The sum of first 14 terms of an AP is 1050 and its 14th terms 140. Find the 20th term.

Answer

To solve the problem step by step, we will use the formulas for the sum of an arithmetic progression (AP) and the nth term of an AP.

### Step 1: Identify the given information
We are given:
- The sum of the first 14 terms, \( S_{14} = 1050 \)
- The 14th term, \( A_{14} = 140 \)

### Step 2: Use the formula for the sum of the first n terms
The formula for the sum of the first n terms of an AP is:
\[ S_n = \frac{n}{2} \times (2A + (n-1)D) \]
For \( n = 14 \):
\[ S_{14} = \frac{14}{2} \times (2A + 13D) \]
This simplifies to:
\[ 1050 = 7 \times (2A + 13D) \]
Dividing both sides by 7 gives:
\[ 150 = 2A + 13D \quad \text{(Equation 1)} \]

### Step 3: Use the formula for the nth term
The formula for the nth term of an AP is:
\[ A_n = A + (n-1)D \]
For \( n = 14 \):
\[ A_{14} = A + 13D \]
Substituting the given value:
\[ 140 = A + 13D \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
We have two equations:
1. \( 150 = 2A + 13D \)
2. \( 140 = A + 13D \)

Let's subtract Equation 2 from Equation 1:
\[ (150 - 140) = (2A + 13D) - (A + 13D) \]
This simplifies to:
\[ 10 = A \]
Thus, we find:
\[ A = 10 \]

### Step 5: Substitute A back to find D
Now substitute \( A = 10 \) into Equation 2:
\[ 140 = 10 + 13D \]
Subtracting 10 from both sides gives:
\[ 130 = 13D \]
Dividing by 13 gives:
\[ D = 10 \]

### Step 6: Find the 20th term
Now that we have \( A = 10 \) and \( D = 10 \), we can find the 20th term using the nth term formula:
\[ A_{20} = A + (20 - 1)D \]
Substituting the values:
\[ A_{20} = 10 + 19 \times 10 \]
Calculating this gives:
\[ A_{20} = 10 + 190 = 200 \]

### Final Answer
The 20th term of the AP is **200**.

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