A certain 300-term geometric sequence has first term 1337 and common ratio -1/2. How many terms of this sequence are greater than 1?

1337 * (-1/2)^n <= 1

Since
2^10 = 1024
2^11 = 2048

|Tn|>1 for n<=11
But, Tn < 0 for n even, so only T1,3,5,7,9,11 are greater than 1. So, just 6 terms.

6 is correct

Well, this sequence sure knows how to flip things around! With a common ratio of -1/2, it's like watching a reverse magic trick. But let me do some math juggling for you!

To find out how many terms of this sequence are greater than 1, we need to determine when our decreasing sequence finally dips below the number 1. So, let me perform some calculations... Ta-da!

By applying the formula for a geometric sequence, we can see that the nth term is given by:

an = a1 * r^(n-1)

With a first term of 1337 and a common ratio of -1/2, we can plug these values into the formula:

1337 * (-1/2)^(n-1)

Okay, let's make sure this sequence stays above 1. Solving the inequality:

1337 * (-1/2)^(n-1) > 1

After a little bit of number crunching, the result is:

n > 9.63

Ah, looks like our secret reveals itself! Since we can't have fractional terms in this sequence, we need to round up. So, drumroll please... the answer is 10 terms!

Which just goes to show, even in a topsy-turvy sequence, there's always something greater waiting around the corner!

To find the number of terms in the geometric sequence that are greater than 1, we need to determine the terms of the sequence and count how many of them are greater than 1.

The formula for the nth term of a geometric sequence is given by:

an = a1 * r^(n-1)

Where:
an = nth term
a1 = first term
r = common ratio
n = term number

In this case, the first term (a1) is 1337 and the common ratio (r) is -1/2. Let's find the terms of the sequence:

a1 = 1337
a2 = 1337 * (-1/2)^1 = -1337/2
a3 = 1337 * (-1/2)^2 = 668.5
a4 = 1337 * (-1/2)^3 = -334.25

From this observation, we can see that the terms alternate between positive and negative values. Since the common ratio (-1/2) is less than 1 in absolute value, the magnitude of the terms will keep decreasing.

To find the number of terms greater than 1, we need to determine when the magnitude of the terms becomes less than or equal to 1. Looking at the sequence, we see that the magnitude becomes less than or equal to 1 for the 4th term, which is -334.25.

Since the magnitude of the terms keep decreasing, we know that the terms after the 4th term will also be less than or equal to 1. Therefore, there are only 3 terms in the sequence that are greater than 1.

Therefore, the answer is 3 terms.

To determine how many terms of the given geometric sequence are greater than 1, we need to find the value of the 300th term and then count the number of terms that satisfy the condition.

To find the value of the 300th term of the geometric sequence, we use the formula for the nth term of a geometric sequence:

a_n = a_1 * r^(n-1)

Where:
a_n = nth term
a_1 = first term
r = common ratio

Given that the first term (a_1) is 1337 and the common ratio (r) is -1/2, we substitute these values into the formula:

a_300 = 1337 * (-1/2)^(300-1)

Now we can calculate the value of the 300th term:

a_300 = 1337 * (-1/2)^299

To determine the number of terms greater than 1, we need to compare the value of each term in the sequence to 1. We can do this by iterating from the first term to the 300th term and counting the terms that are greater than 1.

Let's write a loop to calculate the number of terms greater than 1:

count = 0
for n = 1 to 300:
a_n = 1337 * (-1/2)^(n-1)
if a_n > 1:
count += 1

Finally, the value of the variable "count" will represent the number of terms in the sequence that are greater than 1.