Let S n be the sum to n -terms of an arithmetic progression 3, 7, 11,..... If 40

Question

Let $S_{n}$ be the sum to $n$ -terms of an arithmetic progression 3, 7, 11,..... If $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42,$ then $n$ equals _____ . [2024]

Smart Achievers · Accepted Answer

(9)

Let $S_{k} = 3 + 7 + 11 + \dots$ upto $k$ terms

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

$= k (2 k + 1) = 2 k^{2} + k$

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

Since, $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42$

$\Rightarrow 40 < 2 (2 n + 1) + 3 < 42 \Rightarrow 40 < 4 n + 5 < 42$

$\Rightarrow 35 < 4 n < 37 \Rightarrow 8.75 < n < 9.25$

$∴ n = 9$

Solution

Q.
Let $S_{n}$ be the sum to $n$ -terms of an arithmetic progression 3, 7, 11,..... If $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42,$ then $n$ equals _____ . [2024]

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Solution

Q. Let Sn be the sum to n-terms of an arithmetic progression 3, 7, 11,..... If 40<(6n(n+1)∑k=1nSk)<42, then n equals _____ . [2024]

Ans. (9) Let Sk=3+7+11+… upto k terms =k2[2×3+(k-1)(4)]=k2[6+4k-4] =k(2k+1)=2k2+k ∴ ∑k=1n(2k2+k)=2n(n+1)(2n+1)6+n(n+1)2 Since, 40<(6n(n+1)∑k=1nSk)<42 ⇒40<2(2n+1)+3<42⇒40<4n+5<42 ⇒35<4n<37⇒8.75<n<9.25 ∴ n=9

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Q.
Let $S_{n}$ be the sum to $n$ -terms of an arithmetic progression 3, 7, 11,..... If $40 < (\frac{6}{n (n + 1)} \sum_{k = 1}^{n} S_{k}) < 42,$ then $n$ equals _____ . [2024]