# 1)Find the 20th term of the arithmetic sequence in which a1=3 and d=7

a.143
b.136
c.140
d.133
an=a1+(n-1)d
a20=3+(20-1)7
a20=3+(19)7
a20=3+133

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
an=a1+(n-1)d
an=3+(n-1)-6
an=3-6n+6

3)Find the two arithmetic means between 4 and 22
a.10,16
b.8,16
c.8,12
d.13,13
an=a1+(n-1)d
a6=4+(6-1)d
22=4+6d

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
Sn=n/2(a1+an)
Sn=15/2(3+15)

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
40-3(3)=40-9=31
40-3(4)=40-12=28
40-3(5)=40-15=25
40-3(6)=40-18=22
40-3(7)=40-21=19
40-3(8)=40-24=16

## 2. Isn't a1=-3, and d=+6 ?

3--I have no idea what the question is asking.
5.

Sum= sigma(k=3,8) of (40-3k)

(40-3*3)+(40-3*4) + ... as you wrote the terms.
It will break down to
40*(8-2) - 3(3+4+5+6+7+8)=240-3*33=240-99= 141 as upi determined.
=

## Why don't mathematicians ever go to the circus? Because they have enough clowns in their textbooks! But don't worry, I'm here to provide all the amusement you need. Let's tackle those questions with a touch of laughter!

1) To find the 20th term of the arithmetic sequence, we use the formula an = a1 + (n-1)d. So, a20 = 3 + (20-1)7. Simplifying, we get a20 = 3 + 19*7 = 3 + 133 = 136. So the answer is b! And here I thought arithmetic couldn't be exciting!

2) Ah, the joy of finding equations! For the nth term of the arithmetic sequence -3, 3, 9, 15..., we use an = a1 + (n-1)d. In this case, a1 = -3 and d = 6 (not 7, they went on vacation without me!). Plugging those into the formula, we get an = -3 + (n-1)6. Simplifying, an = -3 + 6n - 6. Voila! The answer is b! 6n + 9, like a little dance of arithmetic happiness!

3) Time to find those arithmetic means! We start with a1 = 4 and a3 = 22. Using the formula an = a1 + (n-1)d, we find a3 = 4 + (3-1)d. Simplifying, we get 22 = 4 + 2d. Subtracting 4 from both sides, we have 18 = 2d. Dividing by 2, d = 9. Now we use the formula to find the first mean: a2 = a1 + (2-1)d. Plugging in the values, we get a2 = 4 + (2-1)9 = 4 + 9 = 13. Next, we find the second mean: a4 = a1 + (4-1)d. Again, plugging in the values, we get a4 = 4 + (4-1)9 = 4 + 3*9 = 4 + 27 = 31. Ta-dah! The answer is d! Two means of 13 and 13, like a pair of identical twins.

4) Let's simplify and find Sn! We have a1 = 3, d = 1/2, and an = 15. Using the formula Sn = n/2(a1 + an), we get Sn = (n/2)(3 + 15). Simplifying, we have Sn = (n/2)(18), which equals 9n. Hooray! The answer is b! 9, like a non-stop roller coaster of arithmetic excitement.

5) I can't see the specific notation you're referring to, but let me attempt to bring the chuckles anyway! It seems like you need to evaluate the expression 40 - 3k, with k ranging from 3 to 8. So, let's plug in those values one by one:

40 - 3(3) = 40 - 9 = 31
40 - 3(4) = 40 - 12 = 28
40 - 3(5) = 40 - 15 = 25
40 - 3(6) = 40 - 18 = 22
40 - 3(7) = 40 - 21 = 19
40 - 3(8) = 40 - 24 = 16

Ah, the answer is d! 141, like a delightful circus performance. Now go grab some cotton candy and enjoy the show!

## 1) To find the 20th term of an arithmetic sequence, we can use the formula an = a1 + (n-1)d, where an represents the nth term, a1 is the first term, and d is the common difference. In this case, a1 is 3 and d is 7. Plugging these values into the formula, we get a20 = 3 + (20-1)7 = 3 + 19*7 = 3 + 133 = 136. Therefore, the answer is b) 136.

2) To write an equation for the nth term of an arithmetic sequence, we can look for patterns in the sequence. In this case, the sequence starts with -3 and each term increases by 6. So, the common difference (d) is 6. By plugging in the first term (a1) and the common difference (d) into the formula an = a1 + (n-1)d, we get an = -3 + (n-1)6 = -3 + 6n - 6 = 6n - 9. Therefore, the answer is b) 6n + 9.

3) To find the two arithmetic means between 4 and 22, we need to find the values of a3 and a4 in the arithmetic sequence. Using the formula an = a1 + (n-1)d where a1 is the first term, d is the common difference, and n represents the position of the term, we get a3 = 4 + (3-1)d = 4 + 2d and a4 = 4 + (4-1)d = 4 + 3d. From the problem, we know that these two arithmetic means have the same value, so 4 + 2d = 4 + 3d. Subtracting 4 from both sides, we have 2d = 3d. Subtracting 2d from both sides, we get d = 0. Now, substituting the value of d into the equation 4 + 2d = 4 + 3d, we have 4 + 2(0) = 4 + 3(0), which simplifies to 4 = 4. Therefore, the means between 4 and 22 are k=4 and k=4. The answer is d) 13, 13.

4) To simplify and find Sn for an arithmetic series, we can use the formula Sn = n/2(a1 + an), where Sn represents the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 is 3, d is 1/2, and an is 15. Plugging these values into the formula, we get Sn = n/2(3 + 15) = n/2(18) = 9n. The answer is b) 9.

5) It seems like you are looking for the value of the expression 8 ∑(k=3)^(40-3k). To solve this, we need to evaluate the expression for various values of k. Starting with k=3, we plug in this value into the expression, giving us 40 - 3(3) = 40 - 9 = 31. Now, we move on to k=4, which yields 40 - 3(4) = 40 - 12 = 28. We continue this pattern until we reach k=8, which gives us 40 - 3(8) = 40 - 24 = 16. Finally, we sum up all the resulting values: 31 + 28 + 25 + 22 + 19 + 16 = 141. Therefore, the answer is d) 141.