Let S 1 be the sum of first 2 n terms of an arithmetic progression. Let S2 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 6 n terms of the arithmetic progression is equal to:

Question

Let S 1 be the sum of first 2 n terms of an arithmetic progression. Let S2 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 6 n terms of the arithmetic progression is equal to: (A) 1000 (B) 7000 (C) 5000 (D) 3000

Tardigrade Admin · Accepted Answer

S2 n=(2 n/2)[2 a+(2 n-1) d], S4 n=(4 n/2)[2 a+(4 n-1)d] ⇒ S 2- S 1=(4 n /2)[2 a +(4 n -1) d ]-(2 n /2)[2 a +(2 n -1)d] =4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n =2 na + nd [8 n -2-2 n +1] ⇒ 2 na + nd [6 n -1]=1000 2 a+(6 n-1) d=(1000/n) Now, S6 n =(6 n /2)[2 a +(6 n -1) d ] =3 n ⋅ (1000/ n )=3000

Q. 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:

1000

7000

5000

3000

S2n​=22n​[2a+(2n−1)d],S4n​=24n​[2a+(4n−1)d] ⇒S2​−S1​=24n​[2a+(4n−1)d]−22n​[2a+(2n−1)d] =4an+(4n−1)2nd−2na−(2n−1)dn =2na+nd[8n−2−2n+1] ⇒2na+nd[6n−1]=1000 2a+(6n−1)d=n1000​ Now, S6n​=26n​[2a+(6n−1)d] =3n⋅n1000​=3000

Q. Let $S_{1}$ be the sum of first $2 n$ 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 $6 n$ terms of the arithmetic progression is equal to: