Question

The sum of the first n terms of the arithmetic progression is equal to half the sum of the next n terms of the same progression. Find the ratio of the sum of the first 3n terms of the progression to the sum of its first n terms.

• A. 5 : 2
• B. 6 : 1
• C. 7 : 3
• D. 8 : 5

Answer 1

The correct option is B 6 : 1

Given is the sum of the first $n$ terms of an A.P which is equal to half of the sum of the next 'n' terms. This suggests that the sum of the first $2n$ terms is thrice the sum of the first $n$ terms.

Let the first term of the A.P ber $a$ and the common difference be $d$.

$\therefore \frac{S_{2n}}{S_n}=3=>\frac{n\times [2a+(2n-1)d]}{\frac{n}{2}\times [2a+(n-1)d]}=3=>4a+4nd-2d=6a+3nd-3d=>2a=nd+d=>\frac{2a}{d+nd}=1$

Now,

$\frac{S_{3n}}{S_n}=\frac{\frac{3n}{2}\times [2a+(3n-1)d]}{\frac{n}{2}\times [2a+(n-1)d]}=\frac{6a+9nd-3d}{2a+nd-d}=3+\frac{6nd}{2a+nd-d}=3+\frac{6nd}{2nd}=3+3=6$