Let a n be a sequence such that a 1 + a 2 + . . . + a n = n 2 + 3 n ( n + 1 ) ( n + 2 ) . If 28 k = 1 10 1 a k = p 1 p 2 p 3 . . . p m , where p 1 , p 2 , . . . . , p m are the first m prime numbers, then m is equal to [2023]

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Solution

Q.
Let $< a_{n} >$ be a sequence such that $a_{1} + a_{2} + . . . + a_{n} = \frac{n^{2} + 3 n}{(n + 1) (n + 2)}$ . If $28 \sum_{k = 1}^{10} \frac{1}{a_{k}} = p_{1} p_{2} p_{3} . . . p_{m},$ where $p_{1}, p_{2}, . . . ., p_{m}$ are the first $m$ prime numbers, then $m$ is equal to [2023]

1 5

2 6

3 7

4 8

Ans.
(2)

$a_{1} + a_{2} + \dots + a_{n} = \frac{n^{2} + 3 n}{(n + 1) (n + 2)}$

$S_{n} = \frac{n^{2} + 3 n}{(n + 1) (n + 2)}$

$a_{n} = S_{n} - S_{n - 1} = \frac{n^{2} + 3 n}{(n + 1) (n + 2)} - [\frac{{(n - 1)}^{2} + 3 (n - 1)}{n (n + 1)}]$

$= \frac{n^{2} + 3 n}{(n + 1) (n + 2)} - \frac{(n - 1) (n + 2)}{n (n + 1)}$

$a_{n} = \frac{4}{n (n + 1) (n + 2)}$

$28 \sum_{k = 1}^{10} \frac{1}{a_{k}} = 28 \sum_{k = 1}^{10} \frac{k (k + 1) (k + 2)}{4} = 7 \sum_{k = 1}^{10} k (k + 1) (k + 2)$

$= 7 [{(\frac{10 \times 11}{2})}^{2} + \frac{3 \times 10 \times 11 \times 21}{6} + \frac{2 \times 10 \times 11}{2}]$

$= 7 [{(5 \times 11)}^{2} + 5 \times 11 \times 21 + 10 \times 11]$

$= 7 \times 5 \times 11 [55 + 21 + 2] = 7 \times 5 \times 11 \times 78$

$= 2 \times 3 \times 5 \times 7 \times 11 \times 13 = p_{1} \cdot p_{2} \cdot p_{3} \dots p_{m} \Rightarrow m = 6$

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