Let $S_n$ be the sum of the first n terms of an arithmetic progression. If $S_{3n} = 3S_{2n}$, then the value of $\frac{S_{4n}}{S_{2n}}$ is :
Show Hint
For ratio problems in A.P., if the relation holds for all $n$, it must hold for $n=1$. Testing with $n=1$ gives $S_3 = 3S_2 \implies 3a + 3d = 3(2a + d) \implies 3a = 0 \implies a = 0$. For $a=0$, $S_k \propto k(k-1)$. Then $S_4 / S_2 = (4 \cdot 3) / (2 \cdot 1) = 6$.
Step 1: Understanding the Concept:
The sum of an Arithmetic Progression (A.P.) depends on the first term (\(a\)) and the common difference (\(d\)).
The problem provides a relationship between sums of different numbers of terms, which allows us to find a ratio between \(a\) and \(d\). Step 2: Key Formula or Approach:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
Step 3: Detailed Explanation:
Given \(S_{3n} = 3 S_{2n}\):
\[ \frac{3n}{2} [2a + (3n-1)d] = 3 \cdot \frac{2n}{2} [2a + (2n-1)d] \]
Divide both sides by \(\frac{3n}{2}\):
\[ [2a + 3nd - d] = 2 [2a + 2nd - d] \]
\[ 2a + 3nd - d = 4a + 4nd - 2d \]
Rearrange to group terms of \(a\) and \(d\):
\[ 2a + nd - d = 0 \Rightarrow 2a = (1 - n)d \]
Now, calculate the ratio \(\frac{S_{4n}}{S_{2n}}\):
\[ S_{4n} = \frac{4n}{2} [2a + (4n-1)d] = 2n [ (1-n)d + (4n-1)d ] = 2n [ 3nd ] = 6n^2 d \]
\[ S_{2n} = \frac{2n}{2} [2a + (2n-1)d] = n [ (1-n)d + (2n-1)d ] = n [ nd ] = n^2 d \]
The ratio is:
\[ \frac{S_{4n}}{S_{2n}} = \frac{6n^2 d}{n^2 d} = 6 \] Step 4: Final Answer:
The value of the ratio is 6.
