The first three terms of a geometric series are 5, 15, 455,15,45. If the nth term of this series is 1093510935, what is the sum of the first n terms?

2 Answers
Mar 30, 2017

16400

Explanation:

This is a geometric series with r = 3. First, find what term 1093510935 is.

5 * (3^(n-1)) = 109355(3n1)=10935

(3^(n-1)) = 2187(3n1)=2187

3^(n-1) = 3^73n1=37

n-1 = 7n1=7

n = 8n=8

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

S_n = (a_1(1 - r^n))/(1-r)Sn=a1(1rn)1r

S_8 = (5(1 - 3^8)) / (1 - 3) = 16400S8=5(138)13=16400

Final Answer

Mar 30, 2017

The sum is 16,40016,400.

Explanation:

Step 1: Classify the sequence

Since t_2 = 3t_1t2=3t1 and t_3 = 3t_2t3=3t2, this sequence is geometric with r = 3r=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)tn=a(r)n1, we have:

10935 = 5(3)^(n - 1)10935=5(3)n1

2187 = 3^(n - 1)2187=3n1

3^7 = 3^(n - 1)37=3n1

7 = n - 17=n1

n = 8n=8

Step 3: Evaluate the sum

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

s_8 = (5(1 - 3^8))/(1 - 3)s8=5(138)13

s_8 = (-32800)/(-2)s8=328002

s_8 = 16,400s8=16,400

Practice Exercises

11. Find the sum:

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

Solution

1.

699,050

Hopefully this helps!