JEE Main 2023 — Sequences And Series Question with Solution

Given that $$a+b+c+d=11$$ and the maximum value of $$a^5{b}^3{c}^{2}d$$ is $$3750\beta$$, you assumed the numbers to be $$\frac{a}{5},\frac{a}{5},\frac{a}{5},\frac{a}{5},\frac{a}{5},\frac{b}{3},\frac{b}{3},\frac{b}{3},\frac{c}{2},\frac{c}{2},d$$.

Applying the AM-GM inequality:

$$\frac{\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{a}{5}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{2}+\frac{c}{2}+d}{11}\ge {\left(\frac{\left(a^5{b}^3{c}^{2}d\right)}{5^5{3}^3{2}^{2}1}\right)}^{\frac{1}{11}}$$

Since $$a+b+c+d=11$$, we have:

$$1\ge {\left(\frac{\left(a^5{b}^3{c}^{2}d\right)}{5^5{3}^3{2}^{2}1}\right)}^{\frac{1}{11}}$$

Now, raising both sides to the power of 11:

$$1^{11}\ge \frac{a^5{b}^3{c}^{2}d}{5^5{3}^3{2}^{2}1}$$

From the given information, we know that $$a^5{b}^3{c}^{2}d\ge 3750\beta$$:

$$5^5{3}^3{2}^2\ge 3750\beta$$

Now, we can solve for $$\beta$$:

$$\beta \le \frac{1}{3750}\cdot 5^5{3}^3{2}^2$$

Since we are looking for the maximum value of $$\beta$$, we take the equality case:

$$\beta =\frac{1}{3750}\cdot 5^5{3}^3{2}^2$$

Calculating the value, we find that:

$$\beta =90$$

So, the value of $$\beta$$ is 90.