The first three terms of a geometric series are #5, 15, 45#. If the nth term of this series is #10935#, what is the sum of the first n terms?

2 Answers
Mar 30, 2017

Answer:

16400

Explanation:

This is a geometric series with r = 3. First, find what term #10935# is.

#5 * (3^(n-1)) = 10935#

#(3^(n-1)) = 2187#

#3^(n-1) = 3^7#

#n-1 = 7#

#n = 8#

Now use the formula for the sum of the first #n# terms of a geometric series:

#S_n = (a_1(1 - r^n))/(1-r)#

#S_8 = (5(1 - 3^8)) / (1 - 3) = 16400#

Final Answer

Mar 30, 2017

Answer:

The sum is #16,400#.

Explanation:

Step 1: Classify the sequence

Since #t_2 = 3t_1# and #t_3 = 3t_2#, this sequence is geometric with #r = 3#.

Step 2: Find the number of terms

There is no formula we can use to evaluate the sum without knowing the number of terms. By the formula #t_n = a(r)^(n - 1)#, we have:

#10935 = 5(3)^(n - 1)#

#2187 = 3^(n - 1)#

#3^7 = 3^(n - 1)#

#7 = n - 1#

#n = 8#

Step 3: Evaluate the sum

The formula for the sum of a geometric series is #s_n = (a(1 - r^n))/(1 - r)#.

#s_8 = (5(1 - 3^8))/(1 - 3)#

#s_8 = (-32800)/(-2)#

#s_8 = 16,400#

Practice Exercises

#1#. Find the sum:

#2 + 8 + 32 +128 + ... + 524,288#

Solution

#1.#

#699,050#

Hopefully this helps!