All the terms of an AP are natural numbers and the sum of the first 20 terms is greater than 1072 and lss than 1162.If the sixth term is 32, then

Options

A. first term is 7
B. first term is 12
C. common difference is 4
D. common difference is 5

Explanation

To solve the problem step by step, we will follow the mathematical reasoning provided in the video transcript.

### Step 1: Define the terms of the AP
Let the first term of the Arithmetic Progression (AP) be \( A \) and the common difference be \( D \). The \( n \)-th term of an AP can be expressed as:
\[ T_n = A + (n - 1)D \]

### Step 2: Use the information about the sixth term
According to the problem, the sixth term is given as 32. Therefore, we can write:
\[ T_6 = A + 5D = 32 \quad \text{(Equation 1)} \]

### Step 3: Use the information about the sum of the first 20 terms
The sum \( S_n \) of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
For the first 20 terms, we have:
\[ S_{20} = \frac{20}{2} \times (2A + 19D) = 10 \times (2A + 19D) = 20A + 190D \]
According to the problem, the sum of the first 20 terms is greater than 1072 and less than 1162:
\[ 1072 < 20A + 190D < 1162 \quad \text{(Equation 2)} \]

### Step 4: Substitute \( A \) from Equation 1 into Equation 2
From Equation 1, we can express \( A \) in terms of \( D \):
\[ A = 32 - 5D \]
Now substitute this into Equation 2:
\[ 20(32 - 5D) + 190D > 1072 \quad \text{and} \quad 20(32 - 5D) + 190D < 1162 \]

### Step 5: Simplify the inequalities
Calculating the left side:
\[ 640 - 100D + 190D > 1072 \quad \text{and} \quad 640 - 100D + 190D < 1162 \]
This simplifies to:
\[ 640 + 90D > 1072 \quad \text{and} \quad 640 + 90D < 1162 \]

### Step 6: Isolate \( D \)
Now, we will isolate \( D \) in both inequalities:
1. For the first inequality:
\[ 90D > 1072 - 640 \]
\[ 90D > 432 \quad \Rightarrow \quad D > \frac{432}{90} = 4.8 \]

2. For the second inequality:
\[ 90D < 1162 - 640 \]
\[ 90D < 522 \quad \Rightarrow \quad D < \frac{522}{90} = 5.8 \]

### Step 7: Determine the possible values of \( D \)
Since \( D \) must be a natural number, the only possible value for \( D \) that satisfies \( 4.8 < D < 5.8 \) is:
\[ D = 5 \]

### Step 8: Substitute \( D \) back to find \( A \)
Now substitute \( D = 5 \) back into Equation 1 to find \( A \):
\[ A = 32 - 5 \times 5 = 32 - 25 = 7 \]

### Conclusion
The first term \( A \) is 7, and the common difference \( D \) is 5.

### Final Answer
- First term \( A = 7 \)
- Common difference \( D = 5 \)

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