The population of cattle in a farm increases | vedclass

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(B) Let $P_n$ be the population in year $n$. According to the problem,$P_{n+2} - P_n = k P_{n+1}$,where $k$ is a constant.Given populations:$P_{2010}

Vedclass · Accepted Answer

$84$

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(B) Let $P_n$ be the population in year $n$. According to the problem,$P_{n+2} - P_n = k P_{n+1}$,where $k$ is a constant.Given populations:$P_{2010} = 39$$P_{2011} = 60$$P_{2013} = 123$Let $P_{2012} = x$.For $n = 2010$:$P_{2012} - P_{2010} = k P_{2011}$ $\Rightarrow x - 39 = k(60)$ $\Rightarrow k = \frac{x - 39}{60} \quad (i)$For $n = 2011$:$P_{2013} - P_{2011} = k P_{2012}$ $\Rightarrow 123 - 60 = k(x)$ $\Rightarrow 63 = kx$ $\Rightarrow k = \frac{63}{x} \quad (ii)$Equating $(i)$ and $(ii)$:$\frac{x - 39}{60} = \frac{63}{x}$$x(x - 39) = 60 	imes 63$$x^2 - 39x - 3780 = 0$Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{39 \pm \sqrt{(-39)^2 - 4(1)(-3780)}}{2}$$x = \frac{39 \pm \sqrt{1521 + 15120}}{2}$$x = \frac{39 \pm \sqrt{16641}}{2}$$x = \frac{39 \pm 129}{2}$Since population must be positive,$x = \frac{39 + 129}{2} = \frac{168}{2} = 84$.Thus,the population in $2012$ was $84$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

The population of cattle in a farm increases such that the difference between the population in year $n+2$ and that in year $n$ is proportional to the population in year $n+1$. If the populations in years $2010, 2011$ and $2013$ were $39, 60$ and $123$ respectively,then the population in $2012$ was:

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