Question:medium

The sum of the first n terms of two AP's are in the ratio \((2n+3):(3n-1)\). The ratio of their 5th terms is

Show Hint

Check proportional constants carefully in ratio problems.
Updated On: Apr 18, 2026
  • 11:6
  • 21:26
  • 13:16
  • 8:5
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Given the ratio of sums \( S_n / S'_n \), the ratio of the \( k\text{-th} \) terms \( a_k / a'_k \) can be found by substituting \( n = 2k - 1 \).
Step 2: Detailed Explanation:
Let the ratio of sums be \( \frac{S_n}{S'_n} = \frac{2n + 3}{3n - 1} \).
We know that \( \frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n-1)d]}{\frac{n}{2}[2A + (n-1)D]} = \frac{a + (\frac{n-1}{2})d}{A + (\frac{n-1}{2})D} \).
We want the ratio of the 5th terms: \( \frac{a_5}{A_5} = \frac{a + 4d}{A + 4D} \).
Comparing the expressions, we need \( \frac{n-1}{2} = 4 \).
\[ n - 1 = 8 \implies n = 9 \]
Substitute \( n = 9 \) into the given ratio:
\[ \text{Ratio} = \frac{2(9) + 3}{3(9) - 1} \]
\[ \text{Ratio} = \frac{18 + 3}{27 - 1} = \frac{21}{26} \].
Step 3: Final Answer:
The ratio of the 5th terms is 21:26.
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