Question

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

Answer 1

Here in the given question we have two different cases, one is the given series $a,x,y,z,b$ are in arithmetic progression and the other is they are harmonic progression. In the first case $xyz=\dfrac{15}{2}$ and in the second case $xyz=\dfrac{18}{5}$ . We have been asked to find the value of $ab$ . We know that in the arithmetic progression all the successive terms have a common difference and in harmonic progression all the reciprocals lie in arithmetic progression.

Complete step-by-step solution:
Now considering from the question we have two different cases, one is the given series $a,x,y,z,b$ are in arithmetic progression and the other is they are harmonic progression. In the first case $xyz=\dfrac{15}{2}$ and in the second case $xyz=\dfrac{18}{5}$ . We have been asked to find the value of $ab$ .
From the basic concepts of progression we know that in the arithmetic progression all the successive terms have a common difference and in harmonic progression all the reciprocals lies in arithmetic progression.
If we consider $a$ as the first term and $d$ as the common difference in arithmetic progression then the ${{n}^{th}}$ term is mathematically given as ${{a}_{n}}=a+\left( n-1 \right)d$ . Similarly If we consider $\dfrac{1}{a}$ as the first term and $d$ as the common difference in harmonic progression then the ${{n}^{th}}$ term is mathematically given as $\dfrac{1}{{{a}_{n}}}=\dfrac{1}{a+\left( n-1 \right)d}$ .
Now by applying these concepts in the first case the value of each term will be given as $x=a+d,y=a+2d,z=a+3d,b=a+4d$. As we need to find the value of $ab$ let us express all terms in terms of $a,b$ . From $b=a+4d$ we will have $d=\dfrac{b-a}{4}$ . By substituting this value in other terms we will have $x=\dfrac{3a+b}{4},y=\dfrac{a+b}{2},z=\dfrac{a+3b}{2}$ . By multiplying all these terms we will have $\left( \dfrac{3a+b}{4} \right)\left( \dfrac{a+b}{2} \right)\left( \dfrac{a+3b}{4} \right)=\dfrac{15}{2}$ …….(1)
Now by applying these concepts in the second case we can say that $d=\dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{4}$ then we will have
$\begin{aligned} & \dfrac{1}{x}=\dfrac{1}{a}+\dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{4} \\\ & \Rightarrow \dfrac{1}{x}=\dfrac{4+a\left( \dfrac{a-b}{ab} \right)}{4a} \\\ & \Rightarrow \dfrac{1}{x}=\dfrac{4b+a-b}{4ab} \\\ & \Rightarrow x=\dfrac{4ab}{3b+a} \\\ \end{aligned}$
Similarly $y=\dfrac{2ab}{a+b},z=\dfrac{4ab}{3a+b}$ . By multiplying these terms we will have $\left( \dfrac{4ab}{3b+a} \right)\left( \dfrac{2ab}{a+b} \right)\left( \dfrac{4ab}{3a+b} \right)=\dfrac{18}{5}$…….(2)
Now if we multiply both the equations then we will have $\begin{aligned} & \Rightarrow \left( \dfrac{3a+b}{4} \right)\left( \dfrac{a+b}{2} \right)\left( \dfrac{a+3b}{4} \right)\left( \dfrac{4ab}{3b+a} \right)\left( \dfrac{2ab}{a+b} \right)\left( \dfrac{4ab}{3a+b} \right)=\dfrac{18}{5}\times \dfrac{15}{2} \\\ & \Rightarrow {{a}^{3}}{{b}^{3}}=27 \\\ & \Rightarrow ab=3 \\\ \end{aligned}$
Therefore we can conclude that the value of $ab$ is 3.

Note: While answering questions we should be sure with our concepts that we apply and the calculations that we perform. To solve these questions easily the key is practice more. Someone can get confused and in the harmonic progression take the common difference wrongly as $\dfrac{1}{d}=\dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{4}$ then we will get confused and our solution will get complex and end up having a mess and will be unable to conclude the answer.