Find the `99^(th)` term of the series ` : 7 (3)/(4) , 9 (1)/(2) , 11(1)/(4) ,…….`

Answer

To find the `99th` term of the series `7 3/4, 9 1/2, 11 1/4, ...`, we first need to determine whether this series is an arithmetic progression (AP) and then find the common difference.

### Step 1: Convert the mixed numbers to improper fractions
- The first term \( a_1 = 7 \frac{3}{4} = \frac{31}{4} \)
- The second term \( a_2 = 9 \frac{1}{2} = \frac{19}{2} = \frac{38}{4} \)
- The third term \( a_3 = 11 \frac{1}{4} = \frac{45}{4} \)

### Step 2: Find the common difference
Now, we will find the common difference \( D \):
\[ D = a_2 - a_1 = \frac{38}{4} - \frac{31}{4} = \frac{7}{4} \]
\[ D = a_3 - a_2 = \frac{45}{4} - \frac{38}{4} = \frac{7}{4} \]
Thus, the common difference \( D = \frac{7}{4} \).

### Step 3: Use the formula for the nth term of an AP
The formula for the nth term of an arithmetic progression is given by:
\[ a_n = a + (n-1)D \]
Here, \( a = \frac{31}{4} \), \( D = \frac{7}{4} \), and \( n = 99 \).

### Step 4: Substitute the values into the formula
\[ a_{99} = \frac{31}{4} + (99 - 1) \cdot \frac{7}{4} \]
\[ = \frac{31}{4} + 98 \cdot \frac{7}{4} \]
\[ = \frac{31}{4} + \frac{686}{4} \]
\[ = \frac{31 + 686}{4} = \frac{717}{4} \]

### Step 5: Convert the improper fraction back to a mixed number
To convert \( \frac{717}{4} \) to a mixed number:
- Divide \( 717 \) by \( 4 \):
\[ 717 \div 4 = 179 \quad \text{(whole number part)} \]
- The remainder is \( 1 \):
\[ 717 - 4 \cdot 179 = 1 \]
Thus, \( \frac{717}{4} = 179 \frac{1}{4} \).

### Final Answer
The `99th` term of the series is \( 179 \frac{1}{4} \).

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