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In a geometric sequence $u_1=125$ and $u_6 =1/25$.

  • a) Find the value of $r$ (the common ratio).
  • b) Find the largest value of $n$ for which $S_n < 156.22$.
  • c) Explain why there is no value of $n$ for which $S_n > 160$.

Can someone please help me figure this out?

Joe
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    Welcome to Mathematics Stack Exchange. Since we are dealing with a geometric sequence, $u_k = 125r^{k-1}$. Can you use this to solve part a)? – Joe Mar 02 '21 at 19:27

1 Answers1

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hint

$$u_6=u_1(r)^{6-1}\implies$$ $$ \frac{1}{5^2}=5^3r^5 \implies$$ $$r^5=(\frac 15)^5$$ $$S_n=u_1+u_2+...+u_n=u_1\frac{r^n-1}{r-1}$$

$$\implies \lim_{n\to+\infty}S_n=\frac{625}{4}=156,25$$ $$\implies (\forall n\ge 1)\;\; S_n\le 156,25<160$$ because $ (S_n) $ is increasing.