1. Find the 11th term of the geometric sequence 64, -32, 16, -8, … .
2. Find the 1st term of a geometric sequence with a 10th term -1024 and r = -2.
3. Find the sum of each infinite geometric series, if possible.
a. 10 + 2 + 2/5 + .....
b. 1.2 – 2.4 + 4.8 – 9.6 + .....
c. 3 – 1 + 1/3 – 1/9 + 1/27 + ...
1. Oh the drama of geometric sequences, always multiplying and dividing! Alright, let's find the 11th term of this sequence. The first term is 64 and the common ratio is -1/2. We can use the formula for geometric sequences:
nth term = first term × common ratio^(n - 1)
Plugging in the values, we get:
11th term = 64 × (-1/2)^(11 - 1)
11th term = 64 × (-1/2)^10
Now, let me grab my calculator... drumroll, please! (tap, tap, tap) The 11th term is 0.125! Ta-da!
2. Ah, another geometric sequence trying to keep us on our toes. This time, we know the 10th term is -1024 and the common ratio is -2. We can use the same formula as before:
nth term = first term × common ratio^(n - 1)
Plugging in the given values, we get:
-1024 = first term × (-2)^(10 - 1)
-1024 = first term × (-2)^9
Okay, calculator to the rescue once again... (tap, tap, tap) The first term is... -2,147,483,648! Whoa, that's a BIG number!
3. Ah, infinite geometric series, a never-ending story! Let's see if we can calculate them.
a. The first term is 10 and the common ratio is 2/5. Applying the infinite geometric series formula:
sum = first term / (1 - common ratio)
Plugging in the numbers, we get:
sum = 10 / (1 - 2/5)
sum = 10 / (3/5)
sum = 50/3, which is approximately 16.6666667. Ah, the sweet taste of infinite fractions!
b. The first term is 1.2, but hold onto your funny bone because this sequence alternates positive and negative! The common ratio is -2. Using the same formula:
sum = first term / (1 - common ratio)
Plugging in the values, we get:
sum = 1.2 / (1 - (-2))
sum = 1.2 / 3
sum = 0.4. Voila! The sum is 0.4. It may be a bit confusing with the dancing positives and negatives, but we made it!
c. Brace yourself, this is an infinite geometric series with the first term being 3 and the common ratio being 1/3. Applying the infinite geometric series formula again:
sum = first term / (1 - common ratio)
Plugging in the numbers, we get:
sum = 3 / (1 - 1/3)
sum = 3 / (2/3)
sum = 9/2. Oh, look at that! The sum is 4.5. Fantastic!
I hope these answers tickled your funny bone!
1. In a geometric sequence, the nth term can be found using the formula: a₁₀ = a₁ * r^(n-1), where a₁₀ represents the nth term, a₁ is the first term, r is the common ratio, and n is the position of the term.
In the given geometric sequence 64, -32, 16, -8, ..., the first term is 64 and the common ratio is -2. We want to find the 11th term, so n = 11.
Using the formula, we can calculate the 11th term:
a₁₁ = 64 * (-2)^(11-1)
= 64 * (-2)^10
= 64 * 1024
= 65536
Therefore, the 11th term of the geometric sequence is 65536.
2. We can use the same formula, a₁₀ = a₁ * r^(n-1), to find the first term of a geometric sequence.
In this case, we are given the 10th term as -1024 and the common ratio as -2. We want to find the first term, so n = 1.
Using the formula, we can rearrange it to solve for a₁:
a₁ = a₁₀ / r^(n-1)
= -1024 / (-2)^(1-1)
= -1024 / (-2)^0
= -1024 / 1
= -1024
Therefore, the first term of the geometric sequence is -1024.
3. To find the sum of an infinite geometric series, we can use the formula: S = a / (1 - r), where S represents the sum, a is the first term, and r is the common ratio.
a. In the series 10 + 2 + 2/5 + ..., the first term is 10 and the common ratio is 2/5.
Using the formula, we can calculate the sum:
S = 10 / (1 - 2/5)
= 10 / (5/5 - 2/5)
= 10 / (3/5)
= 10 * (5/3)
= 50/3
Therefore, the sum of the series is 50/3.
b. In the series 1.2 - 2.4 + 4.8 - ..., the first term is 1.2 and the common ratio is -2.
Using the formula, we can calculate the sum:
S = 1.2 / (1 - (-2))
= 1.2 / (1 + 2)
= 1.2 / 3
= 0.4
Therefore, the sum of the series is 0.4.
c. In the series 3 - 1 + 1/3 - ..., the first term is 3 and the common ratio is -1/3.
Using the formula, we can calculate the sum:
S = 3 / (1 - (-1/3))
= 3 / (1 + 1/3)
= 3 / (4/3)
= 3 * (3/4)
= 9/4
Therefore, the sum of the series is 9/4.
To find the nth term of a geometric sequence, we can use the formula:
a_n = a_1 * r^(n-1)
where a_n is the nth term, a_1 is the first term, r is the common ratio, and n is the position of the term in the sequence.
Now, let's solve each of the questions step by step.
1. Find the 11th term of the geometric sequence 64, -32, 16, -8, … .
For this sequence, we can observe that the first term (a_1) is 64, and the common ratio (r) is -0.5.
To find the 11th term (a_11), we substitute the given values into the formula:
a_11 = 64 * (-0.5)^(11-1)
= 64 * (-0.5)^10
= 64 * 0.0009765625
= 0.0625
Therefore, the 11th term of the sequence is 0.0625.
2. Find the 1st term of a geometric sequence with a 10th term -1024 and r = -2.
In this case, we are given the 10th term (a_10) as -1024, and the common ratio (r) is -2. We need to find the 1st term (a_1).
Using the formula, we substitute the given values:
-1024 = a_1 * (-2)^(10-1)
-1024 = a_1 * (-2)^9
-1024 = a_1 * (-512)
a_1 = -1024 / (-512)
a_1 = 2
Therefore, the 1st term of the sequence is 2.
3. Find the sum of each infinite geometric series, if possible.
a. 10 + 2 + 2/5 + ......
To determine if the sum of an infinite geometric series exists, we need to check the common ratio (r) and ensure that it lies between -1 and 1. If it does, the formula for the sum of infinite geometric series can be used:
S = a / (1 - r)
In this example, the common ratio (r) is 2/5, which lies between -1 and 1. Therefore, the sum of this infinite geometric series exists.
Using the formula, we substitute the given values:
S = 10 / (1 - 2/5)
S = 10 / (3/5)
S = 50/3
Therefore, the sum of the infinite geometric series is 50/3.
b. 1.2 – 2.4 + 4.8 – 9.6 + ......
In this series, the common ratio (r) is -2. The absolute value of -2 is greater than 1, so the sum of this infinite geometric series does not exist.
c. 3 – 1 + 1/3 – 1/9 + 1/27 + ...
In this series, the common ratio (r) is -1/3. The absolute value of -1/3 is less than 1, so the sum of this infinite geometric series exists.
Using the formula, we substitute the given values:
S = 3 / (1 - (-1/3))
S = 3 / (1 + 1/3)
S = 3 / (4/3)
S = 9/4
Therefore, the sum of the infinite geometric series is 9/4.
#1
clearly, r = -1/2
T11 = ar^10
#2
a(-2)^9 = -1024
#3
Just recall that S = a/(1-r)
then plug in the numbers