Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

807,752 users

Algebra help please!!?

0 votes
I'm really struggling with the geometric sequences:/ If you could help me and explain I would really appreciate it:)
1.Determine if the sequence is arithmetic or geometric. Then find the next term in the sequence. Select all that apply.

8, 5, 2, -1, -4, …

A.arithmetic
B.geometric
C.-10
D.-7

2.Find the common ratio, r, in the following geometric sequence.

27, 9, 3, 1, …

A.3
B.-3
C.1/3
D.-9
asked Jul 2, 2013 in ALGEBRA 1 by angel12 Scholar

2 Answers

0 votes

1)Given series is 8, 5, 2, -1, -4, …

The common difference between any two terms is -3.

=> d =-3

First term t1 = a =8

We have to find the 6th term => image         [Formula for nth term]

                                                         image

                                                               = 8 + 5(-3)

                                                               = 8 - 15

                                                               = -7

Therefore the next term in the series is -7

Therefore the answer is D

answered Jul 2, 2013 by joly Scholar
edited Jul 2, 2013 by joly
0 votes

2) Given geometric series is 27, 9, 3, 1, …

     Common ratio r = t2 / t1  where t1 = first term and t2 = second term

                                   = 9 / 27

                                   = 1/3

     Therefore the common ratio is 1/3.

      Therefore the answer is C.

 

answered Jul 2, 2013 by joly Scholar

Related questions

asked Sep 19, 2014 in CALCULUS by anonymous
asked Jun 3, 2015 in CALCULUS by anonymous
asked Jun 1, 2015 in CALCULUS by anonymous
asked Jun 1, 2015 in CALCULUS by anonymous
asked Jun 1, 2015 in CALCULUS by anonymous
asked Feb 11, 2015 in CALCULUS by anonymous
asked Sep 18, 2014 in CALCULUS by anonymous
asked Sep 18, 2014 in CALCULUS by anonymous
...