Question

if (a,b,c) is a geometric sequence and x is an arithmetic mean between a and b, and y is an arithmetic mean between b and c , prove that a/x +c/y=2

Answer 1

Lets see we have
\[\frac{a+b}{2}=x \text{ and } \frac{b+c}{2}=y\]
We want to prove \[\frac{a}{x}+\frac{c}{y}=2\]

Answer 2

yup

Answer 3

We also have (a,b,c) is a geometric sequence.....
That means a*m=b and b*m=c for some number m

Answer 4

Thinking...

Answer 5

What happens if we try to solve both those in my first post for b and set them equal
what would we get...

Answer 6

\[a+b=2x \text{ and } b+c=2y \]
=>
\[b=2x-a \text{ and } b=2y-c\]
=>
\[2x-a=2y-c\]

Answer 7

Now can you solve this for 2?

Answer 8

\[2x-2y=a-c\]

Answer 9

still i tried that

Answer 10

\[2(x-y)=a-c =>2 =\frac{a-c}{x-y}\]
hmmm.....

Answer 11

well u can say that a is a , b is ar and c is ar^2

Answer 12

since its GS

Answer 13

and plug in new values in both first equations then u get r= y/x

Answer 14

True

Answer 15

and since its GS so b is a geometric mean of a and c

Answer 16

b^2=ac

Answer 17

\[2=\frac{a-c}{x-y}\]
\[2=\frac{a-ar^2}{x-xr} \text{ since } c=ar^2 \text{ & } y=xr\]

So we have
\[2=\frac{a(1-r^2)}{x(1-r)}\]
\[2=\frac{a(1-r)(1+r)}{x(1-r)}\]
\[2=\frac{a(1+r)}{x}\]
\[2=\frac{a+ar}{x}\]
\[2=\frac{a}{x}+\frac{ar}{x}\]
guess what.........:)

Answer 18

\[2=\frac{a}{x}+\frac{b}{x}\]
oh i thought i had it

Answer 19

haha but nice try keep going

Answer 20

oh but wait

Answer 21

we do have it

Answer 22

\[\frac{c}{y}=\frac{ar^2}{xr}=\frac{ar}{x}\]

Answer 23

right?

Answer 24

yup

Answer 25

\[2=\frac{a}{x}+\frac{ar}{x}=\frac{a}{x}+\frac{c}{y}\]

Answer 26

Wow @hossam1122 I like this problem

Answer 27

It is very cute.

Answer 28

lol tnx :D

Answer 29

do you have any questions? There were some things you mentioned that i didn't think of that we used to show this so really great job. I love it when askers want to engage. Thanks so much.

Answer 30

you r welcome ;)