Prove $\frac{a}c = \frac{a-b}{b-c}$

Question

Suppose $\frac{1}a,\frac{1}b,\frac{1}c$ are three consecutive terms in an arithmetic sequence. Show that:

$$\frac{a}c = \frac{a-b}{b-c} $$

and that:

$$\frac{2ac}{a+c} = b$$ How would I prove this?

Hint: if $x,y,z$ are three consecutive terms of an arithmetic sequence, then $y-x = z-y$. — MY USER NAME IS A LIE, Mar 02 '15 at 04:38
Hint for the second one:if x,y,z are three consecutive terms of an arithmetic sequence, then x + z = 2y. — Mick, Mar 02 '15 at 04:44

score 2 · Accepted Answer · answered Mar 02 '15 at 04:43

2

You have ${1 \over b} = {1 \over 2} ({1 \over a} + {1 \over c})$. This leads to $$b = {2ac \over a + c}$$ Substitute this in...

answered Mar 02 '15 at 04:43

Zarrax

Boom. Nicely handled; +1. Simplest method for this part of the question I guess... – Panglossian Oporopolist Mar 02 '15 at 04:51

score 2 · Answer 2 · 2015-03-02T04:50:07.900

2

$\dfrac{a}{c}=\dfrac{a-b}{b-c}\impliedby ab-ac=ac-bc\impliedby \color{red}{ab+bc=2ac}\impliedby\dfrac{1}{c}+\dfrac{1}{a}=\dfrac{2}{b}$

The second part follows from the colored part.

edited Mar 02 '15 at 04:50

answered Mar 02 '15 at 04:44

Proofs usually don't first assume the result given is true, then work backwards to show it holds true... – Panglossian Oporopolist Mar 02 '15 at 04:49
Ah, I got until the colored part, but I didn't divide by $abc$ – user1843145 Mar 02 '15 at 04:49
@user1843145: In fact a one line argument could be given by just saying that since $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., $a,b,c$ are in H.P and hence the expression for $b$ directly follows. – Mar 02 '15 at 04:52
Yeah, @ user 170039 sorry I have not looked into harmonic progression – user1843145 Mar 02 '15 at 04:56
^Agreed @user170039 – Panglossian Oporopolist Mar 02 '15 at 04:56

score 2 · Answer 3 · answered Mar 02 '15 at 04:50

2

This is the answer to the first part; second part has been adequately answer by Zarrax. enter image description here

answered Mar 02 '15 at 04:50

3 Answers3