Geometric progressions and the general term

In an arith­metic pro­gres­sion we add the same num­ber at each step. In a geo­met­ric pro­gres­sion we mul­ti­ply by the same num­ber. Money grow­ing at com­pound inter­est, a pop­u­la­tion grow­ing by a fixed per­cent­age, a ball los­ing a fixed frac­tion of its height at each bounce: all of these fol­low geo­met­ric pro­gres­sions. This les­son defines them and finds a for­mula for any term.

Def­i­n­i­tion#

A sequence is called a geo­met­ric pro­gres­sion (G.P.) if the ratio of any two con­sec­u­tive terms is the same. This fixed ratio is called the com­mon ratio, writ­ten $r$:

$$\displaystyle r = \frac{t_2}{t_1} = \frac{t_3}{t_2} = \frac{t_4}{t_3} = \cdots$$

If the first term is $a$ and the com­mon ratio is $r$, the G.P. is

$$a, \quad ar, \quad ar^2, \quad ar^3, \quad \ldots$$

Nei­ther $a$ nor $r$ may be zero; oth­er­wise the ratios could not be formed.

When $r > 1$ (and $a > 0$) the terms grow; when $0 < r < 1$ they shrink towards 0; when $r$ is neg­a­tive the signs alter­nate. Always find $r$ as a term divided by the term just before it.

The sequence $2, 4, 6, 8$ is not a G.P.: the ratios $\displaystyle \frac{4}{2} = 2$ and $\displaystyle \frac{6}{4} = \frac{3}{2}$ are dif­fer­ent. (It is an A.P.)

The gen­eral term of a G.P.#

Each term is the pre­vi­ous one mul­ti­plied by $r$:

The power of $r$ is one less than the posi­tion. So the gen­eral term of a G.P. is

$$t_n = ar^{n-1}$$

Exam­ple 1#

Find the 10th term of the G.P. $3, 6, 12, \ldots$

$a = 3$, $r = 2$: $t_{10} = 3 \times 2^9 = 3 \times 512 = 1536$.

Exam­ple 2#

Find the 6th term of the G.P. $64, -32, 16, \ldots$

$a = 64$, $\displaystyle r = \frac{-32}{64} = -\frac{1}{2}$: $\displaystyle t_6 = 64 \times \left(-\frac{1}{2}\right)^5 = 64 \times \left(-\frac{1}{32}\right) = -2$.

Exam­ple 3#

Which term of the G.P. $2, 6, 18, \ldots$ is 1458?

$a = 2$, $r = 3$. Set $t_n = 1458$:

$$2 \times 3^{n-1} = 1458 \implies 3^{n-1} = 729 = 3^6 \implies n - 1 = 6 \implies n = 7$$

So 1458 is the 7th term.

Exam­ple 4#

The 3rd term of a G.P. is 12 and the 6th term is 96. Find the G.P.

$$ar^2 = 12, \qquad ar^5 = 96$$

Divide the sec­ond by the first: $r^3 = 8$, so $r = 2$. Then $a \times 4 = 12$, so $a = 3$. The G.P. is $3, 6, 12, 24, \ldots$

Divid­ing one equa­tion by the other is the stan­dard move for a G.P., just as sub­tract­ing is for an A.P.

Exam­ple 5#

Is 500 a term of the G.P. $4, 12, 36, \ldots$?

$4 \times 3^{n-1} = 500$ gives $3^{n-1} = 125$. But 125 is not a power of 3 ($3^4 = 81$ and $3^5 = 243$). So 500 is not a term.

Count­ing from the end#

If a G.P. has last term $l$, read­ing it back­wards gives a G.P. with first term $l$ and com­mon ratio $\displaystyle \frac{1}{r}$. So the $n$th term from the end is

$$\displaystyle l \left(\frac{1}{r}\right)^{n-1}$$

Exam­ple 6#

Find the 5th term from the end of the G.P. $3, 6, 12, \ldots, 3072$.

$l = 3072$ and $r = 2$, so the 5th term from the end is $\displaystyle 3072 \times \left(\frac{1}{2}\right)^4 = \frac{3072}{16} = 192$.

Growth by a fixed per­cent­age#

If a quan­tity grows by a fixed per­cent­age each year, it is mul­ti­plied by the same num­ber each year, so its val­ues form a G.P. An increase of 10% means mul­ti­ply­ing by 1.1; a decrease of 20% means mul­ti­ply­ing by 0.8.

Exam­ple 7#

The pop­u­la­tion of a town is 50000 and it grows by 10% every year. What will it be after 3 years?

After 1 year the pop­u­la­tion is $50000 \times 1.1$, after 2 years $50000 \times 1.1^2$, and after 3 years

$$50000 \times 1.1^3 = 50000 \times 1.331 = 66550$$

Notice that after 3 years the pop­u­la­tion is the 4th term of the G.P. $50000, 55000, 60500, 66550, \ldots$

Exam­ple 8#

If $x$, $x + 3$ and $x + 9$ are con­sec­u­tive terms of a G.P., find $x$.

The ratios must be equal: $\displaystyle \frac{x + 3}{x} = \frac{x + 9}{x + 3}$. Cross-mul­ti­ply­ing,

$$(x + 3)^2 = x(x + 9) \implies x^2 + 6x + 9 = x^2 + 9x \implies 3x = 9 \implies x = 3$$

The terms are $3, 6, 12$, with com­mon ratio 2.

Prac­tice#

1. Which of these are G.P.s? Give $r$ for those that are. (a) $5, 15, 45, 135$ (b) $1, 4, 9, 16$ (c) $48, -24, 12, -6$
2. Find the 8th term of $2, 6, 18, \ldots$
3. Find the 7th term of $729, 243, 81, \ldots$
4. Find the 9th term of $\displaystyle \frac{1}{4}, -\frac{1}{2}, 1, \ldots$
5. Which term of $5, 10, 20, \ldots$ is 640?
6. Is 1000 a term of $2, 8, 32, \ldots$?
7. The 2nd term of a G.P. is 6 and the 5th term is 162. Find the first term and the com­mon ratio.
8. Find the 4th term from the end of $1, 3, 9, \ldots, 2187$.
9. A machine worth ₹125000 loses 20% of its value every year. What is it worth after 3 years?
10. Find $k$ if $k - 1$, $k + 1$ and $k + 5$ are con­sec­u­tive terms of a G.P.

Answers#