Question

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

Answer 1

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=2b`
Move one `b` to the LHS, and move `c` to the RHS.

`color(white)(=>)a+c=2b`
`=>a-b=b-c"   "`Multiply both sides by `-1`.
`=>b-a=c-b`

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

Given: `ab+cd+ad=3bc`

Solve for `d`.

`ab+cd+ad=3bc`

`=>d(c+a)=3bc-ab`

`=>d=(b(3c-a))/(c+a)"    "`Substitute `2b=a+c`

`=>d=(cancel b(3c-a))/(2cancelb)`

`=>2d=3c-a"    "`Subtract `2c` from both sides

`=>2d-2c=3c-a-2c`

`=>2(d-c)=c-a"    "`Subtract and add `b` to the RHS

`=>2(d-c)=c-b+b-a"    "`Substitute `b-a=c-b`

`=>2(d-c)=c-b+c-b`

`=>2(d-c)=2(c-b)"    "`Divide both sides by 2

`=>d-c=c-b`

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