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
Jun 1, 2016

#392160#

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 #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#