Progressions-A.P. and G.P. (Part-6)
© ® Written by Sankar Ghosh
In my previous blog post (Part 5), I discussed the sum of n terms of a Geometric Progression (G.P.) in detail.
Now, in Part 6, we will move further and explore three important topics:
1️⃣ Properties of G.P.
2️⃣ Geometric Mean
3️⃣ Infinite Geometric Series
As always, the post includes both theoretical explanations and problem-solving and concludes with an exercise section to help you strengthen your understanding.
I encourage all readers to attempt the exercises for deeper learning. And of course, feel free to share your doubts in the comment section—I’ll be happy to help you clarify them.
Happy Learning!
Properties of Geometric Progression
Let us discuss some properties of Geometric Progression
(1) Square of any term (except first term and last term) is the product of its two adjacent terms. Suppose `t_1,t_2,t_3,...` etc are in G.P. Then `t_2^2=t_1t_3,t_3^2=t_2t_4,t_4^2=t_3t_5,....` etc.
(2) Square of any term (except first term and last term) is the product of the two terms which are equidistant from both the sides of that term.
Suppose, there are `n` terms in a G.P. and `1<r<n`, then `t_r^2=t_{r-p}\timest_{r+p}`, where `p<=min\left(r-1,n-r\right)`.
(3) If all the terms of a G.P. are multiplied /divided by a non-zero number then the new sequence thus obtained is also a G.P. Suppose, `a_1,a_2,a_3,...,a_n` are in G.P. and for any `\lamda\ne0`, `\lamdaa_1,\lamdaa_2,\lamdaa_3,...,\lamdaa_n` and `\frac{a_1}{\lamda},\frac{a_2}{\lamda},\frac{a_3}{\lamda},...,\frac{a_n}{\lamda}` are also in G.P.
(4) The reciprocals of the terms of a G.P. will also form a G.P. Suppose, `a_1,a_2,a_3,...,a_n` are in G.P. then `\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},...,\frac{1}{a_n}`are also in G.P.
(5) If each term of a G.P. be raised to the same power, then the newly formed sequence will also form a G.P. For any `\lamda\ne0` and `a_1,a_2,a_3,...,a_n` are in G.P. then `a_1^\lamda,a_2^\lamda,a_3^\lamda,...,a_n^\lamda` are also a G.P.
Remark: Readers are advised to verify the above properties.
Geometric Mean(G.M.): If `a,b,c` are in G.P., then `\frac{b}{a}=\frac{c}{b}\impliesb^2=ac`
`\implies` `b=\pmsqrt\{ac}`. Here, `b` is called the geometric mean of `a` and `c`. Thus the geometric mean of two numbers is the square root of the product the numbers . Similarly, the geometric mean of `a_1,a_2,a_3,...,a_n` is `\left(a_1\.a_2\.a_3\. ...\.a_n\right)^{\frac{1}{n}` i.e., `n`th root of the product of the numbers `a_1,a_2,a_3,...,a_n`.
Geometric Means between two numbers: Let `a` and `b` are two given numbers. Suppose `n` numbers of G.M. are inserted between `a` and `b`.
`\therefore` `a,a_1,a_2,a_3,...,a_n,b` are in G.P.
`\therefore` first term `=a`, number of terms `=n+2`, last term `=b`. Let the common ratio be `r`.
`\implies` `t_{n+2}=b\impliesar^{n+1}=b\impliesr=\left(\frac{b}{a}\right)^\frac{1}{n+1}`.
`\therefore` the first mean `\left(a_1\right)=ar=a\left(frac{b}{a}\right)^\frac{1}{n+1}`,
the second mean `\left(a_2\right)=ar^2=a\left(\frac{b}{a}\right)^\frac{2}{n+1}`,
third mean `\left(a_3\right)=ar^3=a\left(\frac{b}{a}\right)^\frac{3}{n+1}`
..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .....
`n`th mean`\left(a_n\right)=ar^n=a\left(\frac{b}{a}\right)^\frac{n}{n+1}`.
Now, the product of these `n` G.M.s is
`a_1\.a_2\.a_3\....\.a_n=a\left(frac{b}{a}\right)^\frac{1}{n+1}\timesa\left(\frac{b}{a}\right)^\frac{2}{n+1}\timesa\left(\frac{b}{a}\right)^\frac{3}{n+1}\times...\timesa\left(\frac{b}{a}\right)^\frac{n}{n+1}`
`=a^n\left(\frac{b}{a}\right)^{\frac{1}{n+1}+\frac{2}{n+1}+\frac{3}{n+1}+...+\frac{n}{n+1}}`
`=a^n\left(\frac{b}{a}\right)^\frac{1+2+3+...+n}{n+1}`
`=a^n\left(\frac{b}{a}\right)^\frac{n\left(n+1\right)}{2\left(n+1\right)` [why?]
`=a^n\left(frac{b}{a}\right)^\frac{n}{2}`
`=\left(ab\right)^\frac{n}{2}=\left(\sqrt{ab}\right)^n`
Thus, the product of `n` geometric mean between two quantities is the `n`th power of the single geometric mean between the two quantities.
Sum of an infinite G.P.
The sum of an infinite G.P. whose first term `a` and common ratio `r\left(-1<r<1\right)` i.e. `r\left(|r|<1\right)` is `S=\frac{a}{1-r}`.
[Proof of the above theorem is not given here. Interested readers are advised to consult any textbook]
Note: If, `r>=1`then the sum of an infinite G.P. tends to infinity.
ILLUSTRATIVE EXAMPLES
Problems on Properties of Geometric Progression.
Example 1: If `a,b,c,d` are in G.P., show that:
(a) `a+b,b+c,c+d` are also in G.P.
(b) `a^2+b^2,b^2+c^2,c^2+d^2` are also in G.P.
Solution:
(a) Given that `a,b,c,d` are in G.P. Let the common difference of the G.P. be `r`. `\therefore\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=r\impliesb=ar,c=br=ar^2,d=cr=ar^3`
Now, `\left(a+b\right)\left(c+d\right)=\left(a+ar\right)\left(ar^2+ar^3\right)=a^2r^2\left(1+r\right)^2=\left(ar+ar^2\right)^2=\left(b+c\right)^2`. `therefore` `a+b,b+c,c+d` are also in G.P.
(b) Here, `\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2+a^2r^2\right)\left(a^2r^4+a^2r^6\right)=a^2\left(1+r^2\right)\.a^2r^4\left(1+r^2\right)`
`=a^4r^4\left(1+r^2)^2={a^2r^2\left(1+r^2\right)}^2=\left(a^2r^2+a^2r^4\right)^2=\left(b^2+c^2\right)^2`
Thus, `a^2+b^2,b^2+c^2,c^2+d^2` are also in G.P.
Example 2: If `a,b,c` are in G.P. then show that
(a) `\frac{1}{a+b},\frac{1}{2b},\frac{1}{b+c}` are in A.P.
(b) `\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{a+b}{b+c}`
Solution:
(a) Given that `a,b,c` are in G.P. `\therefore\frac{b}{a}=\frac{c}{b}\impliesb^2=ac`
Now, `\frac{1}{a+b}+\frac{1}{b+c}=\frac{b+c+a+b}{\left(a+b\right)\left(b+c\right)}`
`=\frac{2b+a+c}{ab+b^2+ac+bc}`
`=\frac{2b+a+c}{2b^2+b\left(a+c\right)}` `[\becauseb^2=ac]`
`=\frac{2b+a+c}{b\left(2b+a+c\right)}`
`=\frac{1}{b}=\frac{2}{2b}`
Thus, `\frac{1}{a+b},\frac{1}{2b},\frac{1}{b+c}` are in A.P.
(b) Here, `\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{a^2+ab+ac}{bc+b^2+ab}` `[\becauseb^2=ac]`
`=\frac{a\left(a+b+c\right)}{b\left(c+b+a\right)`
`=\frac{a}{b}=\frac{1}{r}` [where, `r` is the common ratio]
Again, `\frac{b+a}{c+d}=\frac{ar+a}{ar^2+ar}=\frac{a\left(r+1\right)}{ar\left(r+1\right)}=\frac{1}{r}`
Hence, `\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{a+b}{b+c}` (Proved).
Example 3: If `a,b,c` are in G.P. then show that `a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=a^3+b^3+c^3`.
Solution: `a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{b^2c^2}{a}+\frac{a^2c^2}{b}+\frac{a^2b^2}{c}`
`=\frac{ac^3}{a}+\frac{b^4}{b}+\frac{a^3c}{c}` `[\becauseb^2=ac]`
`=a^3+b^3+c^3`
Thus, `a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=a^3+b^3+c^3`(proved).
Example 4: If `a,b,c` are in G.P., prove that `\frac{1}{\log_am},\frac{1}{\log_bm},\frac{1}{\log_cm}` are in A.P.
Solution: Given that `a,b,c` are in G.P. `\therefore` `b^2=ac`.
Now, `\frac{1}{\log_am}+\frac{1}{\log_cm}`
`=\log_ma+\log_mc=\log_mac` `[\because\log_am+\log_an=\log_amn]`
`=\log_mb^2`
`=2\log_mb` `[\because\log_am^n=n\log_am]`
`=\frac{2}{\log_bm}`
Hence,`\frac{1}{\log_am},\frac{1}{\log_bm},\frac{1}{\log_cm}` are in A.P. (proved).
Example 5: If `\left(a-b\right),\left(b-c\right),\left(c-a\right)` are in G.P., then prove that `\left(a+b+c\right)^2=3\left(ab+bc+ca\right)`.
Solution: Given that, `a-b,b-c,c-a\in G.P.` `\therefore` `\left(b-c\right)^2=\left(a-b\right)\left(c-a\right)`
`\implies` `b^2+c^2-2bc=ac-bc-a^2+ab`
`\implies` `a^2+b^2+c^2=bc+ca+ab`
`\implies` `a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca=3\(ab+bc+ca\right)`
Hence, `\left(a+b+c\right)^2=3\left(ab+bc+ca\right)` proved).
Problems on Geometric Mean
Example 6: Insert 4 Geometric Means between `2` and `486`.
Solution: Here, the first term `\left(t_1\right)=2` and `t_6=486`, let the common ratio be `r`
`\therefore` `r=\left(\frac{b}{a}\right)^\frac{1}{6}=\left(\frac{486}{2}\right)^\frac{1}{5}=\left(243\right)^\frac{1}{5}=3`
Now, the first mean`=ar=2\times3=6`, second mean `=ar^2=2\times3^2=18`, third mean `=ar^3=2\times3^3=54` and fourth mean `=ar^4=2\times3^4=162`.
Thus, the required 4 geometric means are `6,18,54,162`
Example 7: If the Arithmetic Mean of two numbers is twice their Geometric mean, then show that the ratio of the numbers is `\left(2+sqrt3\right):\left(2-sqrt3\right)`
Solution: Let the two numbers are `a` and `b`. Now, by the problem we have `A.M.=2G.M.`
`\implies` `\frac{a+b}{2}=2sqrt{ab}\implies\left(a+b\right)^2=16ab`. Again, `\left(a-b\right)^2=\left(a+b\right)^2-4ab=12ab`
`\therefore` `\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\frac{16ab}{12ab}=\frac{4}{3}\implies\frac{a+b}{a-b}=\frac{2}{\sqrt3}\implies\frac{a}{b}=\frac{2+\sqrt3}{2-\sqrt3}`.
Thus, the required ratio of the numbers is `\left(2+\sqrt3\right):\left(2-sqrt3\right)`.
Example 8: The arithmetic mean and geometric mean of two numbers `x` and `y` are respectively `A` and `G` . Prove that `x^2+y^2=\left(2A^2-G^2\right)`.
Solution: We have, `\frac{a+y}{2}=A` and `xy=G^2`
`implies` `x+y=2A\impliesx^2+y^2+2xy=A^2\impliesx^2+y^2+2G^2=4A^2` `[\becausexy=G^2]`
`implies` `x^2+y^2=4A^2-2G^2=2\left(2A^2-G^2\right)`.
Example 9: Find the value of `n` so that `\frac{a^{n+1}+b^{n+1}}{a^n+b^n}` may be the geometric mean between `a` and `b`.
Solution: According to the problem we have,
`\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab}`
`implies` `a^{n+1}+b^{n+1}=a^{\left(n+\frac{1}{2}\right)}b^\frac{1}{2}+a^\frac{1}{2}b^{\left(n+\frac{1}{2}\right)}`
`implies` `a^{n+1}-a^{\left(n+\frac{1}{2}\right)}b^\frac{1}{2}=a^\frac{1}{2}b^{\left(n+\frac{1}{2}\right)}-b^{n+1}`
`\implies` `a^{\left(n+\frac{1}{2}\right)}\left(a^\frac{1}{2}-b^\frac{1}{2}\right)=b^{\left(n+\frac{1}{2}\right)}\left(a^\frac{1}{2}-b^\frac{1}{2}\right)`
`\implies` `a^{\left(n+\frac{1}{2}\right)}=b^{\left(n+\frac{1}{2}\right)}` `[\becauseaneb\thereforea^\frac{1}{2}-b^\frac{1}{2}ne0]`
`implies` `\left(\frac{a}{b}\right)^{\left(n+\frac{1}{2}\right)}=1=\left(\frac{a}{b}\right)^0\impliesn+\frac{1}{2}=0\impliesn=-\frac{1}{2}`
Thus, the required value of `n` is `-\frac{1}{2}`.
Example 10: If one A.M., `A` and two geometric means `G_1` and `G_2` inserted between two positive numbers, show that `\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=2A`.
Solution: Let the two positive numbers are `x` and `y`
`\therefore` according to the problem we have, `x+y=2A` and `G_1=x\left(\frac{y}{x}\right)^\frac{1}{3},G_2=x\left(\frac{y}{x}\right)^\frac{2}{3}`
Now, `\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=\frac{x^2\left(frac{y}{x}\right)^\frac{2}{3}}{x\left(\frac{y}{x}\right)^\frac{2}{3}}+\frac{x^2\left(\frac{y}{x}\right)^\frac{4}{3}}{x\left(\frac{y}{x}\right)^\frac{1}{3}}=x+y=2A`.
Problems on Sum to infinity of a Geometric Progression.
Example 11: Find the sum of the series, `1+\frac{1}{2}+\frac{1}{2^2}+...\infty`
Solution: `1+\frac{1}{2}+\frac{1}{2^2}+...\infty=\frac{1}{1-\frac{1}{2}}=\frac{2}{2-1}=2`.
Example 12: Find the sum of the series, `\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+...\infty`
Solution: `\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+...\infty`
`=\left(\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...\right)+\left(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+...\right)`
[In the above the first bracket contains an infinite G.P., with a first term `\frac{1}{3}` and common ratio`\frac{1}{3^2}` and the second bracket contains an infinite G.P. with first term `\frac{1}{5^2}` and common ratio `\frac{1}{5^2}`]
`=\frac{\frac{1}{3}}{1-\frac{1}{3^2}}+\frac{\frac{1}{5^2}}{1-\frac{1}{5^2}}=\frac{5}{12}`.
Example 13: The sum of an infinite geometric series is `\frac{3}{2}` and its second term is `\frac{1}{3}`. Tind the first term and common ratio of the series.
Solution: Let the first term be `a` and the common ratio be `r`.
`\therefore` `\frac{3}{2}=\frac{a}{1-r}`..........(1) and `ar=\frac{1}{3}`..........(2)
From (1) and (2) we get, `a=1,r=\frac{1}{3}` and `a=\frac{1}{2},r=\frac{2}{3}`.
Thus, according to the conditions there would be two series. One series with first term `1` and common ratio `\frac{1}{3}` and the other series has first term `\frac{1}{2}` and common difference `\frac{2}{3}`.
Example 14: If `x=1+a+a^2+..........\infty` and `=1+b+b^2+..........\infty` then show that `1+ab+a^2b^2+..........\infty=\frac{xy}{x+y-1}`, when `-1<a<1` and `-1<b<1`.
Solution: `\because-1<a<1,\thereforex=1+a^2+..........\infty=\frac{1}{1-a}\implies1-a=\frac{1}{x}\impliesa=1-\frac{1}{x}=\frac{x-1}{x}`.
`\because-1<b<1,\thereforey=1+b+b^2+..........\infty=\frac{1}{1-b}\implies1-b=\frac{1}{y}\impliesb=1-\frac{1}{y}=\frac{y-1}{y}`.
Now `\because-1<a<1` and `-1<b<1,\therefore-1<ab<1`. `\therefore1+ab+a^2b^2+..........\infty=\frac{1}{1-ab}=\frac{1}{1-\frac{\left(x-1\right)\left(y-1\right)}{xy}}`
`=\frac{xy}{xy-\left(xy-x-y+1\right)}`
`implies` `1+ab+a^2b^2+..........\infty=\frac{xy}{x+y-1}`. (Proved)
Example 15: Express `0.\overset{.}{4}` as an infinite geometric series and hence prove that`\left(0.444..........\right)^\frac{1}{2}=0.666..........`
Solution: `0.\overset{.}{4}=0.444..........\infty=0.4+0.04+0.004+..........\infty` which is an infinite G.P. whose first term `\left(a\right)=\frac{4}{10}` and common ratio `\left(r\right)=\frac{\frac{4}{100}}{\frac{4}{10}}=\frac{1}{10}<1`.
`\therefore` `0.\overset{.}{4}=0.444..........\infty=\frac{4}{10}+\frac{4}{100}+\frac{4}{1000}+..........`
`=\frac{a}{1-r}=\frac{\frac{4}{10}}{1-\frac{1}{10}}`
`=\frac{4}{10}\times\frac{10}{9}=\frac{4}{9}`
`\therefore` `\left(1.444..........\right)^\frac{1}{2}=\left(\frac{4}{9}\right)^\frac{1}{2}=\frac{2}{3}=0.666..........`
Exercise-2(C)
A. Chose thee correct Answer
- If `x,y,z` are in geometric progression, then `\frac{1}{x^2-y^2}+\frac{1}{y^2}=`
(a) `\frac{1}{y^2-x^2}`
(b) `\frac{1}{y^2-z^2}`
(c) `\frac{1}{z^2-y^2}`
(d) `\frac{1}{z^2-x^2}` - If `a,b,c` are in G.P., then `log_ax,log_bx,log_cx` are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) none of these. - If `p`th,`q`th,`r`th and `s`th terms of an A.P. be in G.P., then `p-q`,`q-r`,`r-s`are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) none of these - Let `x` be the A.M. and `y,z` be two G.M.s between two positive numbers, then `\frac{y^3+z^3}{xyz}` is
(a) `1`
(b) `2`
(c) `\frac{1}{2}`
(d) none of these - The two geometric means between the numbers `1` and `64` are
(a) `1` and `64`
(b) `4` and `16`
(c) `2` and `16`
(d) `8` and `16` - The minimum value of `4^x+4^{1-x}, x\inR` is
(a) `2`
(b) `4`
(c) `1`
(d) `0` - The value of `9^\frac{1}{3}\.9^\frac{1}{9}\.9^\frac{1}{27}...` to `\infty`
(a) `1`
(b) `3`
(c) `9`
(d) none of these. - If the sum of first two terms of an infinite G.P. is `1` and very term is twice the sum of all the successive terms, then its first term is
(a) `\frac{1}{3}`
(b) `frac{2}{3}`
(c) `\frac{1}{4}`
(d) `\frac{3}{4}` - If second term of a G.P. is 2 and the sum of its infinite term is `8`, then its first term is
(a) `\frac{1}{4}`
(b) `\frac{1}{2}`
(c) `2`
(d) `4` - If `x` is positive, the sum to infinity of the series `\frac{1}{1+x}-\frac{1-x}{\left(1+x\right)^2}+\frac{\left(1-x\right)^2}{\left(1+x\right)^3}-\frac{\left(1-x\right)^3}{\left(1+x\right)^4}+.....`
(a) `\frac{1}{2}`
(b) `\frac{3}{4}`
(c) `1`
(d) none of these.
- If `a,b,c,d` are in G.P. show that
(i) `\left(a^2+b^2+c^2\right)(b^2+c^2+d^2\right)=\left(ab+bc+cd\right)^2`
(ii) `a^n+b^n,b^n+c^n,c^n+d^n` are also in G.P.
(iii) `\frac{1}{a^2+b^2},\frac{1}{b^2+c^2},\frac{1}{c^2+d^2}` are also in G.P.
(iv) `\frac{1}{a^2-b^2},\frac{1}{b^2-c^2},\frac{1}{c^2-d^2}` are also in G.P.
(v) `\left(a^3+b^3\right)^-1,\left(b^3+c^3\right)^-1,\left(c^3+d^3\right)^-1` are aldo in G.P. - If `a,b,c` are in G.P., prove that `a^2+b^2,ab+bc,b^2+c^2` are also in G.P.
- If 4th 10th and 16th terms of a G.P. are `x,y` and `z` respectively, then prove that `x,y,z` are in G.P.
- If `a,b,c` are in G.P. and `a^\frac{1}{x}=b^\frac{1}{y}=c^\frac{1}{z}`, then show that `x,y,z` are in A.P.
- Insert three numbers between `1` and `256` so that the resulting sequence is a G.P. [Ans: `4,16,64`]
- Determine the sum of six G.M.s between `2` and `\frac{1}{64}` [Ans: `\frac{63}{128}`]
- The arithmetic mean of two numbers `x` and `y` is ` and A` geometric mean is `G`. Prove that `x^2+y^2=2\left(2A^2-G^2\right)`.
- The sum of two numbers is `6` times their geometric mean, show that numbers are in the ratio `\left(3+2sqrt2\right):\left(3-2sqrt2\right)`
- If A.M. and G.M of roots of a quadratic equation are `8` and `5` respectively, then obtain the quadratic equation. [Ans: `x^2-16x+25=0`]
- If `G` be the geometric mean and `p` and `q` be two arithmetic means between two given quantities, then show that `G^2=\left(2p-q\right)\left(2q-p\right)`.
- Find the sum of the following infinite series
(i) `0.7+0.07+0.007+........\infty` [Ans: `\frac{7}{9}`]
(ii) `1+2x+4x^2+8x^3+.....\infty` [Ans: `\frac{1}{1-2x}`]
(iii) `\frac{1}{2}\frac{3}{4}+\frac{1}{8}+\frac{3}{16}+......\infty` [Ans: `\frac{5}{3}`] - Express the recurring decimal `0.\overset{.}{3}` as an infinite geometric series and using the sum find out the value of `0.\overset{.}{9}`
- Prove that, `2^\frac{1}{2}\.4^\frac{1}{4}\.8^\frac{1}{8}\.16^\frac{1}{16}.......\infty=4`
- If `-1<r<1` and `x=a+ar+ar^2+....\infty` and `y=b-br+br^2-.....\infty`, then show that `\frac{a}{x}+\frac{b}{y}=2`
- If `-1<\frac{1}{r}<1`, and `x=a+\frac{a}{r}+\frac{a}{r^2}+....\infty`, `y=b-\frac{b}{r}+\frac{b}{r^2}-.......\infty` and `z=c+\frac{c}{r^2}+\frac{c}{r^4}+......\infty`, show that `\frac{xy}{z}=\frac{ab}{c}`.
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