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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**+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