Let a_{1}, a_{2}, ldots, a_{n} be a given A.P | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The $n$-th term of an $A.P.$ is given by $a_{n} = a_{1} + (n-1)d$.Given $a_{1} = 1$ and $a_{n} = 300$,we have $300 = 1 + (n-1)d$,which implies $(n

Vedclass · Accepted Answer

$(2490, 248)$

Vedclass · Answer

(C) The $n$-th term of an $A.P.$ is given by $a_{n} = a_{1} + (n-1)d$.Given $a_{1} = 1$ and $a_{n} = 300$,we have $300 = 1 + (n-1)d$,which implies $(n-1)d = 299$.The prime factorization of $299$ is $13 	imes 23$.Since $15 \leq n \leq 50$,we have $14 \leq n-1 \leq 49$.The possible values for $(n-1)$ are $23$ or $13$ (not possible as $n-1 \geq 14$).Thus,$n-1 = 23 \Rightarrow n = 24$ and $d = 13$.We need to find $(S_{n-4}, a_{n-4}) = (S_{20}, a_{20})$.$a_{20} = a_{1} + 19d = 1 + 19(13) = 1 + 247 = 248$.$S_{20} = \frac{20}{2}(a_{1} + a_{20}) = 10(1 + 248) = 10(249) = 2490$.Therefore,the ordered pair is $(2490, 248)$.

Let $a_{1}, a_{2}, \ldots, a_{n}$ be a given $A.P.$ whose common difference is an integer and $S_{n} = a_{1} + a_{2} + \ldots + a_{n}$. 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

Similar Questions

How many terms of the $A.P.$ $1, 4, 7, \ldots$ are needed to give the sum $715$?

The sum of the first $20$ terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is?

Maximum value of the sum of the Arithmetic Progression $50, 48, 46, 44, \dots$ is

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$, then the greatest number of the series will be:

Let the sequence $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$ Then $a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2n - 1}^2 - a_{2n}^2 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module