Statistics – Variance – JEE Main 2025 Shift 2

Solution:

The given sequence is an Arithmetic Progression (AP).
First term $a = 8$.
Common difference $d = 21 – 8 = 13$.
Last term $l = 320$.

Find the number of terms $n$:
$$l = a + (n-1)d$$
$$320 = 8 + (n-1)13$$
$$312 = 13(n-1)$$
$$n-1 = \frac{312}{13} = 24$$
$$n = 25$$

The formula for the variance ($\sigma^2$) of the first $n$ terms of an AP is:
$$\sigma^2 = \frac{d^2(n^2 – 1)}{12}$$

Derivation/Explanation of the Formula:
1. The variance of the first $n$ natural numbers ($1, 2, 3, \dots, n$) is $\frac{n^2 – 1}{12}$.
2. Our AP is just like the natural numbers, but scaled by the common difference $d$.
3. When you multiply every number in a set by $d$, the variance gets multiplied by $d^2$.
4. Therefore, the variance of an AP is $d^2 \times \left( \frac{n^2 – 1}{12} \right)$.

Substitute $d=13$ and $n=25$:
$$\sigma^2 = \frac{13^2 (25^2 – 1)}{12}$$
$$\sigma^2 = \frac{169 (625 – 1)}{12}$$
$$\sigma^2 = \frac{169 \times 624}{12}$$
$$\sigma^2 = 169 \times 52$$
$$\sigma^2 = 8788$$

Ans. (8788)