The 11th term of the A.P. `13,15(1)/(2),18,20(1)/(2),.......` is :

Options

A. 38
B. `40(1)/(2)`
C. 43
D. `45(1)/(2)`

Answer

To find the 11th term of the arithmetic progression (A.P.) given by the sequence \(13, 15\frac{1}{2}, 18, 20\frac{1}{2}, \ldots\), we can follow these steps:

### Step 1: Identify the first term (a) and the common difference (d)
The first term \(a\) is the first number in the sequence:
\[ a = 13 \]

To find the common difference \(d\), we subtract the first term from the second term:
\[ d = 15\frac{1}{2} - 13 \]

Convert \(15\frac{1}{2}\) to an improper fraction:
\[ 15\frac{1}{2} = \frac{31}{2} \]

Now calculate \(d\):
\[ d = \frac{31}{2} - 13 = \frac{31}{2} - \frac{26}{2} = \frac{5}{2} \]

### Step 2: Use the formula for the n-th term of an A.P.
The formula for the n-th term \(T_n\) of an A.P. is given by:
\[ T_n = a + (n - 1) \cdot d \]
We need to find the 11th term, so we substitute \(n = 11\):
\[ T_{11} = 13 + (11 - 1) \cdot \frac{5}{2} \]

### Step 3: Simplify the expression
Calculate \(n - 1\):
\[ n - 1 = 11 - 1 = 10 \]
Now substitute this back into the formula:
\[ T_{11} = 13 + 10 \cdot \frac{5}{2} \]

Calculate \(10 \cdot \frac{5}{2}\):
\[ 10 \cdot \frac{5}{2} = 25 \]

Now add this to 13:
\[ T_{11} = 13 + 25 = 38 \]

### Final Answer
The 11th term of the A.P. is:
\[ \boxed{38} \]