Which number should be added to the numbers \[13,15,19\] so that the resulting numbers be consecutive terms of an H.P?
A) $7$
B) $6$
C) $ - 6$
D) $ - 7$
Answer
601.5k+ views
Hint: A sequence is called Harmonic if its terms are reciprocals of terms of an arithmetic progression (AP). We have to check adding which of the numbers in the options will make the given sequence harmonic. So consider adding a variable and find the value of it using the definition of AP.
Complete step-by-step answer:
Given three numbers $13,15,19$.
We have to check which number has to be added with these so that the resultant gives consecutive terms of a Harmonic progression.
Let $x$ be the number adding to each.
Therefore the new numbers are $13 + x,15 + x,19 + x$.
For these numbers to be consecutive terms of a Harmonic progression, its reciprocals must be consecutive terms of an arithmetic progression.
That is, $\dfrac{1}{{13 + x}},\dfrac{1}{{15 + x}},\dfrac{1}{{19 + x}}$ must form an Arithmetic progression.
A sequence is called arithmetic progression if the difference of two consecutive terms is the same everywhere.
Therefore, $\dfrac{1}{{15 + x}} - \dfrac{1}{{13 + x}} = \dfrac{1}{{19 + x}} - \dfrac{1}{{15 + x}}$
Cross-multiplying on both sides we get,
$\Rightarrow$$\dfrac{{(13 + x) - (15 + x)}}{{(15 + x)(13 + x)}} = \dfrac{{(15 + x) - (19 + x)}}{{(19 + x)(15 + x)}}$
We can cancel $15 + x$ from denominators on both sides.
$\Rightarrow$$\dfrac{{(13 + x) - (15 + x)}}{{(13 + x)}} = \dfrac{{(15 + x) - (19 + x)}}{{(19 + x)}}$
Simplifying the numerators we get,
$\Rightarrow$$\dfrac{{13 + x - 15 - x}}{{13 + x}} = \dfrac{{15 + x - 19 - x}}{{19 + x}}$
$ \Rightarrow \dfrac{{ - 2}}{{13 + x}} = \dfrac{{ - 4}}{{19 + x}}$
Cancelling $ - 2$ from both sides we have,
$ \Rightarrow \dfrac{1}{{13 + x}} = \dfrac{2}{{19 + x}}$
Cross multiplying we get,
$\Rightarrow$$19 + x = 2(13 + x)$
Simplifying we get,
$\Rightarrow$$19 + x = 26 + 2x$
Rearranging we get,
$\Rightarrow$$2x - x = 19 - 26$
$ \Rightarrow x = - 7$
So we got the answer as $ - 7$.
$\therefore $ Option D is the correct answer.
Note: We can solve this question also by trial and error method. There are four numbers in the options. We can check each one separately and see whether the reciprocals form an HP.
If $a,b,c...$ form an Arithmetic progression (AP), then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},...$ forms a Harmonic progression (HP).
A sequence is called arithmetic progression if the difference of two consecutive terms is the same everywhere.
Complete step-by-step answer:
Given three numbers $13,15,19$.
We have to check which number has to be added with these so that the resultant gives consecutive terms of a Harmonic progression.
Let $x$ be the number adding to each.
Therefore the new numbers are $13 + x,15 + x,19 + x$.
For these numbers to be consecutive terms of a Harmonic progression, its reciprocals must be consecutive terms of an arithmetic progression.
That is, $\dfrac{1}{{13 + x}},\dfrac{1}{{15 + x}},\dfrac{1}{{19 + x}}$ must form an Arithmetic progression.
A sequence is called arithmetic progression if the difference of two consecutive terms is the same everywhere.
Therefore, $\dfrac{1}{{15 + x}} - \dfrac{1}{{13 + x}} = \dfrac{1}{{19 + x}} - \dfrac{1}{{15 + x}}$
Cross-multiplying on both sides we get,
$\Rightarrow$$\dfrac{{(13 + x) - (15 + x)}}{{(15 + x)(13 + x)}} = \dfrac{{(15 + x) - (19 + x)}}{{(19 + x)(15 + x)}}$
We can cancel $15 + x$ from denominators on both sides.
$\Rightarrow$$\dfrac{{(13 + x) - (15 + x)}}{{(13 + x)}} = \dfrac{{(15 + x) - (19 + x)}}{{(19 + x)}}$
Simplifying the numerators we get,
$\Rightarrow$$\dfrac{{13 + x - 15 - x}}{{13 + x}} = \dfrac{{15 + x - 19 - x}}{{19 + x}}$
$ \Rightarrow \dfrac{{ - 2}}{{13 + x}} = \dfrac{{ - 4}}{{19 + x}}$
Cancelling $ - 2$ from both sides we have,
$ \Rightarrow \dfrac{1}{{13 + x}} = \dfrac{2}{{19 + x}}$
Cross multiplying we get,
$\Rightarrow$$19 + x = 2(13 + x)$
Simplifying we get,
$\Rightarrow$$19 + x = 26 + 2x$
Rearranging we get,
$\Rightarrow$$2x - x = 19 - 26$
$ \Rightarrow x = - 7$
So we got the answer as $ - 7$.
$\therefore $ Option D is the correct answer.
Note: We can solve this question also by trial and error method. There are four numbers in the options. We can check each one separately and see whether the reciprocals form an HP.
If $a,b,c...$ form an Arithmetic progression (AP), then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},...$ forms a Harmonic progression (HP).
A sequence is called arithmetic progression if the difference of two consecutive terms is the same everywhere.
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