The sum of the first n terms of the arithmetic progression is equal to half the sum of the next n terms of the same progression. Find the ratio of the sum of the first 3n terms of the progression to the sum of its first n terms.

Let's solve the problem step by step.

Step 1: Let's assume that the first term of the arithmetic progression is 'a' and the common difference is 'd'.

Step 2: The sum of the first n terms of the arithmetic progression can be calculated using the formula Sn = (n/2)(2a + (n-1)d).

Step 3: According to the problem, the sum of the first n terms is equal to half the sum of the next n terms. Mathematically, this can be written as Sn = (1/2)(2Sn).

Step 4: Substituting the formula for Sn in the equation, we get (n/2)(2a + (n-1)d) = (1/2)(2(n/2)(2a + (2n-1)d)).

Step 5: Simplifying the equation, we get (n/2)(2a + (n-1)d) = (n/2)(2a + (2n-1)d).

Step 6: Canceling out the common terms, we get 2a + (n-1)d = 2a + (2n-1)d.

Step 7: Simplifying further, we get (n-1)d = (2n-1)d.

Step 8: Canceling out the common terms, we get n-1 = 2n-1.

Step 9: Solving the equation, we get n = 1.

Step 10: Now, we need to find the ratio of the sum of the first 3n terms to the sum of the first n terms. Substituting n = 1, we get the ratio as (3n/2n) = 3/2.

Therefore, the ratio of the sum of the first 3n terms of the arithmetic progression to the sum of its first n terms is 3/2.