Answered By Jason R

At the Math 11 level, there will be problems involving two types of series. This series is a geometric one because each term is 3 times the one before.There is a formula for the sum of a geometric series. This formula will usually be given to you on a formula sheet.S_n=t₁[(rⁿ-1)/(r-1)] , (assuming r is not equal to 1)

t₁ is the first term of the series. For this one, it is 4.

r is the common ratio. That is 3 for this one. Each term is 3 times the previous one.

s_n is 4372, that is the sum of the first n terms.

From here, we use algebra to solve the equation for rⁿ (= 3ⁿ).

4372 = 4[(3ⁿ-1)/(3-1)]

4372x2/4=3ⁿ-1

2186+1=3ⁿ

^{2187=3n Now we need a way to find what power of 3 = 2187.}

^{If you haven't learned to use logarithms yet, then maybe trial and error would be quicker.}

^{Keep multiplying 3s until you reach 2187.}

^{It will take 7 of them. 3x3x3x3x3x3x3=2187 so n=7.}

^{I hope this has helped you.}