If positive integers `a_(1),a_(2),a_(3),.....` are in A.P. such that `a_(8)+a_(10)=24` then the value of `a_(9)` is.

Answer

To solve the problem step by step, we will use the properties of an arithmetic progression (A.P.).

### Step 1: Define the terms of the A.P.
Let the first term of the A.P. be \( a_1 \) and the common difference be \( d \). The \( n \)-th term of an A.P. can be expressed as:
\[ a_n = a_1 + (n - 1)d \]

### Step 2: Write expressions for \( a_8 \) and \( a_{10} \)
Using the formula for the \( n \)-th term:
- For \( a_8 \):
\[ a_8 = a_1 + (8 - 1)d = a_1 + 7d \]
- For \( a_{10} \):
\[ a_{10} = a_1 + (10 - 1)d = a_1 + 9d \]

### Step 3: Set up the equation based on the problem statement
According to the problem, we have:
\[ a_8 + a_{10} = 24 \]
Substituting the expressions we found:
\[ (a_1 + 7d) + (a_1 + 9d) = 24 \]
This simplifies to:
\[ 2a_1 + 16d = 24 \]

### Step 4: Simplify the equation
Dividing the entire equation by 2 gives:
\[ a_1 + 8d = 12 \]

### Step 5: Find the expression for \( a_9 \)
Now, we need to find \( a_9 \):
\[ a_9 = a_1 + (9 - 1)d = a_1 + 8d \]
From our previous result, we have:
\[ a_1 + 8d = 12 \]

### Step 6: Conclusion
Thus, the value of \( a_9 \) is:
\[ \boxed{12} \]

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