Sum of a part arithmatic and part geometric sequence

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October 22nd 2009, 09:53 PM
wooram83

Sum of a part arithmatic and part geometric sequence

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?
October 23rd 2009, 03:14 AM
mathaddict

Quote:

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?

HI

Let the 4 positive numbers be a , b , c , d

first three terms in AP : $c-b=b-a$

Then the last three terms are GP : $\frac{d}{c}=\frac{c}{b}$

so $2b=c+a$ --- 1

$db=c^2$ --- 2

Then you also know that : $d-a=30$ ---3

Solve the system for a ,b, c , d then add them up .

Can u recheck the original question , i guess you have missed up some information , since the 3 equations i got is not sufficient to solve for 4 different variables .
October 23rd 2009, 09:41 PM
wooram83

Thank you for your response

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.
October 25th 2009, 11:55 AM
aidan

Quote:

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.

Quote:

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

$A+d = Ag$ and dividing results in $\dfrac{A+d}{A} = g$

Which makes the 4th term
$\dfrac{(A+d)^2}{A}$

The difference between the first & fourth:

$\dfrac{(A+d)^2}{A} - (A - d) = 30$

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 $g = \dfrac{4}{3}$

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.

.
October 25th 2009, 10:53 PM
wooram83

Thank you for your clearification. I'm sorry for the mistake on the Question.