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The population of cattle in a farm increases so 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
$81$
$84$
$87$
$90$
Solution
(b)
Given,
Population in year $2010, 2011$ and $2013$ were $39,60$ and $123$ respectively.
According to problems,
The population of cattle in farm increases such that difference between in year $n+2$ and that in year $n$ is proportional to the year $n+1$
$\because \quad(n+2)-(n)=k(n+1)$
$\because$ Let population in year $2012=x$
| $Year$ | $Population$ |
| $2010$ | $39$ |
| $2011$ | $60$ |
| $2012$ | $x$ |
| $2013$ | $123$ |
From Eqs.$(i)$ and $(ii)$, we get
$\frac{x-39}{60}=\frac{123-60}{x}$
$\Rightarrow x^2-39 x-3780=0$
$\Rightarrow (x-84)(x+40)=0$
$\therefore x=84$
