JEE Mathematics 2023 Question with Solution

Given: $9 = x_1 < x_2 < \cdots < x_7$ are in an A.P. with common difference $d$. The standard deviation is $4$.

Find: $\overline{x} + x_6$.

Given that $x_1 = 9 < x_2 < \cdots < x_7$, the terms of the A.P. are:

$$x_1 = 9,\; x_2 = 9 + d,\; x_3 = 9 + 2d,\; \dots,\; x_7 = 9 + 6d$$

To simplify, subtract $9$ from all terms:

$$0,\; d,\; 2d,\; \dots,\; 6d$$

The mean is:

$$\bar{x}_{\text{new}} = \frac{0 + d + 2d + \cdots + 6d}{7} = \frac{21d}{7} = 3d$$

The variance is:

$$\sigma^2 = \frac{1}{7} \left(0^2 + 1^2 + 2^2 + \cdots + 6^2\right) d^2 - \bar{x}_{\text{new}}^2$$

Using the sum of squares formula:

$$\sum_{k=0}^6 k^2 = \frac{n(n+1)(2n+1)}{6},\; n = 6$$ $$\sum_{k=0}^6 k^2 = \frac{6(7)(13)}{6} = 91$$

Thus:

$$\sigma^2 = \frac{1}{7}(91)d^2 - (3d)^2$$ $$\sigma^2 = \frac{91}{7}d^2 - 9d^2$$ $$\sigma^2 = 13d^2 - 9d^2 = 4d^2$$

The standard deviation is given as $4$:

$$\sqrt{4d^2} = 4 \implies d^2 = 4 \implies d = 2$$

Now, calculate $\bar{x} + x_6$:

$$\bar{x} = 3d + 9 = 3(2) + 9 = 15$$ $$x_6 = 9 + 5d = 9 + 5(2) = 19$$ $$\bar{x} + x_6 = 15 + 19 = 34$$

Therefore, $\overline{x} + x_6 = 34$. The correct option is B. Note: the solution incorrectly labels the correct option as C, but its working gives $34$, which matches option B.

Given: Seven terms of an A.P. starting from $9$ with common difference $d$.

Find: $\overline{x} + x_6$.

For seven terms in an A.P., the mean is the middle term:

$$\overline{x} = x_4 = 9 + 3d$$

Also,

$$x_6 = 9 + 5d$$

So,

$$\overline{x} + x_6 = (9 + 3d) + (9 + 5d) = 18 + 8d$$

Now use the standard deviation. For the shifted sequence

$$0,\; d,\; 2d,\; 3d,\; 4d,\; 5d,\; 6d$$

the deviations from the mean $3d$ are

$$-3d,\; -2d,\; -d,\; 0,\; d,\; 2d,\; 3d$$

Hence,

$$\sigma^2 = \frac{1}{7}(9 + 4 + 1 + 0 + 1 + 4 + 9)d^2 = \frac{28}{7}d^2 = 4d^2$$

Given $\sigma = 4$, we get

$$2d = 4 \implies d = 2$$

Therefore,

$$\overline{x} + x_6 = 18 + 8(2) = 34$$

Therefore, the correct option is B.