Question

If $x,y,z$ are $T_l,T_{2m},T_{3n}$ terms of an H.P. respectively then $\Delta =\left|\begin{array}{ccc}yz& xz& yz\\ l& 2m& 3n\\ 1& 1& 1\end{array}\right|$ is independent of

• A. $m$
• B. $n$
• C. $y$
• D. $x$

Answer 1

Answer: (A), (B), (C), (D)

$x$

Since $x,y,z$ are $l^{th},(2m{)}^{th},(3n{)}^{th}$ terms of an H.P. respectively.

$\therefore \frac{1}{x}=a+(l-1)d$

$\frac{1}{y}=a+(2m-1)d$

$\frac{1}{z}=a+(3n-1)d$

$\therefore \left|\begin{array}{ccc}yz& zx& xy\\ l& 2m& 3n\\ 1& 1& 1\end{array}\right|=xyz\left|\begin{array}{ccc}\frac{1}{x}& \frac{1}{y}& \frac{1}{z}\\ l& 2m& 3n\\ 1& 1& 1\end{array}\right|$

$=xyz\left|\begin{array}{ccc}a+(l-1)d& a+(2m-1)d& a+(3n-1)d\\ l& 2m& 3n\\ 1& 1& 1\end{array}\right|$

Operating
$R_1\to R_1-d{R}_2-(a-d)R_3=0$

$=xyz\left|\begin{array}{ccc}0& 0& 0\\ l& 2m& 3n\\ 1& 1& 1\end{array}\right|$

$\therefore$ Given determinant is independent of $x,y,m,n.$