The sum of the first #4# terms of a geometric series is #20# and the sum of the first #7# terms is #546.5#. What is the common ratio?

1 Answer
Feb 25, 2018

The real solution is #r=3#. There are other non-real solutions.

Explanation:

Given a geometric sequence with sum of the first four terms #20# and sum of the first #7# terms #546.5#, find the common ratio.

The general term of a geometric series can be written:

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

where #a# is the initial term and #r# the common ratio.

We find:

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

#color(white)((1-r) sum_(n=1)^N a r^(n-1)) = sum_(n=1)^N a r^(n-1) - sum_(n=2)^(N+1) a r^(n-1)#

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

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

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

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

In our example, we are told #s_4 = 20# and #s_7 = 546.5#

So:

#{ ((a(1-r^4))/(1-r) = 20), ((a(1-r^7))/(1-r) = 546.5) :}#

Dividing the second equation by the first, we find:

#(1-r^7)/(1-r^4) = 546.5/20 = 27.325#

Note that this is close to #3^3 = 27#, So try putting #r=3# and find:

#(1 - (color(blue)(3))^7)/(1-(color(blue)(3))^4) = 2186/80 = 27.325#

So the real solution is #r=3#

Then:

#20 = (a(1-r^4))/(1-r) = (-80a)/(-2) = 40a#

So:

#a = 20/40 = 1/2#