If the ratio of the $$11^{th}$$ term of an AP to its $$18^{th}$$ 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 of an AP to its 18th term = 2 : 3 

Formula Used:

General Term of an AP:

T_n= a+(n−1)d

where a is the first term and d is the common difference.

Sum of First n Terms in an AP:

​$$S_n = \frac{n}{2} \left(2a + (n-1)d\right)$$​​

Solution:

Let the first term of the AP be a and the common difference be d.

The 11th term: 

​$$T_{11} = a + (11 - 1)d = a + 10d$$ 

The 18th term:

​$$T_{18} = a + (18 - 1)d = a + 17d$$ 

We are given the ratio of the 11th term to the 18th term as 2:3. So, 

​$$\frac{T_{11}}{T_{18}} = \frac{2}{3}$$ 

​$$\frac{T_{11}}{T_{18}} = \frac{a + 10d}{a + 17d} = \frac{2}{3}$$

3( a + 10d ) = 2 ( a + 17d )

3a + 30d = 2a + 34d

a = 4d

For the first 5 terms:

​$$S_5 = \frac{5}{2} \left( 2a + 4d \right)$$

​​Substitute $a=4d$:

​$$S_5 = \frac{5}{2} \left( 2(4d) + 4d \right) = \frac{5}{2} \left( 8d + 4d \right) = \frac{5}{2} \times 12d = 30d$$

​​For the first 10 terms:

​$$S_{10} = \frac{10}{2} \left( 2a + 9d \right)$$

Substitute a =4 d:

​$$S_{10} = 5 \left( 2(4d) + 9d \right) = 5 \left( 8d + 9d \right) = 5 \times 17d = 85d$$

The ratio of the sum of the first 5 terms to the sum of the first 10 terms is:

​$$\frac{S_5}{S_{10}} = \frac{30d}{85d} = \frac{30}{85} = \frac{6}{17}$$

The ratio of the sum of the first 5 terms to the sum of the first 10 terms is 6:17