JEE Mathematics 2024 Question with Solution

Given: The arithmetic progression is $3, 7, 11, \ldots$, so first term is $a = 3$ and common difference is $d = 4$.

Find: The integer value of $n$ satisfying the given inequality.

First find the sum of the first $n$ terms:

$$S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) = \frac{n}{2} \left(4n + 2\right) = n(2n + 1)$$

Now,

$$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} k(2k + 1) = 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$$

Using

$$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}, \quad \sum_{k=1}^{n} k = \frac{n(n + 1)}{2}$$

we get

$$\sum_{k=1}^{n} S_k = \frac{n(n + 1)(4n + 5)}{6}$$

Substituting into the expression used in the solution,

$$\frac{6}{n(n + 1)} \sum_{k=1}^{n} S_k = \frac{6}{n(n + 1)} \left( \frac{n(n + 1)(4n + 5)}{6} \right) = 4n + 5$$

Now solve the inequality shown in the detailed working:

$$40 < 4n + 5 < 42$$

So,

$$35 < 4n < 37$$

and hence,

$$8.75 < n < 9.25$$

Since $n$ is an integer, the only possible value is $9$.

Therefore, the value of $n$ is $9$.

Given: $a = 3$ and $d = 4$ for the arithmetic progression $3, 7, 11, \ldots$.

Find: The integer value of $n$.

For the first $k$ terms,

$$S_k = \frac{k}{2}[2a + (k-1)d]$$

Substitute $a = 3$ and $d = 4$:

$$S_k = \frac{k}{2}[2(3) + (k-1)4] = \frac{k}{2}[6 + 4k - 4] = \frac{k}{2}(4k + 2)$$

Thus,

$$S_k = k(2k + 1) = 2k^2 + k$$

Now sum from $k = 1$ to $n$:

$$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k) = 2\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$$

Using standard formulas,

$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$

and

$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$

So,

$$\sum_{k=1}^{n} S_k = 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2}$$

Take $\frac{n(n+1)}{2}$ common:

$$\sum_{k=1}^{n} S_k = \frac{n(n+1)}{2} \left[ 2\frac{(2n+1)}{3} + 1 \right]$$

Therefore,

$$\sum_{k=1}^{n} S_k = \frac{n(n+1)}{2} \left[ \frac{4n+2+3}{3} \right] = \frac{n(n+1)(4n+5)}{6}$$

The detailed the solution then forms

$$E = \frac{6}{n(n+1)} \sum_{k=1}^{n} S_k$$

Hence,

$$E = \frac{6}{n(n+1)} \left( \frac{n(n+1)(4n+5)}{6} \right) = 4n + 5$$

Now solve

$$40 < E < 42$$

that is,

$$40 < 4n + 5 < 42$$

Subtracting $5$,

$$35 < 4n < 37$$

Dividing by $4$,

$$\frac{35}{4} < n < \frac{37}{4}$$

So,

$$8.75 < n < 9.25$$

The only integer in this interval is $9$.

Therefore, the value of $n$ is $9$.

The solution indicates the effective inequality used is $40 < \frac{6}{n(n+1)}\sum_{k=1}^{n} S_k < 42$, which matches the final answer.