The given terms are:
\(nC_1 x^{n-1} y = 135\) (i)
\(nC_2 x^{n-2} y^2 = 30\) (ii)
\(nC_3 x^{n-3} y^3 = \frac{10}{3}\) (iii)
Step 1: Using (i) and (ii)
\(\frac{nC_1 x}{nC_2 y} = \frac{9}{2}\) (iv)
Step 2: Using (ii) and (iii)
\(\frac{nC_2 x}{nC_3 y} = 9\) (v)
From (iv) and (v), solve for \(nC_2\):
\[ \frac{nC_1 \cdot nC_3}{(nC_2)^2} = \frac{1}{2} \]
Substitute \(nC_1 = n\), \(nC_2 = \frac{n(n-1)}{2}\), \(nC_3 = \frac{n(n-1)(n-2)}{6}\):
\[ \frac{n \cdot \frac{n(n-1)(n-2)}{6}}{\left(\frac{n(n-1)}{2}\right)^2} = \frac{1}{2} \]
\[ \frac{2n^2(n-2)}{6n(n-1)} = \frac{1}{2} \]
\[ 2n(n-2) = 3(n-1) \]
\[ 2n^2 - 7n + 3 = 0 \]
\[ n = 5 \quad \text{(as \(n\) is a positive integer)} \]
Step 3: Solving for \(x\) and \(y\)
From (v):
\[ \frac{x}{y} = 9 \implies x = 9y \]
Substitute in (i):
\[ 5C_1 (9y)^4 y = 135 \]
\[ 5C_1 \cdot 94 y^5 = 135 \]
\[ 5 \cdot 81 \cdot 9 \cdot y^5 = 135 \]
\[ y = \frac{1}{3}, \quad x = 3 \]
Step 4: Calculating the final expression
\[ 6(n^3 + x^2 + y) \]
\[ = 6(5^3 + 3^2 + \frac{1}{3}) \]
\[ = 6(125 + 9 + \frac{1}{3}) \]
\[ = 6(134 + \frac{1}{3}) \]
\[ = 806 \]
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: