Question

Let ' $a$ ' and ' $b$ ' be positive integers. If the value of $x y z$ 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

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