Question

The value of $xyz$ is $\frac{15}{2}$ of $\frac{18}{5}$ 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

The correct option is C $3$

Que. value of xyz = 15/2 , a,x,y,z,b in A.P

xyz = 18/5 a,x,y,z,b in H.P

$\to xyz=15/2$

$a,x,y,z,b\to A.P$

$d=\frac{b-a}{n+1}=\frac{b-a}{4}$

$y=\frac{a+b}{2}$ $\to x=a+\frac{b-a}{4}$

$x=\frac{3a+b}{4}$

$z=a+3d=a+\frac{3b-3a}{4}$

$=\frac{a+3b}{4}$

$xyz=\left(\frac{3a+b}{4}\right)\left(\frac{a+b}{2}\right)\left(\frac{a+3b}{4}\right)=\frac{15}{2}...\left(i\right)$

$\to xyz=18/5$

$\frac{1}{4},\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{b}\to H.P.$

$y=\frac{2ab}{a+b}$

$\frac{1}{y}=\frac{a+b}{2ab}=\frac{1/b+1/a}{2}$

$\frac{1}{x}=\frac{3/a+1/b}{4}$ $z=\frac{4ab}{3a+b}$

$x=\frac{4ab}{3b+a}$

$xyz=18/5$

$\left(\frac{4ab}{3b+a}\right)\left(\frac{2ab}{a+b}\right)\left(\frac{4ab}{3a+b}\right)=\frac{18}{5}...\left(ii\right)$

multiplying equation (i) & (ii)

$\left(\frac{3a+b}{4}\right)\left(\frac{a+b}{2}\right)\left(\frac{a+3b}{4}\right)\left(\frac{4ab}{3b+a}\right)\left(\frac{2ab}{a+b}\right)\left(\frac{4ab}{3a+b}\right)=\frac{15}{2}\times \frac{18}{5}$

$\therefore \left(ab\right)\left(ab\right)\left(ab\right)=27$

$\therefore (ab{)}^3=27$

$\therefore (ab{)}^3=3^3$

$ab=3$