What is the sum of a 6–term geometric series if the first term is 20 and the last term is 336,140?
1 Answer
Explanation:
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
So assuming that the geometric series is of Real numbers, the only possible value of
Given any geometric series and positive integer
#(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
#sum_(n=1)^N a r^(n-1) = (a(r^N-1))/(r-1)#
In our example, we have
#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#