NIMCET 2026 — Mathematics PYQ

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Question

NIMCET 2026 — Mathematics PYQ

NIMCET | Mathematics | 2026

Let $A_{k}$ be the arithmetic mean of squares of $k$ natural numbers. If

$k = 1 \sum n (6 A_{k} - 3 k) = 31$

Find the value of $n$ .

Choose the correct answer:

Explanation

To solve this, we first define $A_{k}$ , the arithmetic mean of the squares of the first $k$ natural numbers:

$A_{k} = \frac{1 ^{2} + 2 ^{2} + \dots + k ^{2}}{k}$

Using the formula for the sum of the squares of the first $k$ natural numbers, $\sum_{i = 1}^{k} i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6}$ , we get:

$A_{k} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6 k} = \frac{( k + 1 ) ( 2 k + 1 )}{6}$

Now, substitute $A_{k}$ into the given expression:

$6 A_{k} - 3 k = 6 [\frac{( k + 1 ) ( 2 k + 1 )}{6}] - 3 k$

$= (k + 1) (2 k + 1) - 3 k$

$= 2 k^{2} + 3 k + 1 - 3 k = 2 k^{2} + 1$

Next, we evaluate the summation:

$k = 1 \sum n (2 k^{2} + 1) = 31$

$2 k = 1 \sum n k^{2} + k = 1 \sum n 1 = 31$

$2 [\frac{n ( n + 1 ) ( 2 n + 1 )}{6}] + n = 31$

$\frac{n ( n + 1 ) ( 2 n + 1 )}{3} + n = 31$

Test the given options for $n$ :

If $n = 3$ : $\frac{3 ( 4 ) ( 7 )}{3} + 3 = 28 + 3 = 31$ .

Since $n = 3$ satisfies the equation, the correct value is $3$ .

Correct Option: (a)