Let the first term of the Arithmetic Progression (AP) be $a$ and the common difference be $d$. The formulas for the $n^{th}$ term and the sum of the first $n$ terms are:

• $n^{th}$ term: $a_n = a + (n-1)d$
• Sum of first $n$ terms: $S_n = \frac{n}{2}[2a + (n-1)d]$

Relating Terms Ratio to AP Variables

We are given that the ratio of the 11th term ($a_{11}$) to the 18th term ($a_{18}$) is 2 : 3.

$ \frac{a_{11}}{a_{18}} = \frac{2}{3} $

Using the formula for the $n^{th}$ term:

$ \frac{a + (11-1)d}{a + (18-1)d} = \frac{2}{3} $

$ \frac{a + 10d}{a + 17d} = \frac{2}{3} $

Solving for First Term ($a$) and Common Difference ($d$)

Cross-multiplying gives:

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

$ 3a + 30d = 2a + 34d $

Rearranging the terms to solve for $a$:

$ 3a - 2a = 34d - 30d $

$ a = 4d $

This shows the first term ($a$) is four times the common difference ($d$).

Calculating Sum of First Five Terms ($S_5$)

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n=5$:

$ S_5 = \frac{5}{2}[2a + (5-1)d] = \frac{5}{2}[2a + 4d] $

Substitute $a = 4d$ into the equation:

$ S_5 = \frac{5}{2}[2(4d) + 4d] = \frac{5}{2}[8d + 4d] = \frac{5}{2}[12d] $

$ S_5 = 5 \times 6d = 30d $

Calculating Sum of First Ten Terms ($S_{10}$)

Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n=10$:

$ S_{10} = \frac{10}{2}[2a + (10-1)d] = 5[2a + 9d] $

Substitute $a = 4d$ into the equation:

$ S_{10} = 5[2(4d) + 9d] = 5[8d + 9d] = 5[17d] $

$ S_{10} = 85d $

Finding the Ratio of Sums ($S_5 : S_{10}$)

Now, find the ratio of $S_5$ to $S_{10}$:

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

Simplify the ratio:

$ \frac{30}{85} = \frac{5 \times 6}{5 \times 17} = \frac{6}{17} $

The ratio of the sum of the first five terms to the sum of the first ten terms is 6 : 17.