the sum of 4 + 12 + 36+ 108+ ... +tn is 4372. How many terms are in the series?

2 years ago

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}=t_{1}[(r^{n}-1)/(r-1)] , (assuming r is not equal to 1)

t_{1} 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 r^{n }(= 3^{n}).

4372 = 4[(3^{n}-1)/(3-1)]

4372x2/4=3^{n}-1

2186+1=3^{n}

^{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.}

2 years ago

## 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}=t_{1}[(r^{n}-1)/(r-1)] , (assuming r is not equal to 1)t

_{1}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 r

^{n }(= 3^{n}).4372 = 4[(3

^{n}-1)/(3-1)]4372x2/4=3

^{n}-12186+1=3

^{n}^{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.}