- Home
- Standard 12
- Mathematics
Let $x, y, z > 0$ are respectively $2^{nd}, 3^{rd}, 4^{th}$ term of $G.P.$ and $\Delta = \left| {\begin{array}{*{20}{c}}
{{X^k}}&{{X^{k + 1}}}&{{X^{k + 2}}}\\
{{Y^k}}&{{Y^{k + 1}}}&{{Y^{k + 2}}}\\
{{Z^k}}&{{Z^{k + 1}}}&{{Z^{k + 2}}}
\end{array}} \right| = {\left( {r - 1} \right)^2}\left( {1 - \frac{1}{{{r^2}}}} \right)$ , (where $r$ is common ratio), then $k=$ .......
$-1$
$ 1$
$0$
None of these
Solution
$\Delta = {{\rm{x}}^k}{{\rm{y}}^k}{{\rm{z}}^{\rm{k}}}\left| {\begin{array}{*{20}{c}}
1&{{\rm{ar}}}&{{{\rm{a}}^2}{{\rm{r}}^2}}\\
1&{{\rm{a}}{{\rm{r}}^2}}&{{{\rm{a}}^2}{{\rm{r}}^4}}\\
1&{{\rm{a}}{{\rm{r}}^3}}&{{{\rm{a}}^2}{{\rm{r}}^4}}
\end{array}} \right|$
$ = {{\rm{a}}^{3{\rm{k}}}} \cdot {{\rm{r}}^{6k}} \cdot {{\rm{a}}^3}{{\rm{r}}^3}\left| {\begin{array}{*{20}{c}}
1&1&1\\
1&{\rm{r}}&{{{\rm{r}}^2}}\\
1&{{{\rm{r}}^2}}&{{{\rm{r}}^4}}
\end{array}} \right|$
$=a^{3(k+1)} \cdot r^{6 k+3}(1-r)\left(r-r^{2}\right)\left(r^{2}-1\right)$
clearly, $\mathrm{k}=-1$
$\therefore \Delta=r^{-2}(1-r)^{2}\left(r^{2}-1\right)=(r-1)^{2}\left(1-\frac{1}{r^{2}}\right)$
