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Find the term of the G.P. $1, 1.2, 1.44, 1.728,... $ which is just greater than $100$.

Please somebody explain me the question? All i know is $a=1$ and $r=1.2$

Help :(

should it be $T_{n} > 100$?

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  • $\begingroup$ If $a_1 = 1$ and $r = 1.2$, what is the formula for the $n$th term $a_n$? $\endgroup$ – TonyK Jul 11 '14 at 9:26
  • $\begingroup$ @TonyK $ar^n-1$ ? $\endgroup$ – Kiara Jul 11 '14 at 9:29
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    $\begingroup$ Yes (except you forgot to put the exponent in curly brackets). And $a=1$, so this is just $r^{n-1}$. Now start with the equation $r^{n-1} > 100$, and take the log of both sides. $\endgroup$ – TonyK Jul 11 '14 at 9:59
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$T_1=1$ and $r=1.2$, we want value of $n$ for which $T_n>100$.

That means $1\times(1.2)^n>100$ or $n>\frac{2}{\log_{10}1.2}=\frac{2}{0.0791812460476248277225056927041}=25.258506273026670863695116167616$

So $n\geq 26$.

Edit:

And $T_{26}=114.47545997288281555215581184$.

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  • $\begingroup$ I did the same thing, but the answer in my book is $114.5$ $\endgroup$ – Kiara Jul 11 '14 at 10:13
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    $\begingroup$ @Kiara $1.2^{26} \approx 114.5$. So the term is $114.5$, while the index (i.e. $n$) is $26$. $\endgroup$ – Arthur Jul 11 '14 at 10:19
  • $\begingroup$ Oh, I was being such a silly girl! Thanks a lot! $\endgroup$ – Kiara Jul 11 '14 at 13:39

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