Question

If na is the greatest term in the sequence 3 n 4 n a ,n 1, 2,3....... n 147 , then is equal to ? Solve this q using AM-GM inequality

Answer 1

5

Answer 2

To find the greatest term in the sequence $a_n = \frac{n^3}{n^4 + 147}$, we can minimize its reciprocal, $\frac{1}{a_n}$.

Let $f(n) = \frac{1}{a_n} = \frac{n^4 + 147}{n^3} = n + \frac{147}{n^3}$. We want to minimize $f(n)$. To apply the AM-GM inequality, we need the product of the terms to be constant.

Consider the terms $\frac{n}{3}$, $\frac{n}{3}$, $\frac{n}{3}$, and $\frac{147}{n^3}$. The sum of these four terms is $\frac{n}{3} + \frac{n}{3} + \frac{n}{3} + \frac{147}{n^3} = n + \frac{147}{n^3}$. The product of these four terms is $\frac{n}{3} \cdot \frac{n}{3} \cdot \frac{n}{3} \cdot \frac{147}{n^3} = \frac{n^3}{27} \cdot \frac{147}{n^3} = \frac{147}{27}$. This product is a constant.

According to the AM-GM inequality, for non-negative numbers $x_1, x_2, ..., x_k$:

$$\frac{x_1 + x_2 + ... + x_k}{k} \ge (x_1 \cdot x_2 \cdot ... \cdot x_k)^{\frac{1}{k}}$$

Applying this to our four terms:

$$\frac{\frac{n}{3} + \frac{n}{3} + \frac{n}{3} + \frac{147}{n^3}}{4} \ge \left( \left(\frac{n}{3}\right)^3 \cdot \frac{147}{n^3} \right)^{\frac{1}{4}}$$ $$\frac{n + \frac{147}{n^3}}{4} \ge \left( \frac{147}{27} \right)^{\frac{1}{4}}$$ $$n + \frac{147}{n^3} \ge 4 \cdot \left( \frac{147}{27} \right)^{\frac{1}{4}}$$

The minimum value of $n + \frac{147}{n^3}$ occurs when all the terms are equal:

$$\frac{n}{3} = \frac{147}{n^3}$$ $$n^4 = 3 \cdot 147$$ $$n^4 = 441$$ $$n = (441)^{\frac{1}{4}}$$ $$n = (21^2)^{\frac{1}{4}}$$ $$n = 21^{\frac{1}{2}} = \sqrt{21}$$

Since $n$ must be an integer ($n = 1, 2, 3, ...$), and $\sqrt{21}$ is approximately $4.58$, the minimum value of $f(n)$ (and thus the maximum value of $a_n$) will occur at an integer $n$ close to $4.58$. We need to check $n=4$ and $n=5$.

Let's evaluate $f(n) = n + \frac{147}{n^3}$ for $n=4$ and $n=5$:

For $n=4$:

$$f(4) = 4 + \frac{147}{4^3} = 4 + \frac{147}{64} = 4 + 2.296875 = 6.296875$$

For $n=5$:

$$f(5) = 5 + \frac{147}{5^3} = 5 + \frac{147}{125} = 5 + 1.176 = 6.176$$

Comparing the values, $f(5)$ is smaller than $f(4)$. Since $a_n = \frac{1}{f(n)}$, a smaller $f(n)$ means a larger $a_n$. Therefore, $a_5$ is greater than $a_4$.

To confirm that $n=5$ is indeed the maximum, we can also check $f(3)$ and $f(6)$:

$$f(3) = 3 + \frac{147}{3^3} = 3 + \frac{147}{27} = 3 + \frac{49}{9} = 3 + 5.444... = 8.444...$$ $$f(6) = 6 + \frac{147}{6^3} = 6 + \frac{147}{216} = 6 + 0.6805... = 6.6805...$$

The values of $f(n)$ are $f(3) = 8.444...$, $f(4) = 6.296875$, $f(5) = 6.176$, $f(6) = 6.6805...$. This shows that $f(n)$ decreases until $n=5$ and then starts increasing. Thus, $f(5)$ is the minimum value of $f(n)$ for integer $n$.

Therefore, $a_5$ is the greatest term in the sequence. The value of $n$ for which $a_n$ is the greatest is $5$.

The final answer is $\boxed{5}$

Explanation of the solution:

1. The sequence is $a_n = \frac{n^3}{n^4 + 147}$. To find the maximum $a_n$, we minimize its reciprocal $f(n) = \frac{1}{a_n} = n + \frac{147}{n^3}$.
2. Apply AM-GM inequality to terms $\frac{n}{3}, \frac{n}{3}, \frac{n}{3}, \frac{147}{n^3}$. Their sum is $n + \frac{147}{n^3}$, and their product is $\left(\frac{n}{3}\right)^3 \cdot \frac{147}{n^3} = \frac{147}{27}$ (a constant).
3. The equality condition for AM-GM gives the minimum value: $\frac{n}{3} = \frac{147}{n^3}$, which leads to $n^4 = 441$, so $n = \sqrt[4]{441}$.
4. Since $n$ must be an integer, and $\sqrt[4]{441} \approx 4.58$, we check the integer values $n=4$ and $n=5$.
5. Calculate $f(4) = 4 + \frac{147}{64} = 6.296875$ and $f(5) = 5 + \frac{147}{125} = 6.176$.
6. Since $f(5) < f(4)$, $a_5$ is greater than $a_4$. Therefore, $n=5$ corresponds to the greatest term.