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Arithmetic and Geometric Progressions

Date: 03/23/98 at 11:24:38
Subject: Arithmetic progression

If the third term of an arithmetic progression is three times the 
seventh term and the ninth term is 1, find the first term, the common 
differences, and the first term less than 0.

The third term of a geometric progression is 36 and the fifth term is 
16.  Find the tenth term.

Date: 03/23/98 at 14:22:47
From: Doctor Rob
Subject: Re: Arithmetic progression

An arithmetic progression has the form

   a, a+d, a+2*d, a+3*d, a+4*d, ...,

where a is the first term, d is the common difference, and the n-th 
term is a + (n-1)*d.

The conditions of your problem translate into:

   a+2*d = 3*(a+6*d)
   a+8*d = 1

You need to solve these two simultaneous linear equations for a and d.
Once you have done that, you need to substitute them into the 

   a+(n-1)*d < 0,

and solve for n.

If I have done this problem right, the simultaneous equations are
inconsistent, that is, there is no common solution (a, d), so the 
second part of the question is meaningless.

A geometric progression has the form

   a, a*r, a*r^2, a*r^3, a*r^4, ...,

where a is the first term, r is the common ratio, and the n-th term 
is a*r^(n-1).

The conditions of your problem translate into:

   a*r^2 = 36
   a*r^4 = 16

and you are asked to find a*r^9.  One way to do that is to solve these
equations for a and r. A good method is to divide the first equation 
into the second, and discover that r^2 = 4/9, giving two values of r. 
Now divide that equation into the first, giving you the value of a. 
Now you can compute a*r^9. There are two answers.

-Doctor Rob,  The Math Forum
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Associated Topics:
High School Sequences, Series

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