Suppose $$a_1, a_2, \ldots, a_n$$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $$5:12$$and $$110 < a_{15} < 120$$, then the sum of the first ten terms of the progression is equal to 

Let the first term of the AP be $$a$$ and the common difference be $$d$$. Since it is an AP of natural numbers, both $$a$$ and $$d$$ are positive integers.

The sum of the first $$n$$ terms of an AP is $$S_n = \dfrac{n}{2}(2a + (n-1)d)$$.

$$S_5 = \dfrac{5}{2}(2a + 4d) = 5(a + 2d)$$

$$S_9 = \dfrac{9}{2}(2a + 8d) = 9(a + 4d)$$

Given $$\dfrac{S_5}{S_9} = \dfrac{5}{12}$$:

$$\dfrac{5(a + 2d)}{9(a + 4d)} = \dfrac{5}{12}$$

$$12(a + 2d) = 9(a + 4d)$$

$$12a + 24d = 9a + 36d$$

$$3a = 12d$$

$$a = 4d$$

Since $$a$$ and $$d$$ are natural numbers, the smallest possible value is $$d = 1$$ and $$a = 4$$.

The AP is $$4, 5, 6, 7, 8, \ldots$$ — all natural numbers, as required.

We can verify: $$S_5 = 5(4 + 2) = 30$$ and $$S_9 = 9(4 + 4) = 72$$. The ratio $$\dfrac{30}{72} = \dfrac{5}{12}$$. Correct.

The sum of the first term and the common difference is $$a + d = 4 + 1 = 5$$.

The answer is Option C: $$5$$.