Question

if1/x,1/y,1/z are in A.P then show that xy.zx,yz are in A.P
pls show me the steps of the sum

Answer 1

Well, what does it mean for these three to be in arithmetic progression?

Answer 2

These three, to be precise:
\[\Large \frac1x \qquad \frac1y \qquad \frac1z\]

Answer 3

pls read the question once more

Answer 4

What can you immediately conclude if you know that those three ^
are in an arithmetic progression?

Answer 5

can u prove wht i have asked for

Answer 6

Maybe... I don't really know...
I haven't gone round to actually doing it yet...
I'd like you to do it with me, so that we'll learn together, though :)

Answer 7

Come one, what does it mean, for things to be in an arithmetic progression (AP)?

Answer 8

i have o time to explain it to right now

Answer 9

Well, just a rough definition of what makes an AP will do...

Answer 10

which class are u in??

Answer 11

LOL
If these three
\[\Large \frac1x \qquad \frac1y \qquad \frac1z\] are in arithmetic progression, then
\[\Large \frac1y - \frac1x = \frac1z - \frac1y\]
correct?

Answer 12

yap

Answer 13

Well then, what I want you to do is simplify both sides of
\[\Large \frac1y - \frac1x = \frac1z - \frac1y\]
so that there is only one fraction on each side...

Answer 14

ok.wait

Answer 15

its not working

Answer 16

What's not working?

Answer 17

wht u asked me to do

Answer 18

Oh you of little faith :P
I've already done it ^_^

Now all that's left is for you to be able to do it too :)

Answer 19

Come on, just trust me ;)

Answer 20

ok.wait

Answer 21

ok i have got y-x/xy=y-z/yz after simplifying it

Answer 22

Nope...
Your left side is off... once you realise that
\[\Large \color{red}{\frac{y-x}{xy}=\frac1x - \frac1y}\color{blue}{\neq \frac{1}y-\frac1x}\]

Answer 23

A rather small error, but significant, nevertheless... you got it backwards...
\[\Large \frac{x-y}{xy}=\frac{y-z}{yz}\]
ok? :P

Answer 24

after that wht should i do

Answer 25

Now, it'd be nice if both sides had the same denominator... a good candidate would be the denominator xyz.
On the left side, we may multiply the fraction \(\Large \frac{z}z\) without any worry, since this is effectively just 1.
On the right side, we may multiply the fraction \(\Large \frac{x}{x}\) with similar results...
\[\Large \frac{z}{z}\cdot\frac{x-y}{xy}=\frac{x}{x }\cdot\frac{y-z}{yz}\]

Answer 26

Please distribute :)

Answer 27

ok

Answer 28

its zx-zy /zxy=xy-xz/xyz

Answer 29

mhmm :)
Let me just rewrite that for you ^_^
\[\Large \frac{zx-yz}{xyz}=\frac{xy-zx}{xyz}\]

Notice that the two sides now have the same denominator, and so they may be cancelled...
\[\Large \frac{zx-yz}{\cancel{xyz}}=\frac{xy-zx}{\cancel{xyz}}\]

\[\Large zx - yz = xy - zx\]

Answer 30

then...........

Answer 31

then......?
If you like, you can multiply -1 to both sides (actually, it's unnecessary, but meh)
\[\Large yz - zx = zx - xy\]

Answer 32

its not done ,dude

Answer 33

And why not?

Answer 34

so from this how can u prove that xy.zx,yz are in A.P

Answer 35

What is the definition of an AP?
That's why that's the first thing I asked you... so that you're clear on your objectives...

the numbers a, b, and c form an arithmetic progression if
b - a = c - b

Answer 36

yes

Answer 37

So...
We have now shown that
yz - zx = zx - xy

What don't you get now?

Answer 38

yes,lol i got it .hey thank u so much dude

Answer 39

no problem