Question

If \(S_n=3+7+11....\) upto \(n\) terms and \(40<\frac {6}{n(n+1)}\displaystyle\sum_{k=1}^n S_k<45\). Then \(n\) is

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The Correct Option is A

To solve this problem, we first need to identify and work with the sequence and summation provided in the question.

The sequence given is \(S_n: 3, 7, 11, \ldots\), which is an arithmetic sequence where the first term \(a = 3\) and the common difference \(d = 7 - 3 = 4\).

The formula for the sum of the first \(n\) terms of an arithmetic sequence is:

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

Substituting the values of \(a\) and \(d\):

S_n = \frac{n}{2} \left[2 \times 3 + (n-1) \times 4\right] = \frac{n}{2} [6 + 4n - 4] = \frac{n}{2} \times (4n + 2)

This simplifies to:

S_n = n(2n + 1)

According to the question, we have the inequality:

40 < \frac{6}{n(n+1)} \sum_{k=1}^n S_k < 45

We calculate \sum_{k=1}^n S_k = S_1 + S_2 + \ldots + S_n, using the formula for each term:

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

The sum of squares formula is \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} and the sum of the first \(n\) natural numbers is \sum_{k=1}^n k = \frac{n(n+1)}{2}.

Substitute these into our equation:

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

Simplifying further:

\sum_{k=1}^n S_k = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}

We need to find \(n\) such that:

40 < \frac{6}{n(n+1)} \left(\frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}\right) < 45

Simplify the inequality. First, multiply across by \(n(n+1)\):

40n(n+1) < 6 \left( \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \right) < 45n(n+1)

Simplify inside the brackets:

\Rightarrow \left[ \frac{2n(n+1)(2n+1)}{1} + 3n(n+1)\right]

Substitute these back and solve:

40n(n+1) < 2n(n+1)(2n+1) + 3n(n+1) < 45n(n+1)

Try substituting values for \(n = 9, 10, 11, 12\) into both parts of the inequality to check if it holds. After substitution, you will find it holds only for:

Correct Answer: 9