Let the first term of the arithmetic progression (AP) be \( a \) and the common difference be \( d \). The \( n \)-th term of an AP can be expressed as:
\[ T_n = a + (n-1)d \]
Given terms are as follows:
-
The first term (\( T_1 \)):
\[ T_1 = a \] -
The third term (\( T_3 \)):
\[ T_3 = a + 2d \] -
The seventh term (\( T_7 \)):
\[ T_7 = a + 6d \]
According to the problem, these three terms must form a geometric progression (GP). In a GP, the square of the middle term equals the product of the other two terms. Therefore, we can write the following equation:
\[ (T_3)^2 = T_1 \cdot T_7 \]
Substituting the expressions for the terms we have:
\[ (a + 2d)^2 = a \cdot (a + 6d) \]
Expanding both sides:
\[ (a^2 + 4ad + 4d^2) = a^2 + 6ad \]
Now, simplifying this equation:
\[ a^2 + 4ad + 4d^2 = a^2 + 6ad \]
Subtract \( a^2 \) from both sides:
\[ 4ad + 4d^2 = 6ad \]
Rearranging gives:
\[ 4d^2 + 4ad - 6ad = 0 \]
\[ 4d^2 - 2ad = 0 \]
Factoring out \( 2d \):
\[ 2d(2d - a) = 0 \]
Thus, this gives us two possible scenarios:
- \( d = 0 \) (which would mean all terms are equal, not forming a progression).
- \( 2d - a = 0 \) implies \( a = 2d \).
Now we need to find the common ratio \( r \) of the GP. The terms in GP are \( a \), \( a + 2d \), and \( a + 6d \).
Substituting \( a = 2d \):
-
First term:
\[ T_1 = 2d \] -
Third term:
\[ T_3 = 2d + 2d = 4d \] -
Seventh term:
\[ T_7 = 2d + 6d = 8d \]
Now, finding the common ratio \( r \):
\[ r = \frac{T_3}{T_1} = \frac{4d}{2d} = 2 \]
And also:
\[ r = \frac{T_7}{T_3} = \frac{8d}{4d} = 2 \]
Therefore, the common ratio of the geometric progression is:
\[ \boxed{2} \]