The sum of a geometric series is 3. The series of squares of its terms has sum 18. Findseries.​

The problem is asking for the terms of a geometric series. A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the ratio.

Let's denote the first term of the geometric series as 'a' and the common ratio as 'r'.

From the problem, we know that the sum of the geometric series is 3. The sum of a geometric series can be calculated using the formula:

S = a / (1 - r)

So, we have:

3 = a / (1 - r) ----(equation 1)

The problem also states that the sum of the squares of the terms is 18. The sum of the squares of a geometric series can be calculated using the formula:

S_square = a^2 / (1 - r^2)

So, we have:

18 = a^2 / (1 - r^2) ----(equation 2)

Now, we have a system of two equations. We can solve this system to find the values of 'a' and 'r'.

Let's divide equation 2 by equation 1:

18/3 = (a^2 / (1 - r^2)) / (a / (1 - r))

6 = a / (1 - r^2)

Solving this equation for 'a' gives:

a = 6(1 - r^2) ----(equation 3)

Substitute equation 3 into equation 1:

3 = 6(1 - r^2) / (1 - r)

Solving this equation for 'r' gives:

r = 1/2 or r = -2

Substitute r = 1/2 into equation 3:

a = 6(1 - (1/2)^2) = 4.5

Substitute r = -2 into equation 3:

a = 6(1 - (-2)^2) = -18

So, the geometric series could be either:

4.5, 2.25, 1.125, ...

or

-18, 36, -72, ...