Question

The ratio of the sums of the first 12 terms and the first 18 terms of an arithmetic progression is 4 : 9. What is the ratio of the 10^th and the 15^th terms?

• A. 3 : 5
• B. 23 : 35
• C. 8 : 27
• D. 19 : 29

Hint:

For arithmetic progression sum problems, remember the formula \( S_n = \frac{n}{2} (2a + (n - 1) d) \) and solve using the given ratio of sums. The ratio of terms is simply the ratio of \( a + (n - 1)d \) for the respective terms.

Answer 1

The Correct Option is D

Let the first term of the arithmetic progression be \( a \) and the common difference be \( d \).
The sum of the first \( n \) terms of an arithmetic progression is given by: \[ S_n = \frac{n}{2} \left( 2a + (n - 1) d \right) \] The given ratio of the sums of the first 12 terms to the first 18 terms is: \[ \frac{S_{12}}{S_{18}} = \frac{\frac{12}{2} \left( 2a + 11d \right)}{\frac{18}{2} \left( 2a + 17d \right)} = \frac{12(2a + 11d)}{18(2a + 17d)} = \frac{2a + 11d}{3(2a + 17d)} = \frac{4}{9} \] Simplifying this equation: \[ 9(2a + 11d) = 12(2a + 17d) \quad \Rightarrow \quad 18a + 99d = 24a + 204d \] \[ 6a = 105d \quad \Rightarrow \quad a = 17.5d \] Now, the ratio of the 10\textsuperscript{th} and 15\textsuperscript{th} terms is given by: \[ \frac{a + 9d}{a + 14d} = \frac{17.5d + 9d}{17.5d + 14d} = \frac{26.5d}{31.5d} = \frac{19}{29} \] Thus, the ratio of the 10\textsuperscript{th} and 15\textsuperscript{th} terms is 19 : 29.