# How to solve geometric progression problem

A cake mix is 3 cm deep in a tray when placed in a microwave oven. After 1 min it has risen to 4.5 cm high. After a further minute it rises to 5 cm and after a further min rises to 5 $\frac{1}{6}$ cm. If the cake continues to rise at the same rate, what is the maximum height it can be expected to reach?

rising rates:

$a_1= (4.5 - 3)/1 = 1.5$ cm/min
$a_2= (5-3)/2= 1$ cm/min
$a3= (5 \frac{1}{6} - 3)/3 = 13/18$ cm/min

$S = \frac{a}{1-r}$

How to find r?

• i have calculated the rising rate......with the rising rate of 1.5, 1 and 13/18, i cannot get the ratio r??? so pls help to coach what is wrong with my rising rate calculation?? – sekling Nov 14 '14 at 5:45

Don't look at height after each cycle! Consider growth in height during each cycle

Growth during $1$st cycle $$a_1=4.5-3=1.5=\frac32$$ Growth during $2$nd cycle $$a_2=5-4.5=0.5=\frac12$$ $$a_3=5\dfrac16-5=\frac{1}{6}$$

So $a=\dfrac32$ and $r=\dfrac13$ Hence $$s=\frac{1}{1-r}=\frac{\frac32}{1-\frac13}=\frac{3}{2}\cdot\frac{3}{2}=\frac94$$

Don't forget to add Initial Height of cake!

Unfortunately, your problem is not directly about geometric progressions, but something related.

Common ratios $4.5/3, 5/4.5, 5.166667/5$ are not constant.

You have used a wrong formula... $S = \frac{a}{1-r}$

You did nothing to add to or increase height or to get new sums of

heights in the oven adding fresh cake material into the oven.

That too, you are calculating common differences, that should be of interest in arithmetic progression, not the common ratios!

In a geometric progression $a, ar, a r^2 , a r^3, ....$ common ratio

$\frac {t_{n+1}} {t_{n}} = r$ is constant where $t_n$ is the nth term.

The common ratio $r, r <1$ of geometric progression should be constant.

$\frac {dy} {dt} = a - b\, y$