Find the sum of the first four terms of the geometric sequence with a = -3 and r = 3.
I have NO idea how to attempt to solve this. I do not want an answer just instuctions. Thanks
The terms in the sequence are presumably
a*r^n.
I rceommend that you read
http://en.wikipedia.org/wiki/Geometric_progression
The first four terms would be (starting with n=0),
a, a r, ar^2 and a r^3.
The sum is a (1 + r + r^2 + r^3)
-3 * (1 + 3 + 9 + 27)
= -120
To find the sum of the first four terms of a geometric sequence, you can use the formula for the sum of a geometric series, which is given by:
Sn = a * (1 - r^n) / (1 - r)
where Sn is the sum of the first n terms, a is the first term of the sequence, r is the common ratio, and n is the number of terms.
In this case, you are given that a = -3 and r = 3. You need to find the sum of the first four terms, so n = 4.
Step 1: Plug in the given values into the formula:
Sn = -3 * (1 - 3^4) / (1 - 3)
Step 2: Simplify the expression inside the parentheses:
Sn = -3 * (1 - 81) / (1 - 3)
Step 3: Simplify the expression further:
Sn = -3 * (-80) / (-2)
Step 4: Divide to get the final answer:
Sn = (-3 * -80) / (-2)
Step 5: Calculate the expression:
Sn = 240 / 2
Step 6: Final step, divide to find the sum:
Sn = 120
Therefore, the sum of the first four terms of the geometric sequence with a = -3 and r = 3 is 120.