JEE Mathematics 2025 Question with Solution

Given: The numbers $8, 21, 34, 47, \dots, 320$ form an arithmetic progression.

Find: The variance of these numbers.

First term, common difference, and last term are:

$$a = 8, \quad d = 21 - 8 = 13, \quad l = 320$$

The number of terms is found from the nth term formula:

$$a_n = a + (n-1)d$$

So,

$$320 = 8 + (n-1) \times 13$$ $$320 - 8 = (n-1) \times 13$$ $$312 = (n-1) \times 13$$ $$n-1 = \frac{312}{13} = 24$$ $$n = 25$$

For an arithmetic sequence, the variance is:

$$\sigma^2 = \frac{1}{12}(n^2-1)d^2$$

Substituting $n = 25$ and $d = 13$,

$$\sigma^2 = \frac{1}{12}(25^2-1) \times 13^2$$ $$\sigma^2 = \frac{1}{12}(624) \times 169$$ $$\sigma^2 = \frac{105456}{12}$$ $$\sigma^2 = 8788$$

Therefore, the variance is $8788$.

Given: The numbers $8, 21, 34, 47, \dots, 320$ are in arithmetic progression.

Find: The variance.

First identify:

$$a = 8, \quad d = 13, \quad l = 320$$

From

$$l = a + (n-1)d$$

we get

$$320 = 8 + (n-1) \cdot 13$$

which gives

$$n = 25$$

Now the mean is

$$\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$$

The sum of the arithmetic progression is

$$S_n = \frac{n}{2}(a+l)$$

Hence,

$$S_{25} = \frac{25}{2}(8+320) = \frac{25}{2} \times 328 = 4100$$

So the mean is

$$\bar{x} = \frac{4100}{25} = 164$$

Using the variance definition,

$$\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2$$

For an arithmetic progression, this simplifies to

$$\sigma^2 = \frac{1}{12}(n^2-1)d^2$$

Thus,

$$\sigma^2 = \frac{1}{12}(25^2-1) \times 13^2 = 8788$$

Therefore, the variance is $8788$.