# Geometric Progression or Geometric Series

• October 8th 2009, 06:49 AM
creatively12
Geometric Progression or Geometric Series
Find the total number of terms in the given geometric progression

1, -3, 9,...., 531441

Full procedure would be highly appreciated

• October 8th 2009, 06:55 AM
Quote:

Originally Posted by creatively12
Find the total number of terms in the given geometric progression

1, -3, 9,...., 531441

Full procedure would be highly appreciated

HI

a=1 , r=-3 and the last term is 531441

Using this formula ,

$T_n=ar^{n-1}$

where $T_n = 531441$ , r =-3 and a=1

Solve for n .
• October 8th 2009, 07:03 AM
creatively12
okay.... but I hav one question, how will you be able to find the power of -3, so that we would equate the bases and then equate their powers, is there a simple way other than LCM and Log to find the power ??
• October 8th 2009, 07:14 AM
Soroban
Hello, creatively12!

Quote:

Find the number of terms in the geometric progression:
. . $1,\;\text{-}3,\;9,\;\text{-}27,\; \hdots \;531,\!441$

The $n^{th}$ term is: . $a_n \:=\:a_1r^{n-1}$
. . where: . $a_1$ = first term, $r$ = common ratio.

We have: . $a_1 = 1,\;r = \text{-}3$

Hence: . $1\cdot(\text{-}3)^{n-1} \:=\:531,\!441 \quad\Rightarrow\quad (\text{-}3)^{n-1} \:=\:531,\!441$

Since $3^{12} \:=\:531,\!441 \:=\:(\text{-}3)^{12}$

. . we have: . $(\text{-}3)^{n-1} \:=\:(\text{-}3)^{12} \quad\Rightarrow\quad n-1 \:=\:12 \quad\Rightarrow\quad n \:=\:13$

There are 13 terms in the geometric progression.

• October 8th 2009, 07:19 AM
Quote:

Originally Posted by creatively12
okay.... but I hav one question, how will you be able to find the power of -3, so that we would equate the bases and then equate their powers, is there a simple way other than LCM and Log to find the power ??

Sorry , i made a mistake . Edited my post ..

$(-3)^{n-1}=531441$

you can see from the series that the positive terms are odd terms .

So n must be odd , and (n-1) would be even .

Since the power of -3 is even , then we can just ignore the -ve sign

$3^{n-1}=531441$

take the logs , then you would find that n=13 .
• October 8th 2009, 07:20 AM
creatively12
really really thanks Soroban, but I still don't understand how to get that 3^12, that raise to the power 12 in an easy and simple way, but thanks for all the stuff
• October 8th 2009, 07:21 AM
creatively12
thats what the problem, i don't know how to take logs, cause m not yet told to do logs :(
• October 8th 2009, 07:25 AM
Quote:

Originally Posted by creatively12
thats what the problem, i don't know how to take logs, cause m not yet told to do logs :(

ok

531441 is $3^{12}$ which is also $(-3)^{12}$

$(-3)^{n-1}=(-3)^{12}$

By comparing the powers , n-1 =12 , n=13
• October 8th 2009, 07:31 AM
creatively12
i just found out that my calculator does the trick of finding the powers, but thnaks anyways for the help, really appreciate it