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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(2480, 249)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(2490, 249)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(2490, 248)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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, $(S_{20}, a_{20}) = (2490, 248)$

Suggest Corrections

Similar questions

Q. In an A.P. a₁, a₂, a₃, ..., a_n, ...., if a₁ = 1, a_n= 20 and S_n= 399, then the value of n is __________.

Q. Let

a_{1}, a_{2}, . . . . . . . . . . . . . a_{100}

be an A.P. with

a_{1} = 3

and

s_{p} = p \sum i = 1 a i, a \leq p \leq 100

for any integer n with

1 \leq n \leq 20

, let

m = \sqrt{n}

. If

\frac{s_{m}}{s_{n}}

does not depend on n then

a_{2}

Q. Let

a_{1}, a_{2}, a_{3}, \dots \dots, a_{100}

be an arithmetic progression with

a_{1} = 3

and

S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100

. For any integer n with

1 \leq n \leq 20

, let m = 5n. If

\frac{S_{m}}{S_{n}}

does not depend on n, then

a_{2}

is equal to___

Q. If

S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n},

where

a_{1} = 1

and

a_{n} = 2 a_{n - 1} + 1,

for

n \geq 2

, then the value

S_{50}

Q. If

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P. with

s_{n}

as the sum of first 'n' terms

(S_{0} = 0), t h e n \sum_{k = 0}^{n}^{n} C_{k} S_{k}

is equal to

Join BYJU'S Learning Program

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