# If a+c=2b and ab+cd+ad=3bc, how do I show that a,b,c,d are in arithmetic progression?

Aug 15, 2017

Play around with the equations. Knowing how to (re-)phrase your goal mathematically can help you deduce it from the given information.

#### Explanation:

$a , b , c , d$ are in arithmetic progression if and only if b-a = c-b = d-c.

Can we show this?

Given: $a + c = 2 b$
Move one $b$ to the LHS, and move $c$ to the RHS.

$\textcolor{w h i t e}{\implies} a + c = 2 b$
$\implies a - b = b - c \text{ }$Multiply both sides by $- 1$.
$\implies b - a = c - b$

Now we just need to show that $d - c$ is equal to either $b - a$ or $c - b$.

Given: $a b + c d + a d = 3 b c$

Solve for $d$.

$a b + c d + a d = 3 b c$

$\implies d \left(c + a\right) = 3 b c - a b$

$\implies d = \frac{b \left(3 c - a\right)}{c + a} \text{ }$Substitute $2 b = a + c$

$\implies d = \frac{\cancel{b} \left(3 c - a\right)}{2 \cancel{b}}$

$\implies 2 d = 3 c - a \text{ }$Subtract $2 c$ from both sides

$\implies 2 d - 2 c = 3 c - a - 2 c$

$\implies 2 \left(d - c\right) = c - a \text{ }$Subtract and add $b$ to the RHS

$\implies 2 \left(d - c\right) = c - b + b - a \text{ }$Substitute $b - a = c - b$

$\implies 2 \left(d - c\right) = c - b + c - b$

$\implies 2 \left(d - c\right) = 2 \left(c - b\right) \text{ }$Divide both sides by 2

$\implies d - c = c - b$

Since b-a = c-b = d-c, we have $a , b , c , d$ in arithmetic progression.