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 \[\Delta =\frac{1}{2}\times 1[2(8n\,+12)-4(4n\,+12)]\] \[=\frac{1}{2}[16n\,+24-16n\,-8]=\frac{1}{2}\times 16=8\]
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