Question

The value of \[xyz = \dfrac{{15}}{2}{\text{ or }}\dfrac{{18}}{5}\] according as the series $a,x,y,z,b$ is in arithmetic progression or harmonic progression. Then $ab$ is equal to?
A) $1$
B) $2$
C) $3$
D) $4$

Answer 1

Hint: A sequence of numbers is called Arithmetic Sequence/Progression if every term is added by a fixed number to get the next term. This fixed number is called the common difference. A sequence is called Harmonic progression, if every term of the progression is reciprocals of the terms of some Arithmetic Progression. So we can equate the two equations for$xyz$. Simplifying we get the solution.
Formula used:
If ${x_1},{x_2},{x_3},{x_4},{x_5}$ form an Arithmetic progression with common difference $d$, then
$i){x_i} = {x_{i - 1}} + d$
$ii)$Middle term formula
${x_3} = \dfrac{{{x_1} + {x_5}}}{2} = \dfrac{{{x_2} + {x_4}}}{2}$
$iii)d = \dfrac{{{\text{term difference}}}}{{{\text{position difference}}}}$
$iv)\dfrac{1}{{{x_1}}},\dfrac{1}{{{x_2}}},\dfrac{1}{{{x_3}}},...$ form an Harmonic progression.

Complete step-by-step answer:
Given that the value of \[xyz = \dfrac{{15}}{2}{\text{ or }}\dfrac{{18}}{5}\] according as the series $a,x,y,z,b$ is in arithmetic progression or harmonic progression.
We need to find $ab$.
So we try to express the given $xyz$ in terms of $a,b$.
The common difference, $d = \dfrac{{{\text{term difference}}}}{{{\text{position difference}}}}$
$ \Rightarrow d = \dfrac{{b - a}}{4}$, since $b$ is the fourth term after $a$.
If ${x_1},{x_2},{x_3},{x_4},{x_5}$ form an Arithmetic progression, then by Middle term formula,
${x_3} = \dfrac{{{x_2} + {x_4}}}{2}$
Since $y$ is the middle term of $a,x,y,z,b$ we have by middle term formula, $y = \dfrac{{a + b}}{2}$
Now let us find $x$.
$x$ is got by adding the common difference to $a$.
$ \Rightarrow x = a + \dfrac{{b - a}}{4} = \dfrac{{4a + b - a}}{4}$
$ \Rightarrow x = \dfrac{{3a + b}}{4}$
Also $z$ is obtained by adding common difference to $y$.
This gives $z = y + d$.
Substituting we get,
$ \Rightarrow z = \dfrac{{a + b}}{2} + \dfrac{{b - a}}{4}$
Simplifying we get,
$ \Rightarrow z = \dfrac{{2a + 2b}}{4} + \dfrac{{b - a}}{4}$
$ \Rightarrow z = \dfrac{{a + 3b}}{4}$
Now consider $xyz$.
$ \Rightarrow xyz = \dfrac{{3a + b}}{4} \times \dfrac{{a + b}}{2} \times \dfrac{{a + 3b}}{4}$
In the question it is given that $xyz = \dfrac{{15}}{2}$
Substituting we get,
$ \Rightarrow xyz = \dfrac{{3a + b}}{4} \times \dfrac{{a + b}}{2} \times \dfrac{{a + 3b}}{4} = \dfrac{{15}}{2} - - - (i)$

Now consider the sequence as Harmonic progression.
This gives $\dfrac{1}{a},\dfrac{1}{{{x_{}}}},\dfrac{1}{y},\dfrac{1}{z},\dfrac{1}{b}$ as an arithmetic progression.
So by middle term formula we have,
\[\dfrac{1}{y} = \dfrac{{\dfrac{1}{a} + \dfrac{1}{b}}}{2}\]
$ \Rightarrow \dfrac{1}{y} = \dfrac{1}{{2a}} + \dfrac{1}{{2b}}$
Simplifying we get,
$ \Rightarrow \dfrac{1}{y} = \dfrac{{2a + 2b}}{{2a \times 2b}}$
$ \Rightarrow \dfrac{1}{y} = \dfrac{{2(a + b)}}{{4ab}}$
Cancelling $2$ from numerator and denominator we have,
$ \Rightarrow \dfrac{1}{y} = \dfrac{{a + b}}{{2ab}}$
Taking reciprocals,
$ \Rightarrow y = \dfrac{{2ab}}{{a + b}}$
Again considering $\dfrac{1}{x}$ as a middle term we have,
$\dfrac{1}{x} = \dfrac{{\dfrac{1}{a} + \dfrac{1}{y}}}{2}$
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{2a}} + \dfrac{1}{{2y}}$
Substituting for $y$ we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{2a}} + \dfrac{1}{{2(\dfrac{{2ab}}{{a + b}})}}$
Simplifying we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{1}{{2a}} + \dfrac{{a + b}}{{4ab}}$
Multiplying numerator and denominator of $\dfrac{1}{{2a}}$ by $2b$ we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{{2b}}{{4ab}} + \dfrac{{a + b}}{{4ab}}$
$ \Rightarrow \dfrac{1}{x} = \dfrac{{2b + a + b}}{{4ab}}$
Simplifying we get,
$ \Rightarrow \dfrac{1}{x} = \dfrac{{a + 3b}}{{4ab}}$
Taking reciprocals we get,
$ \Rightarrow x = \dfrac{{4ab}}{{a + 3b}}$
Now we can find $z$ in a similar way.
$\dfrac{1}{z} = \dfrac{{\dfrac{1}{y} + \dfrac{1}{b}}}{2}$
$ \Rightarrow \dfrac{1}{z} = \dfrac{1}{{2y}} + \dfrac{1}{{2b}}$
Substituting for $y$ we get,
$ \Rightarrow \dfrac{1}{z} = \dfrac{1}{{2(\dfrac{{2ab}}{{a + b}})}} + \dfrac{1}{{2b}}$
$ \Rightarrow \dfrac{1}{z} = \dfrac{{a + b}}{{4ab}} + \dfrac{1}{{2b}}$
Multiplying numerator and denominator of $\dfrac{1}{{2b}}$ by $2a$,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{a + b}}{{4ab}} + \dfrac{{2a}}{{4ab}}$
$ \Rightarrow \dfrac{1}{z} = \dfrac{{a + b + 2a}}{{4ab}}$
Simplifying we get,
$ \Rightarrow \dfrac{1}{z} = \dfrac{{3a + b}}{{4ab}}$
Taking reciprocals we get,
$ \Rightarrow z = \dfrac{{4ab}}{{3a + b}}$
We have $xyz = \dfrac{{18}}{5}$ when the series is in harmonic progression.
Substituting we get,
$xyz = \dfrac{{4ab}}{{a + 3b}} \times \dfrac{{2ab}}{{a + b}} \times \dfrac{{4ab}}{{3a + b}} = \dfrac{{18}}{5} - - - (ii)$
Multiplying $(i)\& (ii)$ we get,
 $\dfrac{{3a + b}}{4} \times \dfrac{{a + b}}{2} \times \dfrac{{a + 3b}}{4} \times \dfrac{{4ab}}{{a + 3b}} \times \dfrac{{2ab}}{{a + b}} \times \dfrac{{4ab}}{{3a + b}} = \dfrac{{15}}{2} \times \dfrac{{18}}{5}$
Simplifying we get,
$ \Rightarrow ab \times ab \times ab = 3 \times 9$
$ \Rightarrow {(ab)^3} = 27$
Taking cube root we get,
$ \Rightarrow ab = \sqrt[3]{{27}} = 3$
Therefore the answer is option C.

Note: We have to be clear about the definitions of arithmetic and harmonic progressions. We calculate the terms by using the middle term formulas. For every three consecutive terms of an arithmetic sequence, the second term is always the mean of first and third.