Find the total number of terms in the given geometric progression

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

Full procedure would be highly appreciated

Please Help (Crying)

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- Oct 8th 2009, 06:49 AMcreatively12Geometric 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

Please Help (Crying) - Oct 8th 2009, 06:55 AMmathaddict
- Oct 8th 2009, 07:03 AMcreatively12
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 ??

- Oct 8th 2009, 07:14 AMSoroban
Hello, creatively12!

Quote:

Find the number of terms in the geometric progression:

. . $\displaystyle 1,\;\text{-}3,\;9,\;\text{-}27,\; \hdots \;531,\!441$

The $\displaystyle n^{th}$ term is: .$\displaystyle a_n \:=\:a_1r^{n-1}$

. . where: .$\displaystyle a_1$ = first term, $\displaystyle r$ = common ratio.

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

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

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

. . we have: .$\displaystyle (\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.

- Oct 8th 2009, 07:19 AMmathaddict
Sorry , i made a mistake . Edited my post ..

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

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

$\displaystyle 3^{n-1}=531441$

take the logs , then you would find that n=13 . - Oct 8th 2009, 07:20 AMcreatively12
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

- Oct 8th 2009, 07:21 AMcreatively12
thats what the problem, i don't know how to take logs, cause m not yet told to do logs :(

- Oct 8th 2009, 07:25 AMmathaddict
- Oct 8th 2009, 07:31 AMcreatively12
i just found out that my calculator does the trick of finding the powers, but thnaks anyways for the help, really appreciate it