The sum of the first three terms of a geometric series is 37/8. The sum of the first six terms is 3367/512. Find the first term and common ratio
a + ar + ar^2 = 37/8 ---> a(1 + r + r^2) = 37/8 , #1
37/8 + ar^3 + ar^4 + ar^5 = 3367/512
ar^3(a + r + r^2) = 3367/512 - 37/8 = 999/512 , #2
divide the 2nd by the 1st :
r^3 = 27/64
r = 3/4
back into #1:
a(1 + 3/4 + 9/16) = 37/8
a(37/16) = 37/8
a = 2
So the first term is 2, and the common ratio is 3/4
Proof:
the terms are ...
2 , 3/2, 9/8, 27/32, 81/128, 243/512, ...
(2 + 3/2 + 9/8) = 37/8
sum of all 6 terms = 37/8 + (27/32 + 81/128 + 243/512) = 3367/512
Looks like this was a bit too hard for the bot.