+0  
 
0
64
1
avatar

1. We have two geometric sequences of positive real numbers: $$6,a,b\text{ and }\frac{1}{b},a,54$$Solve for $a$.

2. An infinite geometric series has a first term of $12$ and a second term of $4.$ A second infinite geometric series has the same first term of $12,$ a second term of $4+n,$ and a sum of four times that of the first series. Find the value of $n.$

Guest May 6, 2018
 #1
avatar
+1

2)

4/12 = 1/3 Common ratio of the first GS

Sum = F / [1 - R], where F= First term, R = Common ratio

Sum =12/ [1 - 1/3]

Sum =12 / (2/3)

Sum = 12 x 3/2

Sum = 36/2

Sum = 18 - Sum of the first GS

 

18 x 4 = 72 - Sum of the second GS

72 = 12 / [1 - R]     divide both sides by 12

6 = 1 / [1 - R]         cross multiply

6 [1 - R] = 1           divide both sides by 6

[1 - R] = 1/6           subtract 1 from both sides

- R = 1/6 - 1

- R = - 5/6             Multiply both sides by -1

R = 5/6 - Common ratio of the second GS

12 x 5/6 = 60 / 6 =10 - this is the 2nd term of the second GS

10 - 4 = 6 value of n of the 2nd term of the second GS.

 

1) Sorry, Can't read the LaTex of your first question.

Guest May 6, 2018

3 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.