The 11th term of an A.P. is 16 and the 21st term is 29 . Find the 16th term of this A.P.

Explanation

To find the 16th term of the arithmetic progression (A.P.) given that the 11th term is 16 and the 21st term is 29, we can follow these steps:

### Step 1: Write the formula for the nth term of an A.P.
The nth term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

### Step 2: Set up equations for the given terms.
From the problem, we know:
- The 11th term \( a_{11} = 16 \)
- The 21st term \( a_{21} = 29 \)

Using the formula for the nth term:
1. For the 11th term:
\[ a + 10d = 16 \] (Equation 1)
2. For the 21st term:
\[ a + 20d = 29 \] (Equation 2)

### Step 3: Solve the equations simultaneously.
We can subtract Equation 1 from Equation 2 to eliminate \( a \):
\[ (a + 20d) - (a + 10d) = 29 - 16 \]
This simplifies to:
\[ 10d = 13 \]
So, we find:
\[ d = \frac{13}{10} = 1.3 \]

### Step 4: Substitute \( d \) back into one of the equations to find \( a \).
We can substitute \( d \) back into Equation 1:
\[ a + 10(1.3) = 16 \]
This simplifies to:
\[ a + 13 = 16 \]
Thus, we find:
\[ a = 16 - 13 = 3 \]

### Step 5: Find the 16th term using \( a \) and \( d \).
Now that we have both \( a \) and \( d \), we can find the 16th term \( a_{16} \):
\[ a_{16} = a + (16-1)d = a + 15d \]
Substituting the values:
\[ a_{16} = 3 + 15(1.3) = 3 + 19.5 = 22.5 \]

### Final Answer:
The 16th term of the A.P. is \( 22.5 \).

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