Final answer:
To solve this problem, we use the formula for the nth term of an arithmetic progression and the formula for the sum of an arithmetic series. By solving the equations, we find that the common difference is equal to the first term, which means the arithmetic progression consists of equal terms. This leads to the conclusion that the sum of the first ten terms is eight times the sum of the first three terms.
Step-by-step explanation:
To solve this problem, let's use the formula for the nth term of an arithmetic progression: an = a1 + (n-1)d, where an is the nth term, a1 is the first term, n is the term number, and d is the common difference.
Given that the tenth term is three times the third term, we can write the equation: a10 = 3a3.
Substituting into the formula, we have: a1 + 9d = 3(a1 + 2d).
Simplifying, we get: a1 - d = 0.
This means that the common difference is equal to the first term (d = a1). Therefore, the arithmetic progression is a sequence of equal terms.
Now, let's consider the sum of the first ten terms and the sum of the first three terms. Using the formula for the sum of an arithmetic series, we have:
S10 = (10/2)(a1 + a10)
S3 = (3/2)(a1 + a3)
Since the arithmetic progression is a sequence of equal terms, we can simplify the expressions to:
S10 = 10a1
S3 = 3a1
Therefore, the sum of the first ten terms is 8 times the sum of the first three terms, so the statement is True.