Find the sum of first 31 terms of an A.P. whose nth term is `(3+(2n)/(3))`.

Options

A. `423(2)/(3)`
B. `413(1)/(3)`
C. `417(2)/(3)`
D. `419(2)/(3)`

Answer

To find the sum of the first 31 terms of the arithmetic progression (A.P.) whose nth term is given by \( T_n = 3 + \frac{2n}{3} \), we will follow these steps:

### Step 1: Identify the first term (a)
The first term \( T_1 \) can be found by substituting \( n = 1 \) into the formula for \( T_n \):
\[ T_1 = 3 + \frac{2 \times 1}{3} = 3 + \frac{2}{3} = \frac{9}{3} + \frac{2}{3} = \frac{11}{3} \]
**Hint:** To find the first term, substitute \( n = 1 \) into the nth term formula.

### Step 2: Identify the common difference (d)
Next, we find the second term \( T_2 \) by substituting \( n = 2 \):
\[ T_2 = 3 + \frac{2 \times 2}{3} = 3 + \frac{4}{3} = \frac{9}{3} + \frac{4}{3} = \frac{13}{3} \]
Now, the common difference \( d \) can be calculated as:
\[ d = T_2 - T_1 = \frac{13}{3} - \frac{11}{3} = \frac{2}{3} \]
**Hint:** The common difference is the difference between the second term and the first term.

### Step 3: Use the sum formula for A.P.
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) \]
For our case, we need to find \( S_{31} \):
\[ S_{31} = \frac{31}{2} \left( 2 \times \frac{11}{3} + (31 - 1) \times \frac{2}{3} \right) \]
**Hint:** Use the sum formula for A.P. to calculate the sum of the first \( n \) terms.

### Step 4: Calculate \( S_{31} \)
Now, we simplify the expression:
\[ S_{31} = \frac{31}{2} \left( \frac{22}{3} + 30 \times \frac{2}{3} \right) \]
Calculating \( 30 \times \frac{2}{3} \):
\[ 30 \times \frac{2}{3} = \frac{60}{3} = 20 \]
Now substitute back:
\[ S_{31} = \frac{31}{2} \left( \frac{22}{3} + 20 \right) = \frac{31}{2} \left( \frac{22}{3} + \frac{60}{3} \right) = \frac{31}{2} \left( \frac{82}{3} \right) \]
**Hint:** Combine the terms inside the parentheses before multiplying.

### Step 5: Final Calculation
Now, we can calculate:
\[ S_{31} = \frac{31 \times 82}{2 \times 3} = \frac{2542}{6} \]
Simplifying \( \frac{2542}{6} \):
\[ \frac{2542 \div 2}{6 \div 2} = \frac{1271}{3} \]
This can also be expressed as a mixed fraction:
\[ 1271 \div 3 = 423 \quad \text{remainder } 2 \quad \Rightarrow \quad 423 \frac{2}{3} \]
Thus, the sum of the first 31 terms of the A.P. is:
\[ \boxed{423 \frac{2}{3}} \]