Solveeit Logo
Login

Question

Mathematics Question on Sequence and series

Let SnS_n denote the sum of the first n terms of an A.P. If S2n=3SnS_{2n} = 3S_n then S3n:SnS_{3n} : S_n is equal to

A

4

B

6

C

8

D

10

Answer

6

Explanation

Solution

S2n=3SnS_{2n} = 3S_{n} ⇒2n2[2a+(2n−1)d]\Rightarrow \frac{2n}{2}\left[2a+\left(2n-1\right)d\right] =3⋅n2[2a+(n−1)d]= 3\cdot\frac{n}{2}\left[2a+\left(n-1\right)d\right] ⇒2[2a+(2n−1)d]=3[2a+1(n−1)d]\Rightarrow 2\left[2a+\left(2n-1\right)d\right] = 3\left[2a+1\left(n-1\right)d\right] ⇒2a=(4n−2−3n+3)d=(n+1)d\Rightarrow 2a = \left(4n-2-3n+3\right)d = \left(n+1\right)d Again S3n:Sn=3n2[2a+(3n−1)d]n2[2a+(n−1)d]S_{3n} : S_{n} = \frac{\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]}{\frac{n}{2}\left[2a+\left(n-1\right)d\right]} =3[2a+(3n−1)d]2a+(n−1)d= \frac{3\left[2a+\left(3n-1\right)d\right]}{2a+\left(n-1\right)d} =3[(n+1)d+(3n−1)d](n+1)d+(n−1)d= \frac{3\left[\left(n+1\right)d + \left(3n-1\right)d\right]}{\left(n+1\right)d +\left(n-1\right)d} =3[4 nd]2 nd=6 = \frac{3\left[4\,nd\right]}{2\,nd} = 6