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

Question

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

Answer 1

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!

Answer 2

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