How many natural numbers upto 150 are divisible by 7?

Options

A. 21
B. 14
C. 22
D. 17

Answer

To find how many natural numbers up to 150 are divisible by 7, we can follow these steps:

### Step 1: Identify the first and last natural numbers divisible by 7 up to 150.
- The first natural number divisible by 7 is **7**.
- To find the last natural number divisible by 7 that is less than or equal to 150, we can perform the following calculation:
\[ \text{Last number} = 7 \times \left\lfloor \frac{150}{7} \right\rfloor \]
Calculating \( \left\lfloor \frac{150}{7} \right\rfloor \):
\[ \frac{150}{7} \approx 21.4286 \implies \left\lfloor \frac{150}{7} \right\rfloor = 21 \]
Therefore, the last number is:
\[ 7 \times 21 = 147 \]

### Step 2: Set up the arithmetic progression (AP).
- The numbers divisible by 7 form an arithmetic progression (AP) where:
- First term \( a = 7 \)
- Common difference \( d = 7 \)
- Last term \( l = 147 \)

### Step 3: Use the formula for the nth term of an AP.
- The formula for the nth term of an AP is given by:
\[ l = a + (n - 1) \times d \]
Substituting the known values:
\[ 147 = 7 + (n - 1) \times 7 \]

### Step 4: Solve for \( n \).
- Rearranging the equation:
\[ 147 - 7 = (n - 1) \times 7 \]
\[ 140 = (n - 1) \times 7 \]
Dividing both sides by 7:
\[ n - 1 = \frac{140}{7} = 20 \]
Therefore:
\[ n = 20 + 1 = 21 \]

### Conclusion:
- The total number of natural numbers up to 150 that are divisible by 7 is **21**.