Let a1-a2-an be a given A-P- whose common difference is an integer and Sn-a1-a2 -an- If a1 - 1- an - 300 and 15 - n - 50- then the ordered pair Sn-4-an-4 is equal to

Let d be the common difference. Given: nbsp;a_n=300 ⇒ 300 = 1 + (n – 1)d ⇒(n – 1) = 299d nbsp; The possible positive factors of d are 1,13,23,299. ⇒ n=300,2 ...

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Standard XII

Mathematics

Greatest Integer Function

Let a1,a2,…,a...

Question

Let $a_{1}, a_{2}, \dots, a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2} + \dots + a_{n}$ . If $a_{1} = 1$ , $a_{n} = 300$ and $15 \leq n \leq 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to

(2480, 248)

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(2480, 249)

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(2490, 249)

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(2490, 248)

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Solution

The correct option is D $(2490, 248)$
Let $d$ be the common difference.
Given: $a_{n} = 300$
$\Rightarrow 300 = 1 + (n - 1) d$
$\Rightarrow (n - 1) = \frac{299}{d}$
The possible positive factors of $d$ are $1, 13, 23, 299.$
$\Rightarrow n = 300, 24, 14, 2$
Since $15 \leq n \leq 50$
$∴ n = 24$ and $d = 13$

Now, $S_{n - 4} = S_{20}$
$\Rightarrow S_{20} = (\frac{20}{2}) (2 + 19 \times 13)$
$∴ S_{20} = 2490$
and $a_{20} = 1 + 19 \times 13$
$∴ a_{20} = 248$
Hence, $(S_{20}, a_{20}) = (2490, 248)$

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Let a1,a2,…,an be a given A.P. whose common difference is an integer and Sn=a1+a2+…+an. If a1=1, an=300 and 15≤n≤50, then the ordered pair (Sn−4,an−4) is equal to

Let $a_{1}, a_{2}, \dots, a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2} + \dots + a_{n}$ . If $a_{1} = 1$ , $a_{n} = 300$ and $15 \leq n \leq 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to