Question 546647
1.If the third term of a Geometric Progression is the square of the first and the fifth term is 64,find the series.
<pre>

This is what we are given, where a<sub>1</sub> represents the first term.

a<sub>1</sub>, ___, a<sub>1</sub>², ___, 64, ...

Let's assume that r represents the common ratio. and so the second term
is the first term times r:

a<sub>1</sub>, a<sub>1</sub>r, a<sub>1</sub>², ___, 64, ...

Since the third term is the second term times r, we have the equation

(a<sub>1</sub>r)(r) = (a<sub>1</sub>)²

a<sub>1</sub>r² = (a<sub>1</sub>)²


We divide both sides by a<sub>1</sub> 
  
    r² = a<sub>1</sub>

The square of the common ratio and the first term are the same, so now the
geometric progression starts with r² and each successive term
is the precding one multiplied by r, so we have:

r<sup>2</sup>, r<sup>3</sup>, r<sup>4</sup>, r<sup>5</sup>, r<sup>6</sup> = 64

So we solve for r:

r<sup>6</sup> = 64

Use the principle of even (sixth) roots, 

r = ±2
 
So there are two solutions one with r = 2 and one with r = -2

Using r = 2

So since 
    
r² = a<sub>1</sub>

The first term a<sub>1</sub> = 2² = 4
 
The nth term is 

a<sub>n</sub> = a<sub>1</sub>r<sup>n-1</sup> = 4(2)<sup>n-1</sup>) = 2²(2)<sup>n-1</sup>) = 2<sup>2+n-1</sup> = 2<sup>n+1</sup>

and the progression is

4, 8, 16, 32, 64, ...

-----------

Using r = -2

So since 
    
r² = a<sub>1</sub>

The first term a<sub>1</sub> = (-2)² = 4
 
The nth term is 

a<sub>n</sub> = a<sub>1</sub>(-2)<sup>n-1</sup> = 4(-2)<sup>n-1</sup>) = 4(-1·2)<sup>n-1</sup>) = 4(-1)<sup>n-1</sup>2<sup>n-1</sup> =  2²(-1)<sup>n-1</sup>2<sup>n-1</sup> =  (-1)<sup>n-1</sup>2<sup>n-1+2</sup> =  (-1)<sup>n-1</sup>2<sup>n+1</sup>

and the progression is

4, -8, 16, -32, 64, -+ ...

So the nth term is either given by

a<sub>n</sub> = 2<sup>n+1</sup>

or by:

a<sub>n</sub> = (-1)<sup>n-1</sup>2<sup>n+1</sup>

Edwin</pre>