# Geometric sequences

• Sep 20th 2012, 03:30 AM
Greymalkin
Geometric sequences
2 problems here:
Attachment 24864

The first one(#74) does not look as if it has an error, the standard formula being a1r(n-1) to find the nth term.

The second problem (#75) completely eludes me, there are answers in the back of my book for both a & b that are all fractions, yet I cannot solve for any.(ie find a way to prove the sequence is either arithmetic or geometric)
Trying to divide to find r gives me a remainder, solving for x does not do much as sequences don't seem to involve zeros. There must be some other way to solve for x that I am failing to see.
• Sep 20th 2012, 04:31 AM
Plato
Re: Geometric sequences
Quote:

Originally Posted by Greymalkin
2 problems here:
Attachment 24864
The first one(#74) does not look as if it has an error, the standard formula being a1r(n-1) to find the nth term.

Without having the text it is hard to know the expected answer.
Some authors define a geometric sequence as $\displaystyle a_n=a_0r^n$. In which case $\displaystyle a_0=4$ and $\displaystyle a_{277}=4(5.1)^{277}$.

But that is just a guess. Check the definition in use the that particular text.
• Sep 20th 2012, 05:03 AM
emakarov
Re: Geometric sequences
Note that 104.04 * 5.1 = 530.604, not 531.6444. This makes the sequence not well-defined. My guess is that it continues as a geometric progression and has a single disturbance in this 4th element.

For the second problem, if this is an arithmetic sequence, subtract the first element from the second one to find the common difference, then equate this number to the difference between the third and second elements to get an equation in x. If this is a geometric sequence, equate the ratio of the second and first elements and the ratio of the third and second elements to get an equation in x.
• Sep 20th 2012, 08:08 AM
Soroban
Re: Geometric sequences
Hello, Greymalkin!

Quote:

75. Consider the sequence: .$\displaystyle x+3,\;x+7,\;4x-2\;\hdots$

(a) If the sequence is arithmetic, find $\displaystyle x$ and determine each of the 3 terms and the 4th term.

$\displaystyle d \:=\:a_2-a_1 \:=\:(x+7)-(x+3) \:=\:4\;\;[1]$

$\displaystyle d \:=\:a_3-a_2 \:=\:(4x-2) - (x+7) \:=\:3x-9\;\;[2]$

Equate [1] and [2]: .$\displaystyle 3x-9 \:=\:4 \quad\Rightarrow\quad x = \tfrac{13}{3}$

We have: .$\displaystyle \begin{array}{cccccccccc} a_1 &=& x+3 &=& \tfrac{13}{3} + 3 &=& \tfrac{22}{3}\\ d &=& 4 \end{array}$

Therefore: .$\displaystyle \begin{Bmatrix}a_1 &=& \frac{22}{3} \\ \\[-3mm] a_2 &=& \frac{34}{3} \\ \\[-3mm] a_3 &=& \frac{46}{3} \\ \\[-3mm] a_4 &=& \frac{58}{3} \end{Bmatrix}$

Quote:

(b) If the sequence is geometric, find $\displaystyle x$ and determine of the 3 terms and the 4th term.

There are two solutions.

$\displaystyle r \,=\,\frac{a_2}{a_1} \quad\Rightarrow\quad r \:=\:\frac{x+7}{x+3}\;\;[1]$

$\displaystyle r \,=\,\frac{a_3}{a_2} \quad\Rightarrow\quad r \:=\:\frac{4x-2}{x+7}\;\;[2]$

Equate [1] and [2]: .$\displaystyle \frac{x+7}{x+3} \:=\:\frac{4x-2}{x+7} \quad\Rightarrow\quad x^2+14x+9 \:=\:4x^2 + 10x - 6$

. . . . . . . . . . . . . .$\displaystyle 3x^2 - 4x - 55 \:=\:0 \quad\Rightarrow\quad (x-5)(3x+11) \:=\:0$

Hence: .$\displaystyle x \:=\:5,\:\text{-}\tfrac{11}{3}$

$\displaystyle \text{If }x = 5,\text{ substitute into [1]: }\:r \:=\:\frac{5+7}{5+3} \quad\Rightarrow\quad r \,=\,\tfrac{3}{2}$

. . $\displaystyle \text{Then: }\,a_1 \:=\:x+3 \:=\:5+3 \:=\:8$

$\displaystyle \text{Therefore: }\:\begin{Bmatrix}a_1 &=& 8 \\ a_2 &=& 12 \\ a_3 &=& 18 \\ a_4 &=& 27 \end{Bmatrix}$

$\displaystyle \text{If }x = \text{-}\tfrac{11}{3},\text{ substitute into [1]: }\:r \:=\:\frac{\text{-}\frac{11}{3} + 7}{\text{-}\frac{11}{3} + 3} \:=\:\text{-}5$

. . $\displaystyle \text{Then: }\,a_1 \:=\:x+3 \:=\:\text{-}\tfrac{11}{3} + 3 \:=\:\text{-}\tfrac{2}{3}$

$\displaystyle \text{Therefore: }\:\begin{Bmatrix}a_1 &=& \text{-}\frac{2}{3} \\ \\[-3mm] a_2 &=& \frac{10}{3} \\ \\[-3mm] a_3 &=& \text{-}\frac{50}{3} \\ \\[-3mm] a_4 &=&\frac{250}{3} \end{Bmatrix}$