Find the total number of terms in the given geometric progression
1, -3, 9,...., 531441
Full procedure would be highly appreciated
Please Help
Hello, creatively12!
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.
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 .