Find `21 ^(st)` term of the `A. P. - 4 (1)/(2), - 3, -1 (1)/(2),...`

Answer

To find the 21st term of the arithmetic progression (A.P.) given by the terms -4 (1/2), -3, -1 (1/2), ..., we can follow these steps:

### Step 1: Identify the first term (A) and the common difference (D)
The first term \( A \) is the first term of the A.P.:
\[ A = -4 \frac{1}{2} = -\frac{9}{2} \]

Next, we find the common difference \( D \) by subtracting the first term from the second term:
\[ D = \text{Second term} - \text{First term} = -3 - \left(-\frac{9}{2}\right) \]
To perform this calculation, we convert -3 into a fraction:
\[ -3 = -\frac{6}{2} \]
Now substituting:
\[ D = -\frac{6}{2} + \frac{9}{2} = \frac{3}{2} \]

### Step 2: Use the formula for the nth term of an A.P.
The formula for the nth term \( T_n \) of an A.P. is given by:
\[ T_n = A + (n - 1) \cdot D \]
We need to find the 21st term, so we set \( n = 21 \):
\[ T_{21} = A + (21 - 1) \cdot D \]

### Step 3: Substitute the values of A, D, and n into the formula
Substituting the values we found:
\[ T_{21} = -\frac{9}{2} + (21 - 1) \cdot \frac{3}{2} \]
This simplifies to:
\[ T_{21} = -\frac{9}{2} + 20 \cdot \frac{3}{2} \]
Calculating \( 20 \cdot \frac{3}{2} \):
\[ 20 \cdot \frac{3}{2} = \frac{60}{2} = 30 \]
So we have:
\[ T_{21} = -\frac{9}{2} + 30 \]

### Step 4: Convert 30 to a fraction and combine
Convert 30 into a fraction with a denominator of 2:
\[ 30 = \frac{60}{2} \]
Now, we can combine the fractions:
\[ T_{21} = -\frac{9}{2} + \frac{60}{2} = \frac{-9 + 60}{2} = \frac{51}{2} \]

### Final Answer
Thus, the 21st term of the A.P. is:
\[ \boxed{\frac{51}{2}} \]