Question:
The largest term in the sequence \( a_n = \frac{n^2}{n^3 + 200} \) is given by
The largest term in the sequence \( a_n = \frac{n^2}{n^3 + 200} \) is given by
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For sequences, convert to function and check nearest integers after differentiation.
Updated On: Apr 23, 2026
- $\frac{529}{49}$
- $\frac{8}{89}$
- $\frac{49}{543}$
- None of these
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The Correct Option is C
Solution and Explanation
Concept:
To find maximum term in a sequence, treat $n$ as continuous and differentiate.
Step 1: Let $f(n) = \frac{n^2}{n^3 + 200}$.
Step 2: Differentiate using quotient rule.
\[ f'(n) = \frac{2n(n^3+200) - n^2(3n^2)}{(n^3+200)^2} \]
Step 3: Simplify numerator.
\[ = \frac{2n^4 + 400n - 3n^4}{(n^3+200)^2} = \frac{-n^4 + 400n}{(n^3+200)^2} \]
Step 4: Set derivative = 0.
\[ -n^4 + 400n = 0 \Rightarrow n(n^3 - 400) = 0 \] \[ n = \sqrt[3]{400} \approx 7.36 \]
Step 5: Check nearest integer.
Maximum occurs near $n = 7$ \[ a_7 = \frac{49}{343 + 200} = \frac{49}{543} \] Conclusion:
Largest term = $\frac{49}{543}$
Step 1: Let $f(n) = \frac{n^2}{n^3 + 200}$.
Step 2: Differentiate using quotient rule.
\[ f'(n) = \frac{2n(n^3+200) - n^2(3n^2)}{(n^3+200)^2} \]
Step 3: Simplify numerator.
\[ = \frac{2n^4 + 400n - 3n^4}{(n^3+200)^2} = \frac{-n^4 + 400n}{(n^3+200)^2} \]
Step 4: Set derivative = 0.
\[ -n^4 + 400n = 0 \Rightarrow n(n^3 - 400) = 0 \] \[ n = \sqrt[3]{400} \approx 7.36 \]
Step 5: Check nearest integer.
Maximum occurs near $n = 7$ \[ a_7 = \frac{49}{343 + 200} = \frac{49}{543} \] Conclusion:
Largest term = $\frac{49}{543}$
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