I'm pretty sure these are right but I just want to check.
1)Find the 20th term of the arithmetic sequence in which a1=3 and d=7
a.143
b.136
c.140
d.133
answer=b
2)Write an equation for the nth term of the arithmetic sequence -3,3,9,15...
a.an=n+6
b.an=6n+9
c.an=6n-9
d.an=n-3
answer=b
3)Find the two arithmetic means between 4 and 22
a.10,16
b.8,16
c.8,12
d.13,13
answer=a
4)Simplify:Find Snfor the arithmetic series in which a1=3,d=1/2,and an=15
a.225
b.9
c.45
d.210
answer=b
5)I don't know how to type this but its like this in my book.
Find; 8 is on top of the sigma notation and k=3 is at the bottom(40-3k) is on the right side
a.45
b.282
c.-90
d.141
answer=d
1) To find the 20th term of an arithmetic sequence, you can use the formula: an = a1 + (n-1)d, where an is the nth term, a1 is the first term, d is the common difference, and n is the term number you want to find.
In this case, a1 = 3 and d = 7. Plugging these values into the formula, we get:
a20 = 3 + (20-1)7
a20 = 3 + 133
a20 = 136
Therefore, the 20th term of the sequence is 136, so the answer is b) 136.
2) To find the equation for the nth term of an arithmetic sequence, you need to identify the pattern in the sequence. In this case, the common difference is 6, as each term is obtained by adding 6 to the previous term.
The first term is -3 (when n = 1), and the second term is 3 (when n = 2). Notice that the difference between these two terms is 6, which is the common difference.
Using this information, we can determine that the equation for the nth term is of the form: an = dn + c, where d is the common difference and c is a constant term.
Since d = 6, the equation becomes: an = 6n + c.
To find the value of c, we can substitute the values of one of the terms. Let's use the third term, which is 9 (when n = 3).
9 = 6(3) + c
9 = 18 + c
c = 9 - 18
c = -9
So the equation for the nth term of the sequence is: an = 6n - 9. Therefore, the answer is b) an = 6n + 9.
3) To find the two arithmetic means between two given terms, we need to determine the common difference and the terms in between.
The given terms are 4 and 22. The common difference can be found by subtracting the first term (4) from the second term (22):
d = 22 - 4
d = 18
For arithmetic means, we have:
First mean = First term + d
Second mean = First term + 2*d
Plugging in the values, we get:
First mean = 4 + 18
First mean = 22
Second mean = 4 + 2*18
Second mean = 40
Therefore, the two arithmetic means between 4 and 22 are 22 and 40. Thus, the answer is a) 10,16.
4) To find the sum (Sn) of an arithmetic series, you can use the formula: Sn = (n/2)(a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the nth term.
In this case, a1 = 3, d = 1/2, and an = 15. We need to determine the value of n, which is the number of terms.
Using the formula for the nth term in an arithmetic sequence (an = a1 + (n-1)d), we can substitute the values:
15 = 3 + (n-1)(1/2)
15 = 3 + (n/2) - (1/2)
15 - 3 = (n/2) - (1/2)
12 = n/2 - 1/2
12 + 1/2 = n/2
25/2 = n/2
n = 25
Now, substituting the values into the sum formula:
Sn = (25/2)(3 + 15)
Sn = (25/2)(18)
Sn = 225
Therefore, the sum of the arithmetic series is 225. So the answer is a) 225.
5) The question implies that you need to evaluate a sigma notation (sum). The given expression is 40 - 3k, and k ranges from 3 to 8 (based on "k=3 is at the bottom").
To find the sum, you need to substitute each value of k into the expression and add them together. Let's calculate:
Sum = (40 - 3(3)) + (40 - 3(4)) + (40 - 3(5)) + (40 - 3(6)) + (40 - 3(7)) + (40 - 3(8))
= (40 - 9) + (40 - 12) + (40 - 15) + (40 - 18) + (40 - 21) + (40 - 24)
= 31 + 28 + 25 + 22 + 19 + 16
= 141
Therefore, the sum of the given sigma notation is 141. Hence, the answer is d) 141.