If the third and sixth terms of a geometric progression are 12 and 96, then

Question

find the number of terms in the progression, which are less than 2000.

Answer 1

clearly, d=28 and the sequence is

-44, -16, 12, 40, 68, 96, ...

hint: 71*28 = 1988

but you didn't start at zero.

Answer 2

in a GS
3rd term = ar^2 = 12
6th term = ar^5 = 96
divide them
r^3 = 96/12 = 8
r = 2
then in ar^2 = 12
4a = 12
a = 3

so you have 3 + 6 + 12 + ... +(the last value before 2000)

so term(n) = ar^(n-1) < 2000
3(2^(n-1) ) = ..
if n = 9, 3(2^8) = 768
n = 10 , 3(2^9) = 1536
n = 11 , 3(2^10) = 3072

so there are 10 terms
Sum(10) = 3 (2^10 - 1)/(2-1) = 3069

Answer 3

Thanks, Reiny. I'm glad someone around here can read!

Answer 4

Thanks Steve and Reiny for helping.