SOLUTION: 1.If the third term of of a Geometric Progression is the square of the first and the fifth term is 64,find the series.

This is what we are given, where a1 represents the first term.

a1, ___, a1², ___, 64, ...

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

a1, a1r, a1², ___, 64, ...

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

(a1r)(r) = (a1)²

a1r² = (a1)²


We divide both sides by a1 
  
    r² = a1

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:

r2, r3, r4, r5, r6 = 64

So we solve for r:

r6 = 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 = a1

The first term a1 = 2² = 4
 
The nth term is 

an = a1rn-1 = 4(2)n-1) = 2²(2)n-1) = 22+n-1 = 2n+1

and the progression is

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

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Using r = -2

So since 
    
r² = a1

The first term a1 = (-2)² = 4
 
The nth term is 

an = a1(-2)n-1 = 4(-2)n-1) = 4(-1·2)n-1) = 4(-1)n-12n-1 =  2²(-1)n-12n-1 =  (-1)n-12n-1+2 =  (-1)n-12n+1

and the progression is

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

So the nth term is either given by

an = 2n+1

or by:

an = (-1)n-12n+1

Edwin