1. ## Geometric sequences

2 problems here:

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.

2. ## Re: Geometric sequences

Originally Posted by Greymalkin
2 problems here:

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 $a_n=a_0r^n$. In which case $a_0=4$ and $a_{277}=4(5.1)^{277}$.

But that is just a guess. Check the definition in use the that particular text.

3. ## 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.

4. ## Re: Geometric sequences

Hello, Greymalkin!

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

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

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

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

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

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

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

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

There are two solutions.

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

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

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

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

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

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

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

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

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

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

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