Problems on geometric progressions

Geo­met­ric pro­gres­sions describe any­thing that is mul­ti­plied by the same fac­tor at each step: bac­te­ria dou­bling, money earn­ing com­pound inter­est, a machine los­ing a fixed per­cent­age of its value, a ball los­ing height at each bounce. This les­son works through prob­lems of these kinds, and some that need a clever rearrange­ment before the G.P. for­mu­las can be used.

The for­mu­las#

$$\displaystyle t_n = ar^{n-1} \qquad S_n = \frac{a(r^n - 1)}{r - 1} \quad (r \neq 1) \qquad S_\infty = \frac{a}{1 - r} \quad (-1 < r < 1)$$

For growth by $p\%$ per step, $\displaystyle r = 1 + \frac{p}{100}$; for decay by $p\%$, $\displaystyle r = 1 - \frac{p}{100}$.

Exam­ple 1#

A cul­ture starts with 500 bac­te­ria and the num­ber dou­bles every hour. How many bac­te­ria are there after 8 hours?

After 8 hours the cul­ture has dou­bled 8 times: $500 \times 2^8 = 500 \times 256 = 128000$ bac­te­ria.

Notice the count: the start­ing amount is the 1st term, so "after 8 hours" is the 9th term, $ar^8$.

Exam­ple 2#

Meena puts ₹1 in a box on the first day, ₹2 on the sec­ond day, ₹4 on the third, dou­bling each day. How much is in the box after 10 days? On which day does she first put in more than ₹1000?

The total after 10 days is $\displaystyle S_{10} = \frac{1(2^{10} - 1)}{2 - 1} = 1023$, so ₹1023.

On day $n$ she puts in $2^{n-1}$ rupees. Since $2^9 = 512$ and $2^{10} = 1024$, the first amount over ₹1000 is on day 11.

Exam­ple 3#

₹10000 is invested at 8% per year com­pound inter­est. Find the amount after 3 years.

Each year the amount is mul­ti­plied by 1.08:

$$10000 \times 1.08^3 = 10000 \times 1.259712 = 12597.12$$

So the amount is ₹12597.12.

Exam­ple 4#

Find the sum of $n$ terms of $7 + 77 + 777 + \cdots$

This is not a G.P. as it stands, but it can be turned into one. Take out 7 and write each term with nines:

$$\displaystyle 7 + 77 + 777 + \cdots = \frac{7}{9}(9 + 99 + 999 + \cdots) = \frac{7}{9}\left[(10 - 1) + (100 - 1) + (1000 - 1) + \cdots\right]$$

There are $n$ terms, so $n$ ones are sub­tracted:

$$\displaystyle = \frac{7}{9}\left[(10 + 10^2 + \cdots + 10^n) - n\right] = \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$$

Check with $n = 3$: $\displaystyle \frac{7}{9}\left[\frac{10 \times 999}{9} - 3\right] = \frac{7}{9}(1110 - 3) = \frac{7}{9} \times 1107 = 861$, and $7 + 77 + 777 = 861$.

Exam­ple 5#

Write $0.1\overline{5} = 0.1555\ldots$ as a frac­tion.

Sep­a­rate the part that does not repeat:

$$\displaystyle 0.1555\ldots = \frac{1}{10} + \left(\frac{5}{100} + \frac{5}{1000} + \cdots\right) = \frac{1}{10} + \frac{5/100}{1 - 1/10} = \frac{1}{10} + \frac{5}{90} = \frac{9 + 5}{90} = \frac{14}{90} = \frac{7}{45}$$

Exam­ple 6#

The sum to infin­ity of a G.P. is 15, and the sum to infin­ity of the squares of its terms is 45. Find the G.P.

$$\displaystyle \frac{a}{1 - r} = 15, \qquad \frac{a^2}{1 - r^2} = 45$$

The squares form a G.P. with first term $a^2$ and ratio $r^2$. Divide the sec­ond equa­tion by the first, using $1 - r^2 = (1 - r)(1 + r)$:

$$\displaystyle \frac{a}{1 + r} = 3$$

So $a = 15(1 - r)$ and $a = 3(1 + r)$. Then $15 - 15r = 3 + 3r$, so $18r = 12$ and $\displaystyle r = \frac{2}{3}$, giv­ing $a = 5$. The G.P. is $\displaystyle 5, \frac{10}{3}, \frac{20}{9}, \ldots$

Exam­ple 7#

The sum of the first three terms of a G.P. is 26 and the sum of the next three terms is 702. Find the G.P.

$$a + ar + ar^2 = 26, \qquad ar^3 + ar^4 + ar^5 = 702$$

The sec­ond sum is $r^3(a + ar + ar^2) = 26r^3$. So $26r^3 = 702$, $r^3 = 27$ and $r = 3$. Then $a(1 + 3 + 9) = 26$ gives $a = 2$. The G.P. is $2, 6, 18, 54, \ldots$

Exam­ple 8#

A ball is dropped from 20 m. After each bounce it rises to $\displaystyle \frac{3}{5}$ of the height it fell from. Find the total dis­tance it trav­els.

The first fall is 20 m. The rises are $\displaystyle 12, \frac{36}{5}, \ldots$, a G.P. with sum $\displaystyle \frac{12}{1 - 3/5} = 30$ m, and the later falls add up to the same. Total dis­tance $= 20 + 2 \times 30 = 80$ m.

Prac­tice#

1. A colony of 800 insects grows by 50% every week. How many insects are there after 4 weeks?
2. A car bought for ₹600000 loses 10% of its value each year. What is it worth after 2 years?
3. A chain let­ter is sent to 3 peo­ple, and each per­son who receives it sends it to 3 new peo­ple. If this hap­pens 6 times, how many let­ters are sent in all?
4. ₹5000 is invested at 10% per year com­pound inter­est. Find the amount after 3 years.
5. Find the sum of $n$ terms of $3 + 33 + 333 + \cdots$, and check your for­mula with $n = 2$.
6. Write $0.2\overline{3}$ as a frac­tion.
7. The sum of the first three terms of a G.P. is 14 and the sum of the next three is 112. Find the G.P.
8. The sum to infin­ity of a G.P. is 6 and its first term is 3. Find the sum of its first 4 terms.
9. The first term of a G.P. is 2 and the sum to infin­ity of the squares of its terms is $\displaystyle \frac{16}{3}$. Find its com­mon ratio, given that it is pos­i­tive.
10. The 4th, 7th and 10th terms of a G.P. are $p$, $q$ and $s$. Show that $q^2 = ps$.

Answers#