# Hi I Need Urgent Help!!! geometric series

• Oct 23rd 2006, 04:06 PM
kassandra
Hi I Need Urgent Help!!! geometric series
A finite geomeric sequence has t1 = 0.1024 and t2 = 0.256. How many terms does sequence have, if its middle term has the value 156.25?
How do I do this question? I am little bit confused!

t1 = 0.1024 t2 = 0.256
r=t2 /t1 = 2.5
t1r^n-1 = 156.25
(0.1024)(2.5 ) ^n-1= 156.25

(0.1024)(2.5 ) ^n-1= 156.25/ (0.1024)
(2.5 ) ^n-1= 1525.878906
how do I get n by itself??? am I doing this wrong? thanks for your help! :)
• Oct 23rd 2006, 04:33 PM
topsquark
Quote:

Originally Posted by kassandra
A finite geomeric sequence has t1 = 0.1024 and t2 = 0.256. How many terms does sequence have, if its middle term has the value 156.25?
How do I do this question? I am little bit confused!

t1 = 0.1024 t2 = 0.256
r=t2 /t1 = 2.5
t1r^n-1 = 156.25
(0.1024)(2.5 ) ^n-1= 156.25

(0.1024)(2.5 ) ^n-1= 156.25/ (0.1024)
(2.5 ) ^n-1= 1525.878906
how do I get n by itself??? am I doing this wrong? thanks for your help! :)

One tiny little gripe first. The line in red is written incorrectly. It should be:
$(2.5)^{n-1} = \frac{156.25}{0.1024}$

But the rest is correct.

So, the problem is to solve $2.5^{n-1}= 1525.878906$

I'm going to assume you know logarithms, otherwise you can't solve this. The tricky part is how to do a log with a base of 2.5. I'm not even going to try! 2.5 = 5/2 so: (Edit: Come to think of it you can leave it as 2.5. I was going to do something with the fraction that I realized later I didn't need to do.)
$ln \left [ \left ( \frac{5}{2} \right )^{n-1} \right ] = ln(1525.878906)$ (Or use log base 10, or your favorite base.)

$(n-1) \cdot ln \left ( \frac{5}{2} \right ) = 7.330325855$

$n - 1 = \frac{7.330325855}{ln \left ( \frac{5}{2} \right )}$

$n = 1 + \frac{7.330325855}{ln \left ( \frac{5}{2} \right )} = 8$
(with a BIG sigh of relief that it's an integer...)

-Dan