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
The real solution is
Explanation:
Given a geometric sequence with sum of the first four terms
The general term of a geometric series can be written:
#a_n = a r^(n-1)#
where
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
#s_N = sum_(n=1)^N a r^(n-1) = (a(1-r^N))/(1-r)#
In our example, we are told
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
#(1 - (color(blue)(3))^7)/(1-(color(blue)(3))^4) = 2186/80 = 27.325#
So the real solution is
Then:
#20 = (a(1-r^4))/(1-r) = (-80a)/(-2) = 40a#
So:
#a = 20/40 = 1/2#