Let S_{1} be the sum of the first 2n terms of | vedclass

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

Question

(D) Let the first term be $a$ and the common difference be $d$.$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$$S_{4n} = \frac{4n}{2}[2a + (4n-

Vedclass · Accepted Answer

$3000$

Vedclass · Answer

(D) Let the first term be $a$ and the common difference be $d$.$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$$S_{4n} = \frac{4n}{2}[2a + (4n-1)d] = 2n[2a + (4n-1)d]$Given $S_{2} - S_{1} = 1000$,where $S_{1} = S_{2n}$ and $S_{2} = S_{4n}$:$2n[2a + (4n-1)d] - n[2a + (2n-1)d] = 1000$$n[4a + 2(4n-1)d - 2a - (2n-1)d] = 1000$$n[2a + (8n - 2 - 2n + 1)d] = 1000$$n[2a + (6n - 1)d] = 1000$$2a + (6n - 1)d = \frac{1000}{n}$Now,the sum of the first $6n$ terms is $S_{6n} = \frac{6n}{2}[2a + (6n-1)d]$$S_{6n} = 3n 	imes \frac{1000}{n} = 3000$

Let $S_{1}$ be the sum of the first $2n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of the first $4n$ terms of the same arithmetic progression. If $(S_{2} - S_{1})$ is $1000$,then the sum of the first $6n$ terms of the arithmetic progression is equal to:

Similar Questions

If the $n^{th}$ terms of two $A.P.$'s are $3n + 8$ and $7n + 15$,then the ratio of their $12^{th}$ terms will be:

Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d=11$. If the maximum value of $a^5 b^3 c^2 d$ is $3750 \beta$,then the value of $\beta$ is

If $a, b, c$ are in $A.P.$,then $\frac{(a - c)^2}{(b^2 - ac)} = $

If $a_1, a_2, a_3, \dots, a_n$ are in $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 114$,then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

Three numbers are in $A.P.$ such that their sum is $18$ and the sum of their squares is $158$. The greatest number among them is

Vedclass Test Series

Exam Paper Generator

Online Exam Module