A variable X takes values 0, 2, 6, 12, 20, ..., n(n-1) with frequencies nC0, nC1, nC2, nC3, nC4, ..., nCn respectively. If the mean of this data is 60, then its median is :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q10 – JEE-MAIN 2026

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Home Maths Binomial TheoremQuestion #41

A variable X takes values

0, 2, 6, 12, 20, \dots, n (n - 1)

with frequencies

^{n} C_{0},^{n} C_{1},^{n} C_{2},^{n} C_{3},^{n} C_{4}, \dots,^{n} C_{n}

respectively. If the mean of this data is 60, then its median is :

Clue 1 of 3: Problem Structure

The problem involves finding the mean and median of a frequency distribution where the frequencies are binomial coefficients. The first step is to determine the value of 'n' using the given mean.

🥷

Ninja Strategy•Symmetry and Estimation

Quickly estimate $n \approx 16$ from the mean formula ( $n (n - 1) = 240$ ). Since the frequencies $^{16} C_{k}$ are symmetric about $k = 8$ , the median value must be $x_{8}$ .

🧠 Full Solution Path

Step 1: Determine the value of n from the Mean✦ Active

The data consists of values $x_{k}$ with frequencies $f_{k} =^{n} C_{k}$ for $k = 0, 1, \dots, n$ . The total frequency is $\sum f_{k} = \sum_{k = 0}^{n}^{n} C_{k} = 2^{n}$ . There is an ambiguity in the problem statement: the listed values $0, 2, 6, 12, \dots$ suggest $x_{k} = k (k + 1)$ , but the last term $n (n - 1)$ and the requirement for an integer solution for $n$ from the mean calculation indicate that the values are $x_{k} = k (k - 1)$ . We proceed with $x_{k} = k (k - 1)$ .

\bar{X} = \frac{\sum_{k = 0}^{n} x_{k} f_{k}}{\sum_{k = 0}^{n} f_{k}} = \frac{\sum_{k = 0}^{n} k (k - 1)^{n} C_{k}}{2^{n}}

Using the binomial identity $\sum_{k = 0}^{n} k (k - 1)^{n} C_{k} = n (n - 1) 2^{n - 2}$ , we get:

\bar{X} = \frac{n (n - 1) 2^{n - 2}}{2^{n}} = \frac{n (n - 1)}{4}

Given that the mean is 60:

\frac{n (n - 1)}{4} = 60 ⟹ n (n - 1) = 240

Solving the quadratic equation $n^{2} - n - 240 = 0$ gives $(n - 16) (n + 15) = 0$ . Since $n$ must be a positive integer, we have $n = 16$ .

💡 Teacher's Secret Hint

The key is to correctly identify the general form of the values $x_{k}$ . The listed sequence is misleading; the formula for the mean and the last term are the real clues.

Step 2: Calculate the Median's Position○ Expand

With $n = 16$ , the total number of observations is $N = \sum_{k = 0}^{16}^{16} C_{k} = 2^{16}$ . Since $N$ is even, the median is the average of the $(N / 2)$ -th and $(N / 2 + 1)$ -th observations.

Median = \frac{1}{2} (value of (\frac{2^{16}}{2}) -th term + value of (\frac{2^{16}}{2} + 1) -th term)

\text{Median} = \frac{1}{2} \left( \text{value of } 2^{15}\text{-th term} + \text{value of } (2^{15}+1)\text{-th term} \right)$$

We find which value $x_{k}$ corresponds to these positions by examining the cumulative frequencies. The sum of the binomial coefficients up to the term before the middle one is:

\sum_{k = 0}^{7}^{16} C_{k} = \frac{1}{2} (\sum_{k = 0}^{16}^{16} C_{k} -^{16} C_{8}) = \frac{1}{2} (2^{16} -^{16} C_{8}) = 2^{15} - \frac{1}{2}^{16} C_{8}

The cumulative frequency up to $k = 7$ is less than $2^{15}$ . The cumulative frequency up to $k = 8$ is $\sum_{k = 0}^{8}^{16} C_{k} = 2^{15} + \frac{1}{2}^{16} C_{8}$ , which is greater than $2^{15}$ . This means both the $2^{15}$ -th and $(2^{15} + 1)$ -th observations correspond to the value $x_{8}$ .

💡 Teacher's Secret Hint

For an even $n$ , the sum of binomial coefficients up to $k = n / 2 - 1$ is $2^{n - 1} - \frac{1}{2}^{n} C_{n / 2}$ . This is a useful property for finding the median in such distributions.

Step 3: Find the Median Value○ Expand

The median is the value $x_{8}$ . Using our determined formula for the values, $x_{k} = k (k - 1)$ :

Median = x_{8} = 8 (8 - 1) = 8 \times 7 = 56

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Maths Question 10 – JEE-MAIN 2026

Clue 1 of 3: Problem Structure

🧠 Full Solution Path