The answer is 1 : 1 : square root of 3
Answer:Answer:

Step-by-step explanation:
Given the sequence -4,-6,-8..., in order to get sigma notation to represent the sum of the first seven terms of the sequence, we need to first calculate the sum of the first seven terms of the sequence as shown;
The sum of an arithmetic series is expressed as ![S_n = \frac{n}{2}[2a+(n-1)d]](https://tex.z-dn.net/?f=S_n%20%3D%20%5Cfrac%7Bn%7D%7B2%7D%5B2a%2B%28n-1%29d%5D)
n is the number of terms
a is the first term of the sequence
d is the common difference
Given parameters
n = 7, a = -4 and d = -6-(-4) = -8-(-6) = -2
Required
Sum of the first seven terms of the sequence
![S_7 = \frac{7}{2}[2(-4)+(7-1)(-2)]\\\\S_7 = \frac{7}{2}[-8+(6)(-2)]\\\\S_7 = \frac{7}{2}[-8-12]\\\\\\S_7 = \frac{7}{2} * -20\\\\S_7 = -70](https://tex.z-dn.net/?f=S_7%20%3D%20%5Cfrac%7B7%7D%7B2%7D%5B2%28-4%29%2B%287-1%29%28-2%29%5D%5C%5C%5C%5CS_7%20%3D%20%20%5Cfrac%7B7%7D%7B2%7D%5B-8%2B%286%29%28-2%29%5D%5C%5C%5C%5CS_7%20%3D%20%20%5Cfrac%7B7%7D%7B2%7D%5B-8-12%5D%5C%5C%5C%5C%5C%5CS_7%20%3D%20%5Cfrac%7B7%7D%7B2%7D%20%2A%20-20%5C%5C%5C%5CS_7%20%3D%20-70)
The sum of the nth term of the sequence will be;
![S_n = \frac{n}{2}[2(-4)+(n-1)(-2)]\\\\S_n = \frac{n}{2}[-8+(-2n+2)]\\\\S_n = \frac{n}{2}[-6-2n]\\\\S_n = \frac{-6n}{2} - \frac{2n^2}{2}\\S_n = -3n-n^2\\\\S_n = -n(3+n)](https://tex.z-dn.net/?f=S_n%20%3D%20%5Cfrac%7Bn%7D%7B2%7D%5B2%28-4%29%2B%28n-1%29%28-2%29%5D%5C%5C%5C%5CS_n%20%3D%20%5Cfrac%7Bn%7D%7B2%7D%5B-8%2B%28-2n%2B2%29%5D%5C%5C%5C%5CS_n%20%3D%20%5Cfrac%7Bn%7D%7B2%7D%5B-6-2n%5D%5C%5C%5C%5CS_n%20%3D%20%20%5Cfrac%7B-6n%7D%7B2%7D%20-%20%20%5Cfrac%7B2n%5E2%7D%7B2%7D%5C%5CS_n%20%3D%20-3n-n%5E2%5C%5C%5C%5CS_n%20%3D%20-n%283%2Bn%29)
The sigma notation will be expressed as
. <em>The limit ranges from 1 to 7 since we are to find the sum of the first seven terms of the series.</em>
Hello!

x² + 5x + 6
Find two numbers that add up to 5 and multiply to 6. We get:
2, 3
Therefore:
(x + 2)(x + 3)
In the matrix, the first row is exactly twice the second. This implies that the determinant is zero.
If the matrix system has determinant zero, the system has either no solutions or infinite solutions.
However, just like in the matrix, in the constant vector the first term is twice the second. This means that the system has infinite solutions.
In fact, if you write this system in expanded form, you can see that it generates two equations, the first being twice the second:
