Let ^ n C_ r-1 =28, ^ n C_ r =56 and ^ n C_ r+1 =70. Let A(4 t, 4… - Vidyadip

MCQ

Let ${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56$ and ${ }^{n} C_{r+1}=70$. Let $\mathrm{A}(4 \cos t, 4 \sin t), \mathrm{B}(2 \sin t,-2 \cos t)$ and $\mathrm{C}\left(3 \mathrm{r}-\mathrm{n}, \mathrm{r}^{2}-\mathrm{n}-1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3 x-1)^{2}+(3 y)^{2}=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals :

✓
20
B
8
C
6
D
18

✓

Answer

Correct option: A.

20

(A)
Sol. ${ }^{n} C_{r-1}=28,{ }^{n} C_{r}=56$
$\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{28}{56}$
$\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2}$
$\frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2}$
$3 \mathrm{r}=\mathrm{n}+1$
$\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{{ }^{{ }^{n} \mathrm{C}_{\mathrm{r}+1}}}=\frac{56}{70}$
$\frac{(\mathrm{r}+1)}{(\mathrm{n}-\mathrm{r})}=\frac{56}{70} \Rightarrow 9 \mathrm{r}=4 \mathrm{n}-5$
By (i) \& (ii)
( $\mathrm{r}=3$ ), $(\mathrm{n}=8)$
A (4cost, 4sint) B(2sint, $-2 \cos t) \mathrm{C}\left(3 \mathrm{r}-\mathrm{n}, \mathrm{r}^{2}-\mathrm{n}-1\right)$
A (4cost, 4sint) $\quad \mathrm{B}(2 \sin t,-2 \cos t) \mathrm{C}(1,0)$
$(3 x-1)^{2}+(3 y)^{2}=(4 \cos t+2 \sin t)^{2}+(4 \sin t-\cos t)^{2}$
$(3 \mathrm{x}-1)^{2}+(3 \mathrm{y})^{2}=20 \quad \therefore$

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