Let a_{n} be the n^{text{th}} term of the ser | vedclass

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(C) The series is $5, 8, 14, 23, 35, 50, \ldots$. Let the differences between consecutive terms be $d_n = a_{n+1} - a_n$. The differences are $3, 6, 9

Vedclass · Accepted Answer

$11290$

Vedclass · Answer

(C) The series is $5, 8, 14, 23, 35, 50, \ldots$. Let the differences between consecutive terms be $d_n = a_{n+1} - a_n$. The differences are $3, 6, 9, 12, 15, \ldots$,which form an arithmetic progression with the $n^{	ext{th}}$ difference being $3n$. Thus,$a_n = a_1 + \sum_{k=1}^{n-1} 3k = 5 + 3 \frac{(n-1)n}{2} = \frac{3n^2 - 3n + 10}{2}$. For $n=40$,$a_{40} = \frac{3(40)^2 - 3(40) + 10}{2} = \frac{4800 - 120 + 10}{2} = 2345$. Now,$S_n = \sum_{k=1}^{n} \frac{3k^2 - 3k + 10}{2} = \frac{3}{2} \sum k^2 - \frac{3}{2} \sum k + 5 \sum 1$. $S_{30} = \frac{3}{2} \left( \frac{30(31)(61)}{6} ight) - \frac{3}{2} \left( \frac{30(31)}{2} ight) + 5(30) = 14182.5 - 697.5 + 150 = 13635$. Finally,$S_{30} - a_{40} = 13635 - 2345 = 11290$.

Let $a_{n}$ be the $n^{\text{th}}$ term of the series $5+8+14+23+35+50+\ldots$ and $S_{n}=\sum_{k=1}^{n} a_{k}$. Then $S_{30}-a_{40}$ is equal to

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