How Do You Solve for b in an Arithmetic Sequence Involving 1/a, 1/b, and 1/c?

[Digital Genius]

Homework Statement

* "/" means divided by *

1/a , 1/b , 1/c are consecutive terms in an AS, where a,b,c ε R\0. (whatever that means haha)

express b in terms of a and c. give your answer in its simplest form.

*thats all it says*

Homework Equations

there are none :)

The Attempt at a Solution

sorry but i have absolutely no clue and that's why I am posting this haha

 

[Digital Genius]

ok nevermind guys sorry, i just found that i did do the question haha...

anyways for those who want to know its like this...

T1, T2, T3
1/a, 1/b, 1/c = AS
T2-T1=T3-T2
T2+T2=T3+T1
1/b + 1/b = 1/c + 1/a
2(1/b)=1/ac
2b=1/(1/a - 1/c)
b= 2 1/(1/a - 1/c)

and that's it i think..

 

[adjacent]

ok nevermind guys sorry, i just found that i did do the question haha...

anyways for those who want to know its like this...

T1, T2, T3
1/a, 1/b, 1/c = AS
T2-T1=T3-T2
T2+T2=T3+T1
1/b + 1/b = 1/c + 1/a
2(1/b)=1/ac
2b=1/(1/a - 1/c)
b= 2 1/(1/a - 1/c)

and that's it i think..

It's wrong.
$\frac{1}{c}+\frac{1}{a} \neq \frac{1}{ac}$
You have some algebra problems there. I get a different answer. Try again
I have coloured the wrong parts in red.

 

[adjacent]

Please post your correct solution too,for other to see

 

[Digital Genius]

so what is the rest of it that you got? because I am stumped haha

 

[adjacent]

Let's start from the original question.
$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ . Express b interms of a and c.
As you showed, $\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$
So $\frac{1}{b}+\frac{1}{b}=\frac{1}{c}+\frac{1}{a}$
$\frac{2}{b}=\frac{1}{c}+\frac{1}{a}$


Simplify $\frac{1}{c}+\frac{1}{a}$ then solve for b.

 

[Digital Genius]

ok thanks haha, i have it now, i even checked with my A- standard student/friend :D