{a_1, a_2, ....., a_n, .....} is a progressio | vedclass

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(B) Let $f(x) = \frac{x^2}{x^3 + 200}$. To find the maximum,we differentiate with respect to $x$:$f'(x) = \frac{(x^3 + 200)(2x) - x^2(3x^2)}{(x^3 + 20

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$a_7$

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(B) Let $f(x) = \frac{x^2}{x^3 + 200}$. To find the maximum,we differentiate with respect to $x$:$f'(x) = \frac{(x^3 + 200)(2x) - x^2(3x^2)}{(x^3 + 200)^2} = \frac{2x^4 + 400x - 3x^4}{(x^3 + 200)^2} = \frac{x(400 - x^3)}{(x^3 + 200)^2}$.For $f(x)$ to be increasing,$f'(x) > 0$,which implies $400 - x^3 > 0$,or $x Since $7^3 = 343$ and $8^3 = 512$,we have $7 Thus,the function increases until $x = (400)^{1/3}$ and decreases thereafter.We compare $a_7$ and $a_8$:$a_7 = \frac{7^2}{7^3 + 200} = \frac{49}{343 + 200} = \frac{49}{543} \approx 0.0902$.$a_8 = \frac{8^2}{8^3 + 200} = \frac{64}{512 + 200} = \frac{64}{712} = \frac{8}{89} \approx 0.0898$.Since $a_7 > a_8$,the largest term is $a_7$.

${a_1, a_2, ....., a_n, .....}$ is a progression where $a_n = \frac{n^2}{n^3 + 200}$. The largest term of this progression is

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