The first term of an A.P. is 20 and the sum of its first seven terms is 2100, find the 31st term of this A.P.

Answer

To find the 31st term of the arithmetic progression (A.P.) where the first term \( A = 20 \) and the sum of the first seven terms \( S_7 = 2100 \), we can follow these steps:

### Step 1: Write down the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
where:
- \( S_n \) = sum of the first \( n \) terms,
- \( A \) = first term,
- \( D \) = common difference,
- \( n \) = number of terms.

### Step 2: Substitute the known values into the formula.
For our problem, we have:
- \( S_7 = 2100 \)
- \( A = 20 \)
- \( n = 7 \)

Substituting these values into the formula:
\[ 2100 = \frac{7}{2} \times (2 \times 20 + (7 - 1)D) \]

### Step 3: Simplify the equation.
Calculating \( 2 \times 20 \):
\[ 2 \times 20 = 40 \]
Now substituting back:
\[ 2100 = \frac{7}{2} \times (40 + 6D) \]

### Step 4: Multiply both sides by 2 to eliminate the fraction.
\[ 4200 = 7 \times (40 + 6D) \]

### Step 5: Divide both sides by 7.
\[ 600 = 40 + 6D \]

### Step 6: Solve for \( D \).
Subtract 40 from both sides:
\[ 600 - 40 = 6D \implies 560 = 6D \]
Now divide by 6:
\[ D = \frac{560}{6} = \frac{280}{3} \]

### Step 7: Write the formula for the \( n \)-th term of the A.P.
The formula for the \( n \)-th term \( A_n \) is:
\[ A_n = A + (n - 1)D \]

### Step 8: Substitute \( n = 31 \) into the formula.
We want to find \( A_{31} \):
\[ A_{31} = 20 + (31 - 1)D = 20 + 30D \]

### Step 9: Substitute the value of \( D \).
\[ A_{31} = 20 + 30 \times \frac{280}{3} \]

### Step 10: Calculate \( 30 \times \frac{280}{3} \).
\[ 30 \times \frac{280}{3} = \frac{8400}{3} = 2800 \]

### Step 11: Add the results.
\[ A_{31} = 20 + 2800 = 2820 \]

### Final Answer:
The 31st term of the A.P. is \( 2820 \).

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