A geometric series has t1 = 3 and r = 2. If tn = 96. What does n equal in this problem?
What is the special name given to the terms between 3 and 96 in this problem?
You are given that a = 3 and r = 2
so
tn = ar^(n-1)
96 = 3(2^(n-1)
32 = 2^(n-1)
2^5 = 2^(n-1)
n-1 = 5
n = 6
3, 6, 12, 24, 48, 96
terms in between are called geometric means
Thanks!
To find the value of n, we can use the formula for the nth term of a geometric series:
tn = t1 * r^(n-1)
Given that t1 = 3, r = 2, and tn = 96, we can substitute these values into the formula:
96 = 3 * 2^(n-1)
Dividing both sides of the equation by 3, we get:
32 = 2^(n-1)
To simplify the equation, we can rewrite 32 as 2^5:
2^5 = 2^(n-1)
Since the bases are the same, we can equate the exponents:
5 = n - 1
Adding 1 to both sides of the equation, we find:
n = 6
Therefore, n equals 6.
The special name given to the terms between 3 and 96 in this problem is the common ratio. In this case, the common ratio is 2.
To find the value of n in this geometric series problem, we first need to understand the formula for the nth term of a geometric series. The formula is given by:
tn = t1 * r^(n-1)
In this case, we are given t1 = 3 and r = 2, and we are trying to find n when tn = 96.
Plugging these values into the formula, we get:
96 = 3 * 2^(n-1)
To solve for n, we need to isolate it on one side of the equation. Let's divide both sides of the equation by 3:
96/3 = 2^(n-1)
32 = 2^(n-1)
Now, we need to solve for the exponent (n-1). Since 2^5 = 32, we can rewrite the equation as:
2^(n-1) = 2^5
By comparing the exponents, we can conclude that:
n - 1 = 5
Adding 1 to both sides of the equation, we find:
n = 6
Therefore, the value of n in this problem is 6.
Now, to answer the second part of your question, the terms between 3 and 96 in this problem are called "intermediate terms" or "interior terms" of the geometric series. These terms lie between the first term (t1 = 3) and the last term (tn = 96) in the series.