# Series and Sequences Part Two

• Jan 9th 2010, 03:33 AM
christina
Series and Sequences Part Two
Insert 4 terms in geometric sequence between 3/8 and 128/81

Find a number which when added to each of -1,1 and 5 gives 3 terms in geometric sequence

The sum of the first 8 terms of a geometric series is 17 times the sum of its first 4 terms. Find the common ratio.

Thank you for any help given on these questions !
• Jan 9th 2010, 07:50 AM
Hello christina
Quote:

Originally Posted by christina
Insert 4 terms in geometric sequence between 3/8 and 128/81

The first term $\displaystyle a = \frac38$. The sixth term $\displaystyle = ar^5 = \frac{128}{81}$

$\displaystyle \Rightarrow r^5 = \frac{128}{81}\cdot\frac{8}{3}$

This will give you the value of $\displaystyle r$, and then you can work out the terms in between.

Spoiler:
$\displaystyle r = \frac{4}{3}$; the terms in between are $\displaystyle \frac{1}{2},\frac{2}{3},\frac{8}{9}, \frac{32}{27}$

Quote:

Find a number which when added to each of -1,1 and 5 gives 3 terms in geometric sequence
Let the number be $\displaystyle x$. Then the terms $\displaystyle (x-1),(x+1)$ and $\displaystyle (x+5)$ are in geometric sequence. In other words:
$\displaystyle \frac{x-1}{x+1}=\frac{x+1}{x+5}$
Solve this equation for $\displaystyle x$.
Spoiler:
x=3

Quote:

The sum of the first 8 terms of a geometric series is 17 times the sum of its first 4 terms. Find the common ratio.
With the usual notation, the sum of the first $\displaystyle n$ terms is:
$\displaystyle \frac{a(r^n-1)}{r-1}$
So:
$\displaystyle \frac{a(r^8-1)}{r-1}=\frac{17a(r^4-1)}{r-1}$

$\displaystyle \Rightarrow r^8-1=17(r^4-1)$
This is easy to solve if you spot that $\displaystyle r^8-1=(r^4-1)(r^4+1)$, and so the equation simplifies considerably!
Spoiler:
$\displaystyle r^4+1 = 17 \Rightarrow r = \pm2$