Question

Let ' $a$ ' and ' $b$ ' be positive integers. If the value of $xyz$ is $\frac{15}{2}$ or $\frac{18}{5}$ according as the series $a,x,y,z,b$ is arithmetic progression or harmonic progression, then compute the value of $\left(a^2+b^2\right)$.

Answer 1

Answer: 10

If $a,x,y,z,b$ are in A.P., then $b=5^{\text{th }}$ term $=a+4d$
( $d=$ common difference)
$\Rightarrow d=\frac{b-a}{4}$
$\therefore xyz=(a+d)(a+2d)(a+3d)=\frac{15}{2}$
$\Rightarrow \left(\frac{b+3a}{4}\right)\left(\frac{2a+2b}{4}\right)\left(\frac{a+3b}{4}\right)=\frac{15}{2}$
$\Rightarrow (b+3a)(a+b)(a+3b)=(55)(16)$....(1)
Again, if a, $x,y,z,b$ are in H.P., then common difference of the corresponding A.P. is $\frac{1}{4}\left(\frac{1}{b}-\frac{1}{a}\right)$
i.e., $\frac{(a-b)}{4ab}$
$\therefore \frac{1}{x}=\frac{1}{a}+\frac{(a-b)}{4ab}\Rightarrow x=\frac{4ab}{a+3b}$
$\frac{1}{y}=\frac{1}{a}+\frac{2(a-b)}{4ab}\Rightarrow y=\frac{4ab}{2a+2b}=\frac{2ab}{a+b}$
$\text{ and }\frac{1}{z}=\frac{1}{a}+\frac{3(a-b)}{4ab}\Rightarrow z=\frac{4ab}{3a+b}$
$\text{ so, }xyz=\left(\frac{4ab}{a+3b}\right)\left(\frac{2ab}{a+b}\right)\left(\frac{4ab}{3a+b}\right)=\frac{18}{5}\text{ (given) }$
$\Rightarrow \frac{32a^3{b}^3}{55\times 16}=\frac{18}{5}\Rightarrow a^3{b}^3=27\Rightarrow ab=3\Rightarrow a=3\text{ or }b=1$
$\therefore \left(a^2+b^2\right)=10$