(1/a)+(1/c)=2/b?

Question

if a=(b+c)/2 & c=(a+b)/2
and b is the mean proportion between a and c

Answer 1

If a=(b + c)/2 then b=2a-c
If c=(a + b)/2 then b=2c -a

2a-c = 2c-a // + a + c

3a = 3c // divide by 3

a=c

and b=2c-a=a=c

1/a + 1/c = 1/b + 1/b = 2/b

Answer 2

Well, I'm not as clever as I used to be, so I used Mathcad to solve your three simultaneous equations. Once you see the answer it's pretty obvious, but I resorted to technology so I can't really explain the mathematics.
a = 1, b = 1, c = 1
Replace every variable with 1 and all three equations are simultaneously true:
(1 / 1) + (1 /1) = 2 / 1
1 = (1 + 1) / 2
1 = (1 + 1) / 2
Get it? Wish I could do a better job of explaining the math's logic. Sorry.

Answer 3

If a=(b + c)/2 then b=2a-c
If c=(a + b)/2 then b=2c -a

2a-c = 2c-a // add + a + c on both sides

3a = 3c // divide by 3

a=c

and b=2c-a=a=c

hence, a=b=c

1/a + 1/c = 1/b + 1/b = 2/b

Answer 4

a = (b+c)/2 ----> 2a = b+c ---->
2a-b-c = 0 -eq1

c = (a+b)/2 ----> 2c = a+b ----->
2c-a-b = 0 -eq2

solving eq1 & eq2:
3a-3c = 0 ----> a = c ----->
b = 2a-c (from eq1)
= 2a-a (since a = c)
= a

so a = b =c
Verifying
(1/a) + (1/c) = 2/b

(1/a) + (1/a) = 2/a
2/a = 2/a
so the eqn is true for the given conditions

Answer 5

a=(b+c)/2 & c=(a+b)/2
2a = b + c (x2) (eq.1)
2c = a + b
-----------------------
4a = 2b + 2c
2c = a + b (Substitude for 2c).
-----------------------
4a = 2b + 2a + 2b (eq.2)
2c = a + b (eq.3)
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4a = 2b + 2a + 2b (eq.2)
6a = 4b
a = 4/6b
a = 2/3b (eq.4)
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2a = b + c (eq.1)
2(2/3b) = b + c (Substituding eq.4)
4/3b = b + c
1/3b = c (x 2)
2/3b = 2c (eq.5)
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Compare equations 4 and 5.
a = 2/3b and 2/3b = 2c
→ a = 2/3b = 2c

Answer 6

a=b=c
(weird)