Step 1: Determine the first term (a) and common difference (d).
In this sequence, the first term (a) is -15. The common difference (d) is 2 because each term is increasing by 2.
Step 2: Determine the last term (l).
The last term can be found using the formula:
l = a + (n - 1)d
where n is the number of terms.
Substitute the values we know:
13 = -15 + (n - 1)2
Simplify and solve for n:
28 = 2n - 2
30 = 2n
n = 15
Therefore, the number of terms in the sequence is 15.
Find the number of terms in a sequence; -15,-13,-11,-9...,13
5 answers
Find the summation of 16 when r=1(5r-1).
To find the summation of 16 when r=1(5r-1), we need to plug in the values of r and add up the resulting terms:
r = 1: 5(1) - 1 = 4
16(4) = 64
So when r = 1, the term is 64.
The problem doesn't specify what values of r to sum over, so we'll assume it means to sum over all positive integers less than infinity (in other words, an infinite series). We can write this as:
Σ[16(5r-1)] from r=1 to infinity
To determine whether this series converges or diverges, we can use the ratio test:
a_n = 16(5n-1)
a_{n+1} = 16(5(n+1)-1) = 16(5n+4) = 80n + 64
|a_{n+1}/a_n| = |(80n + 64) / (16(5n-1))| = |(5n + 4)/(n - 1)|
As n approaches infinity, this ratio approaches 5. Since the ratio is greater than 1, the series diverges.
Therefore, the summation of 16 when r=1(5r-1) does not converge.
r = 1: 5(1) - 1 = 4
16(4) = 64
So when r = 1, the term is 64.
The problem doesn't specify what values of r to sum over, so we'll assume it means to sum over all positive integers less than infinity (in other words, an infinite series). We can write this as:
Σ[16(5r-1)] from r=1 to infinity
To determine whether this series converges or diverges, we can use the ratio test:
a_n = 16(5n-1)
a_{n+1} = 16(5(n+1)-1) = 16(5n+4) = 80n + 64
|a_{n+1}/a_n| = |(80n + 64) / (16(5n-1))| = |(5n + 4)/(n - 1)|
As n approaches infinity, this ratio approaches 5. Since the ratio is greater than 1, the series diverges.
Therefore, the summation of 16 when r=1(5r-1) does not converge.
Evaluate Σr=1^16(r-1)
We can expand the sum as follows:
Σr=1^16(r-1) = (1-1) + (2-1) + (3-1) + ... + (15-1) + (16-1)
= 0 + 1 + 2 + ... + 14 + 15
This is an arithmetic series with a common difference of 1. We can use the formula for the sum of an arithmetic series to find the answer:
S = (n/2)(a + l)
where S is the sum, n is the number of terms, a is the first term, and l is the last term.
In this case, n = 16, a = 0, and l = 15. Plugging in the values, we get:
S = (16/2)(0 + 15) = 8(15) = 120
Therefore, Σr=1^16(r-1) = 120.
Σr=1^16(r-1) = (1-1) + (2-1) + (3-1) + ... + (15-1) + (16-1)
= 0 + 1 + 2 + ... + 14 + 15
This is an arithmetic series with a common difference of 1. We can use the formula for the sum of an arithmetic series to find the answer:
S = (n/2)(a + l)
where S is the sum, n is the number of terms, a is the first term, and l is the last term.
In this case, n = 16, a = 0, and l = 15. Plugging in the values, we get:
S = (16/2)(0 + 15) = 8(15) = 120
Therefore, Σr=1^16(r-1) = 120.