What is the sum of a 6–term geometric series if the first term is 22 and the last term is 1,299,078?

1 Answer
Jun 15, 2016

Sum of the series is #1,461,460#

Explanation:

In a geometric series #{a,ar,ar^2,ar^3,.........}# whose first term is #a# and ratio between a term and its preceeding tetm is #r#, the #n^(th)# term is given by #ar^(n-1)# and sum of first #n# terms is given by #axx(r^n-1)/(r-1)#.

Here as first term is #22# and sixth term is #1299078#, as such #a=22# and #ar^5=1299078# or #r^5=1299078/22=9^5# i.e. #r=9#.

Hence sum of the series is #22xx(9^6-1)/(9-1)=22xx66,430=1,461,460#