What would be the formula for finding the number of sequences needed to reach some value when we know that the start value doubles on each sequence. So, for example for a problem where we know that the start value is $2.5$ and we want to find out in how many sequences if the number doubles on each sequence $(2.5, 5, 10, 20, 40...)$ it will reach $1000$?
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If the sequence is $a, 2a, 2^2a, \dots$, then the $n$-th term is $2^{n-1}a$.
Given a number $b$, you want to find the smallest $n$ such that $2^{n-1}a \ge b$, that is $$ 2^{n-2}a < b \le 2^{n-1}a $$ or equivalently $$ n-2 < \log_2 \frac{b}{a} \le n-1 $$
This means that $n=1+\lceil \log_2 \frac{b}{a} \rceil$.
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Note that if the starting term of the sequence is $a$ and the common ratio is $r$, then the $n^{\text{th}}$ term of a geometric progression is given by $G_n = ar^{n-1}$.
In this case, we have, $a= 2.5, r=2$. Thus, we need: $$[2.5](2)^{n-1}>1000 \implies 2^{n -1} >400 \implies n-1 >9 \implies n >10$$