Question

The $S_n$ denote the sum of the first $n$ terms of an A.P., if $S_{2n} = 3S_n$. Then $S_{3n}:S_n$ is equal to

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

Answer: (B)

6

Answer 2

Let $S_k$ be the sum of the first $k$ terms of an arithmetic progression (A.P.) with first term $a$ and common difference $d$. The formula for $S_k$ is given by: $S_k = \frac{k}{2} [2a + (k-1)d]$

We are given the condition $S_{2n} = 3S_n$. Using the formula for $S_k$: $S_{2n} = \frac{2n}{2} [2a + (2n-1)d] = n [2a + (2n-1)d]$ $S_n = \frac{n}{2} [2a + (n-1)d]$

Substituting these into the given condition: $n [2a + (2n-1)d] = 3 \left( \frac{n}{2} [2a + (n-1)d] \right)$

Assuming $n \neq 0$, we can divide both sides by $n$: $2a + (2n-1)d = \frac{3}{2} [2a + (n-1)d]$

Multiply by 2 to clear the fraction: $2 [2a + (2n-1)d] = 3 [2a + (n-1)d]$ $4a + 2(2n-1)d = 6a + 3(n-1)d$ $4a + (4n-2)d = 6a + (3n-3)d$

Rearranging the terms to find a relationship between $a$ and $d$: $(4n-2)d - (3n-3)d = 6a - 4a$ $(4n-2 - 3n + 3)d = 2a$ $(n+1)d = 2a$

This relation must hold for the given $n$.

Now we need to find the ratio $S_{3n} : S_n$. $S_{3n} = \frac{3n}{2} [2a + (3n-1)d]$

The ratio is: $\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2} [2a + (3n-1)d]}{\frac{n}{2} [2a + (n-1)d]}$

Cancel out $\frac{n}{2}$: $\frac{S_{3n}}{S_n} = \frac{3 [2a + (3n-1)d]}{[2a + (n-1)d]}$

Substitute $2a = (n+1)d$ into the expression: Numerator: $2a + (3n-1)d = (n+1)d + (3n-1)d = (n+1+3n-1)d = 4nd$ Denominator: $2a + (n-1)d = (n+1)d + (n-1)d = (n+1+n-1)d = 2nd$

So, the ratio becomes: $\frac{S_{3n}}{S_n} = \frac{3 [4nd]}{[2nd]}$

Assuming $n \neq 0$ and $d \neq 0$ (if $d=0$, then $a=0$, leading to trivial sums), we can cancel $nd$: $\frac{S_{3n}}{S_n} = \frac{3 \times 4}{2} = \frac{12}{2} = 6$

Thus, $S_{3n} : S_n = 6$.

Alternative Method: Let $U_1 = S_n$. Let $U_2 = S_{2n} - S_n$. Let $U_3 = S_{3n} - S_{2n}$.

The sums of consecutive blocks of $n$ terms in an A.P. form another A.P. We are given $S_{2n} = 3S_n$. So, $U_2 = 3S_n - S_n = 2S_n$. Thus, $U_1 = S_n$ and $U_2 = 2S_n$. Since $U_1, U_2, U_3$ form an A.P., the common difference is $U_2 - U_1 = 2S_n - S_n = S_n$. Therefore, $U_3 = U_2 + (U_2 - U_1) = 2S_n + S_n = 3S_n$.

Now, $S_{3n} = U_1 + U_2 + U_3 = S_n + 2S_n + 3S_n = 6S_n$. The ratio $S_{3n} : S_n = 6S_n : S_n = 6$.