1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4

A)15
B)15/2
C)-15
D)1/15
I chose A

2)Find the sum of the infinite geometric series. Sigma sign with infinity symbol above and n=1 below. To the right 20(-1/4)n-1
A)25
B)80/3
C16
D)does not exist
I chose A

3)Find the sum of the infinite geometric series:4+3+9/4+:...
A)16/7
B)16
C)-12
D)does not exist
I chose B

4)Write 0.72 repeating as a fraction.
A)7/9
B)8/11
C)18/25
D)7 and 2/9
I chose B but my book confuses me on how to work it out. I just divided 8 and 11 and I got .72 repeating

5)Find the fifth term of the sequence in which a1=-3,and aN+1=3aN-n
A)-301
B)-99
C)-193
D)-341
I don't know

Please note Damon's last post.

math - Jon, Wednesday, December 12, 2007 at 5:43pm
thanks for the help. are you possitive they are correct?

math - Damon, Wednesday, December 12, 2007 at 5:48pm

Oh good heavens, always check anything I do!

3. 4+3+9/4+:... = 4(1 + (3/4) + (3/4)^2^3 + ...]

= 4*[1/(1-3/4)] = 16

4. There is nothing wrong with trying the multiple choices and seeing which one works. That is what you did. There is a direct way also.

Let X = 0.727272...
0.727272...-(0.01)X = 0.727272..-0.0072727..)
= 0.720000...
Therefore
X - 0.01X = 0.72
0.99 X = 0.72
X = 72/99 = 8/11

1) To find a1 in a geometric series, we can use the formula Sn = a1 * (1 - r^n) / (1 - r), where Sn is the sum of the series, a1 is the first term, r is the common ratio, and n is the number of terms. In this case, Sn is given as 300, r is -3, and n is 4. Plugging the values into the formula:

300 = a1 * (1 - (-3)^4) / (1 - (-3))

300 = a1 * (1 - 81) / (1 + 3)

300 = a1 * (-80) / 4

To solve for a1, we can multiply both sides by 4 and divide by -80:

a1 = (300 * 4) / (-80) = -15

Therefore, the first term is -15. So, option C) -15 is the correct answer.

2) To find the sum of an infinite geometric series, we can use the formula S = a1 / (1 - r), where S is the sum, a1 is the first term, and r is the common ratio. In this case, a1 is 20 and r is -1/4. Plugging the values into the formula:

S = 20 / (1 - (-1/4))

S = 20 / (1 + 1/4)

S = 20 / (5/4)

S = 20 * (4/5) = 80/5 = 16

Therefore, the sum of the infinite geometric series is 16. So, option C) 16 is the correct answer.

3) To find the sum of an infinite geometric series, we can use the formula S = a1 / (1 - r), where S is the sum, a1 is the first term, and r is the common ratio. In this case, a1 is 4 and r is 3/4. Plugging the values into the formula:

S = 4 / (1 - 3/4)

S = 4 / (1/4) = 4 * 4 = 16

Therefore, the sum of the infinite geometric series is 16. So, option B) 16 is the correct answer.

4) To write 0.72 repeating as a fraction, we can use an algebraic technique. Let's call the repeating decimal x:

x = 0.727272...

Now, if we multiply x by 100, we get:

100x = 72.727272...

Subtracting these two equations, we can eliminate the repeating part:

100x - x = 72.727272... - 0.727272...

99x = 72

Dividing both sides by 99:

x = 72/99

Simplifying the fraction by dividing both numerator and denominator by 9:

x = 8/11

Therefore, 0.72 repeating is equal to 8/11. So, option B) 8/11 is the correct answer.

5) To find the fifth term of the sequence, we can use the given recurrence relation aN+1 = 3aN - n. We know that a1 equals -3. We can use this relation iteratively to find the subsequent terms.

a2 = 3*a1 - 1 = 3*(-3) - 1 = -10

a3 = 3*a2 - 2 = 3*(-10) - 2 = -32

a4 = 3*a3 - 3 = 3*(-32) - 3 = -99

a5 = 3*a4 - 4 = 3*(-99) - 4 = -301

Therefore, the fifth term of the sequence is -301. So, option A) -301 is the correct answer.