# Geometric progression sum of terms and to infinity

Given that the first two terms of a geometric progression is twice the value of the fifth term , I want to find the least value of $n$ such that $S_n$ is within 5% of S. First, I let $a+ar=2ar^4$ which gives $r=-0.648$. Then equating $a(1-r^n)/(1-r)<0.95a/(1-r)$ leads to $r^n < 0.05$ i.e. $(-0.648)^n <0.05$ which I don't know how to solve. Did I miss something?

I would have made this a comment but I do not have 50 rep ... What do you mean when you say "the first two terms of a geometric terms is twice the value of the fifth term?"

• $g_0 + g_1 = 2g_5$ – gt6989b Jul 13 '17 at 14:51

Why does $r^n < 0.05$ make no sense to solve? $r = 0.648 > 0.05$ so as you increase $n$, $r^n$ will decrease, eventually leading to a solution.

Also, I get $r = -0.6478$ negative, not positive as you suggest. In that case, your inequalities are wrong. With $r<0$, you need to constrain the 95\% on both sides: $$\left| \frac{a(1-r^n)}{1-r} - \frac{a}{1-r}\right| < \frac{0.05a}{1-r}$$ which after cancellation of $a/(1-r)$ (assuming $a>0$) implies $$\left|1 - r^n - 1 \right| < 0.05$$ which is equivalent to $|r|^n<0.05$, which happens when $n=7$.

• Yes i meant negative. And -0.648 to the power of 1 is already less than 0.05? – Homaniac Jul 13 '17 at 14:52
• @Homaniac see update – gt6989b Jul 13 '17 at 14:55
• yes thank you that i already figured..and my problem here is precisely how to solve modulus 1-r^n < 0.95 to get the final answer – Homaniac Jul 13 '17 at 14:59
• @Homaniac see update for methodology change and solution – gt6989b Jul 13 '17 at 15:32
• Are you sure answer is n=2 ? – Homaniac Jul 13 '17 at 15:41

The absolute value of $r$ can be used to determine $n$. See chart below which is self-explanatory. 