Question

If x, y and z are respectively 1th, 2mth ,3nth terms of a H.P., then \[\Delta = \left| {\begin{array}{*{20}{c}}
  {yz}&{zx}&{xy} \\
  l&{2m}&{3n} \\
  1&1&1
\end{array}} \right|\] is independent of
A.x and y
B.m and n
C.x, y, z, m, n
D.x, y, z

Answer 1

Hint: First, simplify the determinant \[\Delta \] and use the above given information to solve the determinant and find the value of \[\Delta \] . Check if any of the variables x, y, z, m or n are present in the value of \[\Delta \] . If any, choose the correct option which does not include that variable.

Complete step-by-step answer:
Here, it is given that \[\Delta = \left| {\begin{array}{*{20}{c}}
  {yz}&{zx}&{xy} \\
  l&{2m}&{3n} \\
  1&1&1
\end{array}} \right|\]
First, we need to simplify the determinant and then solve further
 \[\Delta = \left| {\begin{array}{*{20}{c}}
  {yz}&{zx}&{xy} \\
  l&{2m}&{3n} \\
  1&1&1
\end{array}} \right|\]
Taking \[xyz\] common from \[{R_1}\]
 \[\therefore \Delta = xyz\left| {\begin{array}{*{20}{c}}
  {\dfrac{1}{x}}&{\dfrac{1}{y}}&{\dfrac{1}{z}} \\
  l&{2m}&{3n} \\
  1&1&1
\end{array}} \right|\]
Now, it is also given that x, y and z are respectively 1^th, 2m^th and 3n^th terms of a H.P.
So, using the formula \[{T_n} = a + \left( {n - 1} \right)d\] , we will find x, y and z.
 \[\dfrac{1}{x} = a + \left( {l - 1} \right)d\]
 \[\dfrac{1}{y} = a + \left( {2m - 1} \right)d\]
 \[\dfrac{1}{z} = a + \left( {3n - 1} \right)d\]
Now, substituting the values of \[\dfrac{1}{x}\] , \[\dfrac{1}{y}\] and \[\dfrac{1}{z}\] in \[\Delta \]
 \[\therefore \Delta = xyz\left| {\begin{array}{*{20}{c}}
  {a + \left( {l - 1} \right)d}&{a + \left( {2m - 1} \right)d}&{a + \left( {3n - 1} \right)d} \\
  l&{2m}&{3n} \\
  1&1&1
\end{array}} \right|\]
Now, applying \[{R_1} = {R_1} + a{R_3}\]
 \[\therefore \Delta = xyz\left| {\begin{array}{*{20}{c}}
  {\left( {l - 1} \right)d}&{\left( {2m - 1} \right)d}&{\left( {3n - 1} \right)d} \\
  l&{2m}&{3n} \\
  1&1&1
\end{array}} \right|\]
Now, applying \[{R_2} = {R_2} - {R_3}\]
 \[\therefore \Delta = xyz\left| {\begin{array}{*{20}{c}}
  {\left( {l - 1} \right)d}&{\left( {2m - 1} \right)d}&{\left( {3n - 1} \right)d} \\
  {l - 1}&{2m - 1}&{3n - 1} \\
  1&1&1
\end{array}} \right|\]
Taking d common from \[{R_1}\]
 \[\therefore \Delta = xyzd\left| {\begin{array}{*{20}{c}}
  {l - 1}&{2m - 1}&{3n - 1} \\
  {l - 1}&{2m - 1}&{3n - 1} \\
  1&1&1
\end{array}} \right|\]
Now, we can see that, \[{R_1} = {R_2}\] . So, it is clear that the value of determinant becomes 0.
 \[\therefore \Delta = 0\]
The value of \[\Delta \] is 0.
Hence, \[\Delta \] does not depend on the values of x, y, z, m or n.
Hence option C is correct answer
Note: In this question, the value of \[\Delta \] becomes 0, because \[{R_1} = {R_2}\] .
If either of any two rows or any two columns in a determinant become equal, then the value of that determinant becomes 0. In this case the first two rows become equal. So, the value of determinant became 0.
Harmonic Progression (H.P.) is a sequence of real numbers formed by taking the reciprocals of the terms of an Arithmetic Progression (A.P.). Since they are mostly similar, they have the same formulas for nth term, sum of n terms, etc.