find the number of terms in a gp given that its first and last terms are 5 1/3k and 243/256k respectively and that its common ratio is 3/4?
3 answers
Please solve this question please
ar^(n-1) = (243/256)k , but a = (16/3)k
(16/3)k r^(n-1) = (243/256)k
r^(n-1) = 729/4096
but r = 3/4
(3/4)^(n-1) = 729/4096
on my calculator I found (3/4)^6 = 729/4096
so n-1 = 6
n = 7
we have 7 terms
(16/3)k r^(n-1) = (243/256)k
r^(n-1) = 729/4096
but r = 3/4
(3/4)^(n-1) = 729/4096
on my calculator I found (3/4)^6 = 729/4096
so n-1 = 6
n = 7
we have 7 terms
please I don't understand how you got 6, please put more light to the question