question_answer

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}}...,{{a}_{r}}\] are in GP, prove that the determinant   is independent of r.

OR

Evaluate

Answer:

Let A = first term of a GP and R = Common ratio of GP. Then, rth term \[({{a}_{r}})=A{{R}^{r-1}}\]

Now,

[ taking \[A{{R}^{r}},\] \[A{{R}^{r+6}}\] and \[A{{R}^{r+10}}\] common from \[{{R}_{1}},\,{{R}_{2}}\] and \[{{R}_{2}}\] respectively]

= 0        [since, \[{{R}_{1}}\] and \[{{R}_{2}}\] are identicals]

Hence, given determinant is independent of r.

Hence proved.

OR

Let

\[\left[ \because \,\,{{\,}^{n}}{{C}_{r}}=\frac{n!}{r!(n-r)!};0\le r\le n \right]\]

Applying \[{{R}_{3}}\to 2{{R}_{3}},\] we get

Applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}},\] we get

\[\Rightarrow \]

Now, expanding along \[{{R}_{1}},\] we get

\[\Delta =\frac{1}{2}\times 1[2(8n\,+12)-4(4n\,+12)]\]

\[=\frac{1}{2}[16n\,+24-16n\,-8]=\frac{1}{2}\times 16=8\]