Chesley

2021-08-20

Find the first term of the sequence 12, 6, 3, 1.5, .... which is less than 0.0001.

faldduE

Consider the sequence, 12,6,3,1.5…..
Let ${u}_{n}$ denote the nth term of the sequence.
We have that $\frac{6}{12}=\frac{3}{6}=\frac{1.5}{3}=\frac{1}{2}$ which gives us that the sequence whose the first term is
${u}_{1}=12$ and common ratio is $r=\frac{1}{2}$ and therefore ${u}_{n}=12{\left(\frac{1}{2}\right)}^{n-1}=\frac{12}{{2}^{n-1}}=\frac{3{\left(2\right)}^{2}}{{2}^{n-1}}=3\left({2}^{3-n}\right).$
In order to find the first term of the sequence which is less than 0.0001 we need to find the first n such that $3\left({2}^{3-n}\right)<0.0001$
Observe that the function $f\left(x\right)=3\left({2}^{3-n}\right)$ is decreasing. Then, in particular we have that
Using the formula we obtained we have that

${u}_{17}=3\left({2}^{3-17}\right)=3\left({2}^{-14}\right)$

$=3\left(0.00006103515625\right)=0.0001831054875>0.0001$

which gives us that

We have also that ${u}_{18}=3\left({2}^{3-18}\right)=3\left({2}^{-15}\right)$

$=3\left(0.000030517578125\right)=0.000030517578125=0.00009155273475<0.0001.$
Hence, the first term of the sequence which is less than 0.0001 is
${u}_{18}=0.00009155273475$

Jazz Frenia

To find the first term of the sequence $12,6,3,1.5,\dots$ that is less than $0.0001$, we can set up the inequality:
$12×{\left(\frac{1}{2}\right)}^{n}<0.0001$ where $n$ represents the number of terms in the sequence. We can solve this inequality using logarithms:
$\mathrm{log}\left(12×{\left(\frac{1}{2}\right)}^{n}\right)<\mathrm{log}\left(0.0001\right)$
$\mathrm{log}12+n·\mathrm{log}\left(\frac{1}{2}\right)<\mathrm{log}\left(0.0001\right)$
Simplifying further:
$n·\mathrm{log}\left(\frac{1}{2}\right)<\mathrm{log}\left(0.0001\right)-\mathrm{log}12$
$n·\mathrm{log}\left(\frac{1}{2}\right)<\mathrm{log}\left(\frac{0.0001}{12}\right)$
$n>\frac{\mathrm{log}\left(\frac{0.0001}{12}\right)}{\mathrm{log}\left(\frac{1}{2}\right)}$
Evaluating the right-hand side of the inequality using a calculator, we find:
$n>13.2877$
Therefore, the smallest value of $n$ that satisfies the inequality is $n=14$. Thus, the first term of the sequence that is less than $0.0001$ is:
$12×{\left(\frac{1}{2}\right)}^{14}\approx 0.0001220703125$

Andre BalkonE

$9.1552734375×{10}^{-5}$
Explanation:
The given sequence is: $12,6,3,1.5,\dots$
We can observe that each term is obtained by dividing the previous term by 2. Therefore, we can express the terms of the sequence as:
${a}_{1}=12$
${a}_{2}=\frac{{a}_{1}}{2}=\frac{12}{2}=6$
${a}_{3}=\frac{{a}_{2}}{2}=\frac{6}{2}=3$
${a}_{4}=\frac{{a}_{3}}{2}=\frac{3}{2}=1.5$
We can see that the terms are decreasing by a factor of 2 in each step. We can continue this pattern to find the subsequent terms, but for this problem, we only need to find the first term that is less than 0.0001.
Let's find the value of $n$ for which ${a}_{n}<0.0001$.
${a}_{n}=\frac{{a}_{n-1}}{2}$
$0.0001=\frac{{a}_{n-1}}{2}$
Multiplying both sides by 2:
$0.0001×2={a}_{n-1}$
$0.0002={a}_{n-1}$
Now we can determine the value of $n$ by finding the position of ${a}_{n-1}$ in the sequence.
To calculate the position, we start from ${a}_{4}$ and keep dividing by 2 until we reach $0.0002$.
${a}_{4}=1.5$
${a}_{5}=\frac{{a}_{4}}{2}=\frac{1.5}{2}=0.75$
${a}_{6}=\frac{{a}_{5}}{2}=\frac{0.75}{2}=0.375$
${a}_{7}=\frac{{a}_{6}}{2}=\frac{0.375}{2}=0.1875$
${a}_{8}=\frac{{a}_{7}}{2}=\frac{0.1875}{2}=0.09375$
${a}_{9}=\frac{{a}_{8}}{2}=\frac{0.09375}{2}=0.046875$
${a}_{10}=\frac{{a}_{9}}{2}=\frac{0.046875}{2}=0.0234375$
${a}_{11}=\frac{{a}_{10}}{2}=\frac{0.0234375}{2}=0.01171875$
${a}_{12}=\frac{{a}_{11}}{2}=\frac{0.01171875}{2}=0.005859375$
${a}_{13}=\frac{{a}_{12}}{2}=\frac{0.005859375}{2}=0.0029296875$
${a}_{14}=\frac{{a}_{13}}{2}=\frac{0.0029296875}{2}=0.00146484375$
${a}_{15}=\frac{{a}_{14}}{2}=\frac{0.00146484375}{2}=0.000732421875$
${a}_{16}=\frac{{a}_{15}}{2}=\frac{0.000732421875}{2}=0.0003662109375$
${a}_{17}=\frac{{a}_{16}}{2}=\frac{0.0003662109375}{2}=0.00018310546875$
${a}_{18}=\frac{{a}_{17}}{2}=\frac{0.00018310546875}{2}=9.1552734375×{10}^{-5}$
Thus, ${a}_{18}$ is the first term of the sequence that is less than $0.0001$.
Therefore, the first term of the sequence $12,6,3,1.5,\dots$ which is less than $0.0001$ is $9.1552734375×{10}^{-5}$.

fudzisako

Step 1:
A geometric sequence is a sequence in which each term is found by multiplying the previous term by a constant factor, called the common ratio. In this case, the common ratio is $\frac{1}{2}$ because each term is half of the previous term.
Let's assume the first term of the sequence is $a$. Then, the terms of the sequence can be expressed as:

We want to find the first term that is less than 0.0001, which can be represented as:
$\frac{a}{{2}^{n}}<0.0001$ where $n$ represents the number of terms needed to reach a value less than 0.0001.
To solve this inequality, we can express 0.0001 as $\frac{1}{10000}$ and substitute it into the inequality:
$\frac{a}{{2}^{n}}<\frac{1}{10000}$
To simplify the equation, we can multiply both sides by ${2}^{n}$:
$a<\frac{1}{10000}·{2}^{n}$
Step 2:
Since we are looking for the first term that satisfies this inequality, we want to find the smallest value of $n$ that makes the right side of the inequality less than 0.0001.
Let's solve for $n$:
$\frac{1}{10000}·{2}^{n}<0.0001$
Multiplying both sides by 10000:
${2}^{n}<1$
To solve for $n$, we can take the logarithm base 2 of both sides:
${\mathrm{log}}_{2}\left({2}^{n}\right)<{\mathrm{log}}_{2}\left(1\right)$
Simplifying:
$n<0$
Since $n$ represents the number of terms in the sequence, it cannot be negative. Therefore, there is no value of $n$ that satisfies the inequality.
Hence, there is no term in the sequence 12, 6, 3, 1.5, ... that is less than 0.0001.

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