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(A) The given arithmetic progression is $3, 7, 11, \ldots$ with first term $a = 3$ and common difference $d = 4$.The sum of the first $n$ terms is $S_

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$9$

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(A) The given arithmetic progression is $3, 7, 11, \ldots$ with first term $a = 3$ and common difference $d = 4$.The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[6 + (n-1)4] = \frac{n}{2}[4n + 2] = 2n^2 + n$.Now,we calculate $\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k) = 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$.Using the standard summation formulas,$\sum_{k=1}^{n} S_k = 2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.Factoring out $\frac{n(n+1)}{6}$,we get $\sum_{k=1}^{n} S_k = \frac{n(n+1)}{6} [2(2n+1) + 3] = \frac{n(n+1)(4n+5)}{6}$.Substituting this into the given inequality: $40 This simplifies to $40 Subtracting $5$ from all parts: $35 Dividing by $4$: $8.75 Since $n$ must be an integer,$n = 9$.

Let $S_n$ be the sum of the first $n$ terms of an arithmetic progression $3, 7, 11, \ldots$. If $40 < \left(\frac{6}{n(n+1)} \sum_{k=1}^{n} S_{k}\right) < 42$,then $n$ equals

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