- #1

- 416

- 52

An increasing sequence that is made of 4 positive numbers, The first three of it are arithmetic series. and the last three are geometric series. The last number minus the first number is equal to 30. Find the sum of this series.

So here is my steps:

Here is the arithmetic series: ##a-d , a, a+d##

The geometric one is: ## a , a+d, ar^2##

So first in order for 2nd one to be geometric series: ##a+d## must equal to ## ar##

We conclude from this that:

## d = a(r-1)##

and the question gave us an equation which is

##ar^2 - (a-d) = 30##

By putting the two equation in each other we get:

## a(r^2 + r - 2) = 30##

Now the sum of the whole sequence is just the sum of the arthmetic plus the last number:

## \text{Sum} = 3a + ar^2 ##

## \text{Sum} = a(r^2 + 3)##and everything pretty much stopped here, I need a third equation to complete this and I didn't find a way to substitute the whole thing is another thing that is solvable. So the question is: Is it solvable in the first place?