What is the sum of a 6–term geometric series if the first term is 20 and the last term is 336,140?

Question

Caleb Alexander · Accepted Answer

#392160#

Given:

#a_1 = 20#

#a_6 = 336140#

Then general form of a term of a geometric series is:

#a_n = a*r^(n-1)#

So #r^5 = (a r^5)/(a r^0) = a_6/a_1 = 336140/20 = 16807 = 7^5#

So assuming that the geometric series is of Real numbers, the only possible value of #r# is #7#.

Given any geometric series and positive integer #N#, we find:

#(r-1) sum_(n=1)^N a r^(n-1)#

#=r sum_(n=1)^N a r^(n-1) - sum_(n=1)^N a r^(n-1)#

#=ar^N + color(red)(cancel(color(black)(sum_(n=2)^(N-1) a r^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^(N-1) a r^(n-1)))) - a#

#=a(r^N-1)#

So dividing both ends by #(r-1)# we find:

#sum_(n=1)^N a r^(n-1) = (a(r^N-1))/(r-1)#

In our example, we have #a=20#, #r=7#, #N=6# and we get:

#sum_(n=1)^N a r^(n-1) = (a(r^N-1))/(r-1)#

#= (20(7^6-1))/(7-1) = (20(117649-1))/6 = 2352960/6 = 392160#

Cora Alexander · Answer

undefined To find the sum of a geometric series, you can use the formula:

$$S_n = \frac{{a(1 - r^n)}}{{1 - r}}$$

Where:
- $S_n$ is the sum of the first $n$ terms,
- $a$ is the first term,
- $r$ is the common ratio, and
- $n$ is the number of terms.

Given $a = 20$ and the last term $336,140$, you can find the common ratio $r$ using the formula for the nth term of a geometric sequence:

$$a_n = a \times r^{(n-1)}$$

Since the last term is the 6th term, you have:

$$336,140 = 20 \times r^{(6-1)}$$

Solving for $r$:

$$r^5 = \frac{336,140}{20}$$
$$r^5 = 16,807$$

$$r = \sqrt[5]{16,807}$$

Now that you have $r$, you can find the sum $S_6$ using the formula for the sum of a geometric series:

$$S_6 = \frac{{20(1 - \sqrt[5]{16,807}^6)}}{{1 - \sqrt[5]{16,807}}}$$

$$S_6 = \frac{{20(1 - 9)}}{{1 - 9}}$$

$$S_6 = \frac{{20 \times (-8)}}{{-8}}$$

$$S_6 = -20 \times 8$$

$$S_6 = -160$$

Therefore, the sum of the 6-term geometric series is $-160$.