Let S_1 be the sum of first 2n terms of an arithmetic progression. Let S_2 be the sum of first 4n terms of the same arithmetic progression. If S_2-S_1…

Question

Let S_1 be the sum of first 2n terms of an arithmetic progression. Let S_2 be the sum of 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:

MARKS Content Team · Accepted Answer

Correct answer: (D) 3000. S_2n=2n/22a+2n-1d, S_4n=4n/22a+4n-1d ⇒S_2-S_1=4n/2[2a+(4n-1)d]-2n/22a+2n-1d =4an+(4n-1)2nd-2na-(2n-1)dn =2na+nd[8n-2-2n+1] ⇒2na+nd[6n-1]=1000 2a+6n-1d=1000/n Now, S_6n=6n/22a+6n-1d =3n·1000/n=3000

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

Solution

Relevant Questions

Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If S2-S1 is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to:

Solution

Relevant Questions

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