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I am Lyosha [343]
2 years ago
9

For some integer n, the first, the third and the fifth terms of an arithmetic sequence are respectively 3n, 5n – 6, and 11n + 8.

What is the fourth term?
Mathematics
1 answer:
Mazyrski [523]2 years ago
8 0

Answer:

a₄=8n+1= -39.

Step-by-step explanation:

1) if a₁=3n; a₃=5n-6 and a₅=11n+8, then it is possible to calculate the difference according to 0.5(a₅-a₃)=0.5(a₃-a₁). Then

2) 0.5(11n+8-5n+6)=0.5(5n-6-3n); ⇔ 6n+14=2n-6; ⇔ n= -5.

3) if n=-5, then the 4th term is:

a_4=\frac{a_5+a_3}{2}; \ = > \ a_4=\frac{11n+8+5n-6}{2}=8n+1;

or a₄=-39.

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The lengths of three sides of a quadrilateral are shown below:
julsineya [31]
Answers: 
Part A: 12y² + 10y – 21
Part B: 4y³ + 6y² + 6y – 5 
Part C: See below.

Explanations: 
Part A: 
For this part, you add Sides 1, 2 and 3 together by combining like terms:
Side 1 = 3y² + 2y – 6
Side 2 = 4y² + 3y – 7
Side 3 = 5y² + 5y – 8
3y² + 2y – 6 + 4y² + 3y – 7 + 5y² + 5y – 8
Combine like terms: 
3y² + 4y² + 5y² + 2y + 3y + 5y – 6 – 7 – 8 
12y² + 10y – 21

Part B: 
You have the total perimeter and the sum of three of the sides, so you just need that fourth side value, which we can call d.
P = 4y³ + 18y² + 16y – 26
Sides 1, 2 & 3 = 12y² + 10y – 21 
Create an algebraic expression: 
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
Solve for d: 
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
– 12y²                                – 12y²
10y – 21 + d = 4y³ + 6y² + 16y – 26 
– 10y                               – 10y
– 21 + d = 4y³ + 6y² + 6y – 26 
+ 21                                + 21 
d = 4y³ + 6y² + 6y – 5 

Part C: 
If closed means that the degree that these polynomials are at stay that way, then yes, this is true in these cases because you will notice that each side had a y², y and no coefficient value except for the fourth one. This didn't change, because you only add and subtract like terms.
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