Properties of a G.P. and selecting terms

Geo­met­ric pro­gres­sions have prop­er­ties that mir­ror those of arith­metic pro­gres­sions, with mul­ti­pli­ca­tion in place of addi­tion. Where an A.P. has "sum of the ends", a G.P. has "prod­uct of the ends"; where an A.P. has "add a con­stant", a G.P. has "mul­ti­ply by a con­stant". This les­son sets out five prop­er­ties and shows how to choose unknown terms of a G.P. so that prod­ucts come out sim­ply.

Prop­erty 1: three num­bers in G.P.#

$$\displaystyle a, b, c \text{ are in G.P.} \iff \frac{b}{a} = \frac{c}{b} \iff b^2 = ac$$

The square of the mid­dle term equals the prod­uct of its neigh­bours. This is the quick­est test for three num­bers.

Exam­ple 1#

Are $4$, $10$, $25$ in G.P.? Find $x$ if $2$, $x$, $18$ are in G.P.

$10^2 = 100$ and $4 \times 25 = 100$, so yes (the com­mon ratio is $\displaystyle \frac{5}{2}$).

$x^2 = 2 \times 18 = 36$, so $x = 6$ or $x = -6$. Both work: $2, 6, 18$ has ratio 3, and $2, -6, 18$ has ratio $-3$.

Prop­erty 2: terms equidis­tant from the two ends#

In a finite G.P., the prod­uct of two terms equidis­tant from the begin­ning and the end is always the same, and it equals the prod­uct of the first and last terms.

Why. The $k$th term from the begin­ning is $ar^{k-1}$, and the $k$th term from the end is $\displaystyle l \cdot \frac{1}{r^{k-1}}$. Their prod­uct is $al$, what­ever $k$ is.

For exam­ple, in $2, 4, 8, 16, 32, 64, 128, 256$: $2 \times 256 = 512$, $4 \times 128 = 512$, $8 \times 64 = 512$, $16 \times 32 = 512$.

Prop­erty 3: mul­ti­ply­ing or divid­ing by a con­stant#

If every term of a G.P. is mul­ti­plied (or divided) by the same non-zero num­ber, the result is again a G.P. with the same com­mon ratio. For exam­ple, mul­ti­ply­ing $1, 3, 9, 27$ by 5 gives $5, 15, 45, 135$, still with $r = 3$.

Prop­erty 4: rec­i­p­ro­cals and pow­ers#

The rec­i­p­ro­cals of the terms of a G.P. form a G.P. with com­mon ratio $\displaystyle \frac{1}{r}$. For exam­ple, $2, 6, 18, 54$ gives $\displaystyle \frac{1}{2}, \frac{1}{6}, \frac{1}{18}, \frac{1}{54}$, with ratio $\displaystyle \frac{1}{3}$.

More gen­er­ally, if every term is raised to the same power $k$, the result is a G.P. with com­mon ratio $r^k$. Squar­ing $1, 2, 4, 8$ gives $1, 4, 16, 64$, with ratio 4.

Prop­erty 5: log­a­rithms turn a G.P. into an A.P.#

If $a_1, a_2, a_3, \ldots$ is a G.P. of pos­i­tive terms with com­mon ratio $r$, then $\log a_1, \log a_2, \log a_3, \ldots$ is an A.P. with com­mon dif­fer­ence $\log r$.

Why. $\displaystyle \log a_{n+1} - \log a_n = \log \frac{a_{n+1}}{a_n} = \log r$, the same for every $n$.

For exam­ple, for $2, 6, 18, 54$ the log­a­rithms are $\log 2$, $\log 2 + \log 3$, $\log 2 + 2\log 3$, $\log 2 + 3\log 3$: an A.P. with com­mon dif­fer­ence $\log 3$. This link between the two kinds of pro­gres­sion is one rea­son log­a­rithms are so use­ful.

Select­ing terms in a G.P.#

Choose the unknown terms so that the $r$'s can­cel when you mul­ti­ply:

With three or five terms the prod­uct is $a^3$ or $a^5$; with four terms it is $a^4$. So the prod­uct gives $a$ at once.

Exam­ple 2#

Three num­bers in G.P. have prod­uct 216 and sum 21. Find them.

Let the num­bers be $\displaystyle \frac{a}{r}$, $a$, $ar$. Their prod­uct is $a^3 = 216$, so $a = 6$. Their sum gives

$$\displaystyle \frac{6}{r} + 6 + 6r = 21 \implies \frac{6}{r} + 6r = 15 \implies 6r^2 - 15r + 6 = 0 \implies 2r^2 - 5r + 2 = 0$$

So $(2r - 1)(r - 2) = 0$, giv­ing $r = 2$ or $\displaystyle r = \frac{1}{2}$. Either way the num­bers are 3, 6 and 12. Check: $3 + 6 + 12 = 21$ and $3 \times 6 \times 12 = 216$.

Exam­ple 3#

Three num­bers in G.P. have prod­uct 64 and the sum of their prod­ucts taken two at a time is 56. Find them.

Let the num­bers be $\displaystyle \frac{a}{r}, a, ar$. Then $a^3 = 64$, so $a = 4$. The prod­ucts two at a time are $\displaystyle \frac{a^2}{r}$, $a^2 r$ and $a^2$, so

$$\displaystyle \frac{16}{r} + 16r + 16 = 56 \implies \frac{1}{r} + r = \frac{5}{2} \implies 2r^2 - 5r + 2 = 0$$

So $r = 2$ or $\displaystyle \frac{1}{2}$, and the num­bers are $2, 4, 8$. Check: $2 \times 4 + 4 \times 8 + 2 \times 8 = 8 + 32 + 16 = 56$.

Exam­ple 4#

Four num­bers in G.P. have prod­uct 4096, and the mid­dle two add up to 20. Find them.

Let the num­bers be $\displaystyle \frac{a}{r^3}, \frac{a}{r}, ar, ar^3$. Their prod­uct is $a^4 = 4096$, so $a^2 = 64$; take $a = 8$. (Tak­ing $a = -8$ leads to the same four num­bers.) The mid­dle two give

$$\displaystyle \frac{8}{r} + 8r = 20 \implies 8r^2 - 20r + 8 = 0 \implies 2r^2 - 5r + 2 = 0$$

So $r = 2$ or $\displaystyle r = \frac{1}{2}$. With $r = 2$ the num­bers are $\displaystyle \frac{8}{8}, \frac{8}{2}, 16, 64$, that is $1, 4, 16, 64$, a G.P. with ratio 4. ($\displaystyle r = \frac{1}{2}$ gives the same num­bers in reverse order.) Check: $1 \times 4 \times 16 \times 64 = 4096$ and $4 + 16 = 20$. Notice also Prop­erty 2: $1 \times 64 = 4 \times 16$.

Exam­ple 5#

If $a$, $b$, $c$ are in G.P., show that $a^2 + b^2$, $ab + bc$, $b^2 + c^2$ are also in G.P.

By Prop­erty 1 we must show $(ab + bc)^2 = (a^2 + b^2)(b^2 + c^2)$. Using $b^2 = ac$:

$$(ab + bc)^2 = b^2(a + c)^2 = ac(a + c)^2$$

$$(a^2 + b^2)(b^2 + c^2) = (a^2 + ac)(ac + c^2) = a(a + c) \cdot c(a + c) = ac(a + c)^2$$

The two sides are equal, so the three expres­sions are in G.P.

Prac­tice#

1. Are $3, 12, 48$ in G.P.? Are $2, 5, 12$?
2. Find $x$ if $4$, $x$, $9$ are in G.P. and $x$ is pos­i­tive.
3. In the G.P. $3, 6, 12, 24, 48, 96$, check Prop­erty 2 for all three pairs.
4. Each term of $5, 10, 20, 40$ is divided by 5. Write the new sequence and its ratio.
5. Write the rec­i­p­ro­cals of $3, 9, 27, 81$. What is their com­mon ratio?
6. The terms $1, 3, 9, 27$ are squared. What is the new com­mon ratio?
7. Three num­bers in G.P. have prod­uct 27 and sum 13. Find them.
8. Three num­bers in G.P. have prod­uct 1000 and sum 35. Find them.
9. Five num­bers in G.P. have prod­uct 32. Find the mid­dle num­ber.
10. If $p$, $q$, $r$ are in G.P., show that $\log p$, $\log q$, $\log r$ are in A.P.

Answers#