If the ratio of the 11th term of an AP to its 18th term is 2 : 3, find the ratio of the sum of its first five terms to the sum of its first 10 terms.

Correct option is D

Given:
The ratio of the 11th term to the 18th term of an AP is 2:3.
Formula Used:
The nth term of an AP is given by $$a_n$$​ = a + (n-1)d,

where a is the first term and d is the common difference.
Sum of the first n terms of an AP ($$S_n = \frac{n}{2} \times (2a + (n-1)d)$$​​
Solution:
The 11th term of the AP is $$a_{11} = a + 10d \text { and the 18th term is}\, a_{18} = a + 17d$$​​
Given,

​$$\frac{a + 10d}{a + 17d} = \frac{2}{3}$$​​
3(a + 10d) = 2(a + 17d)
3a + 30d = 2a + 34d
 a = 4d
Now, we need to find the ratio of the sum of the first five terms to the sum of the first ten terms.
Sum of the first 5 terms $$S_5 = \frac{5}{2} \times (2a + 4d)$$​​
Sum of the first 10 terms $$S_{10} = \frac{10}{2} \times (2a + 9d)$$​​
Put a= 4d
​$$S_5 = \frac{5}{2} \times (8d + 4d) = 30d\\S_{10} = 5 \times (8d + 9d) = 85d\\$$​​
The required ratio $$\frac{S_5}{S_{10}} = \frac{30d}{85d} = \frac{6}{17}$$​​