Applications of Geometric Progressions

When a quantity keeps changing in size across fixed length time intervals, so that at the beginning of each new interval the new size is some constant factor multiplied by the previous size, then we say that the change in quantity is geometric, and the sequence of sizes forms a geometric progression with the common ratio equal to that constant factor. 

Maybe this will be more clear in an example. Suppose I spend $10%$10% of my $\$500$$500 birthday money every week. At the beginning of the second week, I'll have spent $\$50$$50, and will only have $\$450$$450 or $90%$90% of the birthday money left. At the beginning of the third week I'll spend slightly less - $\$45$$45 or $10%$10% so at the beginning of the third week $\$405$$405 will remain. The sequence of birthday dollars that remain at the beginning of the first five weeks is given as $500,450,405,364.5,328.05$500,450,405,364.5,328.05 and we can see here that this sequence is geometric with the first term $a=500$a=500 and the common ratio $r=0.9$r=0.9.  

Now suppose that instead of spending $10%$10% of my birthday money, I decide on a plan to double my money using the following strategy. Firstly, I'll add $10%$10% to the birthday money, so that at the beginning of the second week, the additional $\$50$$50 will take the total to $\$550$$550. At the beginning of week $3$3 I'll add a further $10%$10% of this accrued $\$550$$550, so that the new total becomes $\$605$$605. I'll keep adding $10%$10% of what ever is there at the beginning of each week until I exceed $\$1000$$1000.

The geometric progression for this plan becomes $500,550,605,665.5,732.05,...$500,550,605,665.5,732.05,...

This progression has $a=500$a=500 and $r=\frac{550}{500}=1.1$r=550500​=1.1, so that the $n$nth term of the progression is given by $t_n=500\left(1.1\right)^{n-1}$tn​=500(1.1)n−1.

By setting $500\left(1.1\right)^{n-1}=1000$500(1.1)n−1=1000 and dividing by $500$500, I know that I have to find $n$n so that:

$\left(1.1\right)^{n-1}>2$(1.1)n−1>2

I could guess and check for $n$n and would eventually come up with $n=8$n=8. But an algebraic way to proceed is to take logarithms so that:

$\left(n-1\right)\log_{10}\left(1.1\right)>\log_{10}\left(2\right)$(n−1)log10​(1.1)>log10​(2)

By dividing by $\log_{10}\left(1.1\right)$log10​(1.1) and adding one to both sides, we have:

$n-1>\frac{\log_{10}\left(2\right)}{\log_{10}\left(1.1\right)}$n−1>log10​(2)log10​(1.1)​

$\therefore n>\frac{\log_{10}\left(2\right)}{\log_{10}\left(1.1\right)}+1$∴n>log10​(2)log10​(1.1)​+1

Using a scientific calculator I see that $n$n must be greater than $8.27..$8.27.., so after $9$9 weeks my birthday money will exceed $\$1000$$1000.

As a final example, suppose I start with my $\$500$$500, and add to it $90%$90% of $\$500$$500 in the second week. That's an addition of $\$450$$450 and the total becomes $\$950$$950. Then I add $90%$90% of the $\$450$$450 or $\$405$$405 to the new total to bring the savings to $\$1355$$1355. Again I add $90%$90% of $\$405$$405 or $\$364.50$$364.50 to bring the total savings to $\$1719.50$$1719.50. If I keep the pattern of savings going, how much will I have?

The total savings $S$S in dollars is given by the geometric sum:

$S=500+0.9\left(500\right)+0.9\left(0.9\left(500\right)\right)+0.9\left(0.9\left(0.9\left(500\right)\right)\right)+...$S=500+0.9(500)+0.9(0.9(500))+0.9(0.9(0.9(500)))+...

In other words,

$S=500+405+364.50+...$S=500+405+364.50+...

If I continue saving in this manner forever, then, because $S$S is a geometric sum with a ratio between $0$0 and $1$1, then we know it has a limiting sum. No matter how many weeks I continue the savings pattern, I will never exceed $\$5000$$5000, as shown by the following calculation. 

$S_{\infty}=\frac{a}{1-r}=\frac{500}{1-0.9}=5000$S∞​=a1−r​=5001−0.9​=5000

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3