Originally Posted by

**wooram83** Please Help me.

Sequence Question

There are 4 positive numbers in a sequence. The first three are in Arithmatic sequence and the last three are in geometric sequence. If the difference between the last number and the first number is 30. What is the sum of the 4 numbers?

A1 A2 A3 A4

A1 A2 A3 are in arithmatic sequence

A2 A3 A4 are in geometric sequence

A4 - A1 = 30

What is the sum of all four terms?

Mathaddict is correct. With what is given there are many solutions that will work.

Originally Posted by

**wooram83** I have checked the question and that was it. I have tried your method and failed. So I tried to work with b (2nd term) and c (3rd term) being both Arithmatic and Geometric and it put me in circles. I'm not sure how to go around this. Maybe we can just find the sum without actually getting the individual terms? I'm working on that right now but nothing yet.

You've worked with the second & third terms which are both arithmetic and geometric. Something like this.

If d is the arithmetic difference and g is the geometric difference then:

Code:

term1 term2 term3 term4
A-d A A+d
A Ag Ag^2

and dividing results in

Which makes the 4th term

The difference between the first & fourth:

Simplifying that results in a quadratic.

AS STATED:

4 positive numbers, with arithmentic/geometric constraints,

difference between the last number and the first number is 30.

There are lots of numbers that will fit.

HOWEVER, If you restrict the **4 positive numbers** to whole numbers, then you have a limited solution.

There are limited values in the radical that provide whole numbers.

This is one result:

18 27 36 48

The first three are in arithmetic progression, and the last three have a geometric term of

What is the sum of the 4 numbers?

OR

If the question was:

__What is the SMALLEST SUM of the 4 numbers?__

That is easily solved.

.