If the third and sixth terms of a geometric progression are 12 and 96, then
find the number of terms in the progression, which are less than 2000.
clearly, d=28 and the sequence is
-44, -16, 12, 40, 68, 96, ...
hint: 71*28 = 1988
but you didn't start at zero.
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
Thanks, Reiny. I'm glad someone around here can read!
Thanks Steve and Reiny for helping.
To solve this problem, we need to find the common ratio (r) of the geometric progression first. Once we have the common ratio, we can determine if the terms of the progression are less than 2000.
Let's denote the first term of the geometric progression as 'a' and the common ratio as 'r'. Based on the given information, we can write the following equations:
a * r^2 = 12 (Third term)
a * r^5 = 96 (Sixth term)
To find the common ratio, we can divide the second equation by the first equation:
(a * r^5) / (a * r^2) = 96 / 12
r^3 = 8
r = 2
Now that we have the common ratio (r = 2), we can determine the terms of the progression that are less than 2000.
The nth term of the geometric progression can be calculated using the formula:
Tn = a * r^(n-1)
We need to find the highest value of n for which Tn is less than 2000. Let's substitute 2000 in the formula and solve for n:
2000 = a * 2^(n-1)
To simplify the equation, we can divide both sides by 'a':
2000/a = 2^(n-1)
Since 2 is a power of 2, we can rewrite the equation as:
1000/a = 2^n
Take the logarithm of both sides:
log(1000/a) = n * log(2)
Solve for n:
n = log(1000/a) / log(2)
We know that the common ratio (r) is 2, so the third term of the progression is given by:
12 = a * 2^2
12 = 4a
a = 3
Substituting the value of a in the equation for n:
n = log(1000/3) / log(2)
Using a calculator, we can determine the value of n approximately.
Hence, the number of terms in the geometric progression that are less than 2000 is approximately the calculated value of n.