If an = n^3n^4 + 147, n = 1, 2, 3,, and an is the greatest term in the sequence, then an is equal to:

Given: an = n^3n^4 + 147 for n = 1, 2, 3, Find: The greatest term of the sequence. Define f(x) = x^3x^4 + 147 and find its derivative: f'(x) = 3x^2(x^4 + 147) - x^3(4x^3)(x^4 + 147)^2 Set the derivative equal to zero: 3x^6 + 441x^2 - 4x^6 = 0 x^2(441 - x^4) = 0 So, x = [4]441 3.5 Now evaluate the function at the critical point: f(3.5) = (3.5)^3(3.5)^4 + 147 0.158 Therefore, the greatest term is 0.158.

If an = n^3n^4 + 147, n = 1, 2, 3,, and an is the greatest term i… — Solution | JEE Mathematics

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If $a_n = \frac{n^3}{n^4 + 147}$ , $n = 1, 2, 3, \dots$ , and $a_n$ is the greatest term in the sequence, then $a_n$ is equal to:

Answer

Correct answer:0.158

Step-by-step solution

Standard Method

Given: $a_n = \frac{n^3}{n^4 + 147}$ for $n = 1, 2, 3, \dots$

Find: The greatest term of the sequence.

Define

f(x) = \frac{x^3}{x^4 + 147}

and find its derivative:

f'(x) = \frac{3x^2(x^4 + 147) - x^3(4x^3)}{(x^4 + 147)^2}

Critical Point Evaluation

Set the derivative equal to zero:

3x^6 + 441x^2 - 4x^6 = 0 \implies x^2(441 - x^4) = 0

So,

x = \sqrt[4]{441} \approx 3.5

Approximate Maximum Value

Now evaluate the function at the critical point:

f(3.5) = \frac{(3.5)^3}{(3.5)^4 + 147} \approx 0.158

Therefore, the greatest term is $0.158$ .

Common mistakes

Treating the sequence directly as continuous without checking that the maximum term must correspond to an integer $n$ . The derivative helps locate where the maximum occurs, but the sequence is defined only for natural numbers, so nearby integer terms should be considered.
Making an algebraic mistake while simplifying $f'(x) = 0$ . In the numerator, combining terms incorrectly can change the critical point. Carefully simplify to $x^2(441 - x^4) = 0$ .
Using the inconsistent stray line $a = 5$ from the solution. It does not match the preceding working, whereas the computed value of the greatest term is approximately $0.158$ .

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