Question

The determinant \(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0,\) if
1. x, y, z are in A.P.
2. x, y, z are in G.P.
3. x, y, z are in H.P.
4. xy, yz, zx are in A.P.

Answer 1

Correct Answer - Option 2 : x, y, z are in G.P.

\(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0\)

xp + y (y(yp + z) + z(xp + y)) - (yp + z) (x(yp + z) - y(xp + y)) + 0(xz - y²) = 0

xp + y (y²p + yz - xzp - yz) - (yp + z) (xyp + xz - xyp - y²) = 0

(xp + y) (y² p - xzp) - (yp + z) (xz - y²) = 0

(xp + y) p(y² - xz) + (yp + z) (y² - xz) = 0

(y² - xz) ((xp + y)p + (yp + z)) = 0

(y² - xz) (xp² + yp + yp + z) = 0

(y² = xz) (xp2 + 2yp + z) = 0

The determinant will be 0 if (y² - xz) will be 0 which is geometric mean.

Hence x, y and z are in G.P.