfrac{1}{a} + \dfrac{1}{{a + d}} = 37-----(1)
⇒a−d+a+a+d=14\Rightarrow a - d + a + a + d = \dfrac{1}{4}-----(2)
Using equation (2) we find out the value of aa,
3a=14 ⇒a=112 3a = \dfrac{1}{4} \\\ \Rightarrow a = \dfrac{1}{{12}} \\\
Using the value of aa 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}} \\\
⇒d2=125×144 ⇒d=±160 \Rightarrow {d^2} = \dfrac{1}{{25 \times 144}} \\\ \Rightarrow d = \pm \dfrac{1}{{60}} \\\
Hence, d=±160d = \pm \dfrac{1}{{60}}
a−d=115,110 ⇒a=112 ∴a+d=110,115 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 15,12,1015,12,10 or 10,12,1510,12,15.

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.