In an arithmetic progression (AP), the \( n \)-th term can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.
Given:
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The fourth term \( a_4 = 7 \): \[ a + 3d = 7 \quad \text{(1)} \]
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The sixth term \( a_6 = 11 \): \[ a + 5d = 11 \quad \text{(2)} \]
Now, we can solve these two equations simultaneously.
Let's subtract equation (1) from equation (2):
\[ (a + 5d) - (a + 3d) = 11 - 7 \]
This simplifies to:
\[ 2d = 4 \]
Dividing both sides by 2 gives:
\[ d = 2 \]
Now, substitute \( d = 2 \) back into equation (1) to find \( a \):
\[ a + 3(2) = 7 \] \[ a + 6 = 7 \] \[ a = 1 \]
Now we have both \( a \) and \( d \):
- \( a = 1 \)
- \( d = 2 \)
We need to find the ninth term \( a_9 \):
\[ a_9 = a + 8d \] \[ a_9 = 1 + 8(2) \] \[ a_9 = 1 + 16 = 17 \]
Thus, the ninth term of the AP is \( \boxed{17} \).
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