A variable \(X\) takes values \(0,0,2,6,12,20, \ldots\). \(\mathrm{n}(\mathrm{n}-1)\) with frequencies \({ }^{\mathrm{n}} \mathrm{C}_{0},{ }^{\mathrm{n}} \mathrm{C}_{1},{ }^{\mathrm{n}} \mathrm{C}_{2},{ }^{\mathrm{n}} \mathrm{C}_{3},{ }^{\mathrm{n}} \mathrm{C}_{4}\), \({ }^{n} \mathrm{C}_{5}, \ldots . .{ }^{n} \mathrm{C}_{n}\), respectively. If the mean of this data is 60, then its median is:

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Binomial Theorem

A variable

X

takes values

0, 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}

^{n} C_{5}, \dots ..^{n} C_{n}

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

Answer: (A)

The value at index

r

r (r - 1)

with frequency

(r n)

, so the mean is

\frac{\sum _{r} r ( r - 1 ) ( r n )}{2 ^{n}} = \frac{n ( n - 1 ) 2 ^{n - 2}}{2 ^{n}} = \frac{n ( n - 1 )}{4}

. Setting this to

60

gives

n (n - 1) = 240

, so

n = 16

. The total frequency is

2^{16} = 65536

, and

\sum_{r = 0}^{7} (r 16) = 26333

while

\sum_{r = 0}^{8} (r 16) = 39203

; the middle observations (the

32768

th and

32769

th) lie in the class

r = 8

, whose value is

8 \cdot 7 = 56

. The median is

56

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option