   SEARCH HOME Math Central Quandaries & Queries Question from Nathi: Hi I am really struggling with this question please help !!!! a pohutukawa tree is 86 centimetres when it is planted. in the first year after it is planted , the tree grows 42 centimetres in height.Each year the tree grows in height by 95% of the growth of the previous year. assume that the growth in height of the pohutukawa tree can be modelled by a geometric sequence. A)find the height of the tree 5 years after it is planted and figure out the maximum height the pohutukawa tree is expected to reach in centimetres. The maximum height part is not answered. Hi,

In the first year the growth was $42$ centimeters, the growth in the second year was 95% of that or $0.95 \times 42,$ the growth in the third year was thus $0.95 \times 0.95 \times 42 = 0.95^2 \times 42$ and so on. Hence the growth of the tree can be modelled by the geometric sequence

$a, a r, a r^2, a r^3, \cdot \cdot \cdot , a r^{n-1}, \cdot \cdot \cdot$

where $a = 42$ centimeters and $r = 0.95.$

Thus the height after 5 years is

$86 + a + a r + a r^2 + a r^3 + a r^4 \mbox{ centimeters.}$

To solve the maximum height question I would let $S_n$ be the GROWTH after $n$ years and hence

$S_n = a + a r + a r^2 + a r^3 + \cdot \cdot \cdot + a r^{n-1}.$

Multiply both sides of this equation by $r$ to get

$r S_n = a r + a r^2 + a r^3 + \cdot \cdot \cdot + a r^{n-1} + a r^{n}.$

Subtracting the second equation from the first yields

$(1 - r) S_n = a - a r^{n}$

or

$S_n = \frac{a \left( 1 - r^n \right)}{1-r}.$

You have $r = 0.95$ which is positive and less than 1 so $r^n$ is positive for any positive integer $n,$ and the larger $n$ becomes the closer $r^n$ is to zero. Hence as $n$ gets large, $S_n$ gets closer and closer to $\large \frac{a}{1-r}$ but it never quite gets there.

I hope this helps,
Penny       * Registered trade mark of Imperial Oil Limited. Used under license. Math Central is supported by the University of Regina and the Imperial Oil Foundation.