not working out for me! geometric series

**frogsrcool** · October 23rd 2006, 05:48 PM

the third term of a geometric sequence is t3= -75 and the sixth term is t6=9375. determine the first term and the common ratio.

how do I do this question I am trying really hard but I still don't comprehend this question!

**Quick** · October 23rd 2006, 06:01 PM

Originally Posted by frogsrcool

the third term of a geometric sequence is t3= -75 and the sixth term is t6=9375. determine the first term and the common ratio.

how do I do this question I am trying really hard but I still don't comprehend this question!

If you write geometric sequences on a graph, they form lines...

Let us call the geometric term "x"

Let us call the value of the geometric term "y"

Now let's find the slope of that line...

The slope equation is: $m=\frac{y_2-y_1}{x_2-x_1}$

Substitute: $m=\frac{9375-(-75)}{6-3}$

Subtract: $m=\frac{9450}{3}$

Divide: $m=3150$ (this is a very steep line)

All lines can be written in the form: $y=mx+b$

Substitute: $-75=3150(3)+b$

Multiply: $-75=9450+b$

Subtract: $-9525=b$

Go back to equation: $y=mx+b$

Substitute: $y=3150x-9525$

Substitute to find the first term: $y=3150(1)-9525$

Multiply: $y=3150-9525$

Subtract: $\boxed{y=-6365}$

but please, CHECK MY ARITHMETIC!!!!!!!!!!!!

**Soroban** · October 23rd 2006, 06:01 PM

Hello, frogsrcool!

The third term of a geometric sequence is $t_3 = -75$ and the sixth term is $t_6 = 9375$
Determine the first term and the common ratio.

You're expected to know that the $n^{th}$ term is: . $t_n = a\cdot r^{n-1}$
where $a$ is the first term and $r$ is the common ratio.

The $3^{rd}$ term is -75: . $t_3\:=\:a\cdot r^2 \:=\:-75$ [1]

The $6^{th}$ term is 9375: . $t_6\:=\:a\cdot r^5\:=\:9375$ [2]

Divide [2] by [1]: . $\frac{a\cdot r^5}{a\cdot r^2}\:=\:\frac{9375}{-75}\quad\Rightarrow\quad r^3\:=\:-125\quad\Rightarrow\quad\boxed{r = -5}$

Substitute into [1]: . $a(-5)^2 \:=\:-75\quad\Rightarrow\quad\boxed{a = -3}$

**Quick** · October 23rd 2006, 06:11 PM

Originally Posted by Soroban

Hello, frogsrcool!

You're expected to know that the $n^{th}$ term is: . $t_n = a\cdot r^{n-1}$
where $a$ is the first term and $r$ is the common ratio.

The $3^{rd}$ term is -75: . $t_3\:=\:a\cdot r^2 \:=\:-75$ [1]

The $6^{th}$ term is 9375: . $t_6\:=\:a\cdot r^5\:=\:9375$ [2]

Divide [2] by [1]: . $\frac{a\cdot r^5}{a\cdot r^2}\:=\:\frac{9375}{-75}\quad\Rightarrow\quad r^3\:=\:-125\quad\Rightarrow\quad\boxed{r = -5}$

Substitute into [1]: . $a(-5)^2 \:=\:-75\quad\Rightarrow\quad\boxed{a = -3}$

I thought geometric sequences were linear, and it was algebraic that aren't...

Or maybe it was arithmetic are linear and geometric aren't...

Anyway, listen to Soroban before listening to me.

**ThePerfectHacker** · October 23rd 2006, 06:16 PM

Originally Posted by Quick

I thought geometric sequences were linear, and it was algebraic that aren't...

Or maybe it was arithmetic are linear and geometric aren't...

Anyway, listen to Soroban before listening to me.

Arithmetic sequences differ by a konstant:
$a,a+k,a+2k,a+3k,...$

Geometric sequences differ by a product:
$a,ak,ak^2,ak^3,...$

Both these sequenes are popular because there is an easy way to sum them.

Math Help - not working out for me! geometric series

LinkBack

Thread Tools

Display

not working out for me! geometric series

Similar Math Help Forum Discussions

Help working out a Taylor series of cos3x

Taylor Series using Geometric Series.

Exploiting geometric series with power series to find Taylors series

Geometric Progression or Geometric Series

Series and Sequence/Application (sum of a geometric series) - Supperannuation

Search Tags