The sum of the first n terms of an exponential series is 364. The sum of their reciprocal is 364/243. If the first term is 1,find n and the common ratio.

Answer 1

Sum(n) = a(r^n - 1)/(r-1), but a=1
= (r^n - 1)/(r-1)
(r^n - 1)/(r-1) = 364 ---> #1

the sequence of reciprocals is
1 + 1/r + 1/r^2 + ...
now, a=1, common ratio is 1/r
sum(n) = ( (1/r)^n - 1)/( 1/r - 1)
= ( 1/r^n - 1)/(1/r - 1)
= ( (1-r^n)/r^n ) / ( (1-r)/r )
= (1 - r^n)/(r(1-r))
= (r^n - 1)/(r(r-1)) = 364/243 ---> #2

divide #2 by #1
1/r^(n-1) = (364/243) / 364
1/r^(n-1) = 1/243
r^(n-1) = 243

now we know that n has to be a whole number
and I recognize 243 as 3^5
so let's try r = 3, and n = 6

then our sequence is 1, 3, 9, 27, 81, 243, 729, ...
sum(6) = 1(3^6 - 1)/(3-1) = 364 , yeahhh!

the reciprocal series is
1 + 1/3 + 1/9 + ...
sum(6) = ( (1/3)^6 - 1)/(1/3 - 1)
= (-728/729) / (-2/3)
= (728/729)(3/2)
= 364/243

r=3 , n=6