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 formula for the $n^{th}$ term of an $A.P.$ is $a_{n} = a_{1} + (n-1)d$.Given $a_{1} = 1$ and $a_{n} = 300$,we have $300 = 1 + (n-1)d$,which im

Vedclass · Accepted Answer

$(2490, 248)$

Vedclass · Answer

(C) The formula for the $n^{th}$ term of an $A.P.$ is $a_{n} = a_{1} + (n-1)d$.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 factors of $299$ are $1, 13, 23, 299$.For $n-1$ to be in the range $[14, 49]$,the only possible value is $n-1 = 23$,which gives $n = 24$.Then $d = 13$.We need to find $(S_{n-4}, a_{n-4})$. Since $n = 24$,$n-4 = 20$.$a_{20} = a_{1} + 19d = 1 + 19(13) = 1 + 247 = 248$.$S_{20} = \frac{20}{2}(a_{1} + a_{20}) = 10(1 + 248) = 10(249) = 2490$.Thus,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

If the middle term of the $AP$ is $300$,then the sum of its first $51$ terms is

If $x, y, z \in \mathbb{R}^+$ such that $x + y + z = 4$,then the maximum possible value of $xyz^2$ is

If $a, b, c$ are in Arithmetic Progression,then $\frac{1}{\sqrt{b} + \sqrt{c}}, \frac{1}{\sqrt{c} + \sqrt{a}}, \frac{1}{\sqrt{a} + \sqrt{b}}$ are in:

If the roots of the equation $x^3 - 9x^2 + \alpha x - 15 = 0$ are in $A.P.$,then $\alpha$ is:

If $a$,$b$,and $c$ are positive real numbers,then $a/b + b/c + c/a$ is greater than or equal to what value?

Vedclass Test Series

Exam Paper Generator

Online Exam Module