# Find the terms a1 and a3 for the following following geometric sequence: a1, -3, a3, -4/3 , ....

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Let a1, -3, a3, -4/3 be terms of a geometric sequence:

Let r be the common difference;

Then we know that:

a2= a1*r

a3= a1*r^2

a4= a1*r^3

Then we conclude that:

a4= a2*r^2

let us substitute:

==> -4/3 = -3 * r^2

Now we will divide by -3

==> r^2 = (-4/3) / -3

==> r^2 = 4/9

==> r = +-2/3

**Then we have two values for the common difference:**

**r = { -2/3 , 2/3}**

**Then the sequesnces are:**

** 9/2, -3, 2, -4/3**

**-9/2, -3, -2, -4/3**

As this is a geometric series we have a common ratio between the consecutive terms.

Therefore -3/a1 = a3/-3 = (-4/3)/a3.

Now we use a3/-3 = (-4/3)/a3

=> a3^2 = (-4/3)*-3

=> a3^2 = 4

=> a3 = 2 or a3 = -2

Now -3/a1 = a3/-3

Therefore if a3= +2 , a1 = 9/2

If a3 = -2 , a1 = -9/2

**So a1 and a3 can take the values (-9/2 , -2) , (9/2 , 2).**

**Also, the common ratio could be -2/3 or 2/3.**

We'll use the theorem of geometric mean of the terms of a geometric series.

(-3)^2 = a1*a3

We'll use the symmetric property and we'll get:

a1*a3 = 9

We also could write:

a3^2 = (-3)*(-4/3)

We'll simplify and we'll get:

a3^2 = 4

a3 = sqrt 4

a3 = 2 or a3 = -2

For a3 = 2, we'll get:

a1*2 = 9

a1 = 9/2

For a3 = -2, we'll get:

a1 = -9/2

**The geometric series are: {-9/2 , -3 , -2 , -4/3} or {9/2 , -3 , 2 , -4/3}.**