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### (a) How many integers are there between 1 and 1,000 which are divisible by 17?

To find this, we calculate how many multiples of 17 exist between 1 and 1,000.

The multiples of 17 form an arithmetic sequence:
- First term: \( a_1 = 17 \)
- Common difference: \( d = 17 \)
- Last number \( \leq 1000 \): \( a_n \leq 1000 \)

The formula for the \(n\)-th term of an arithmetic sequence is:
\[
a_n = a_1 + (n-1)d
\]
Here, we know \( a_n \leq 1000 \), so:
\[
1000 \geq 17 + (n-1) \times 17
\]
Simplifying:
\[
1000 \geq 17n \\
n \leq \frac{1000}{17} \approx 58.82
\]
Thus, \( n = 58 \).

Therefore, there are **58 integers** between 1 and 1,000 that are divisible by 17.

---

### (b) The 4th and 7th terms of a G.P. are 144 and 18 respectively. Find:
- (i) The common ratio
- (ii) The first term
- (iii) The sum of the first six terms

#### Given:
- \( a_4 = 144 \)
- \( a_7 = 18 \)

The formula for the \(n\)-th term of a geometric progression is:
\[
a_n = a r^{n-1}
\]
For the 4th term:
\[
144 = a r^3 \quad \text{(1)}
\]
For the 7th term:
\[
18 = a r^6 \quad \text{(2)}
\]

Dividing equation (2) by equation (1):
\[
\frac{18}{144} = \frac{a r^6}{a r^3} \\
\frac{1}{8} = r^3
\]
So, the common ratio is:
\[
r = \frac{1}{2}
\]

#### (ii) Finding the first term:
Substitute \( r = \frac{1}{2} \) into equation (1):
\[
144 = a \left( \frac{1}{2} \right)^3 \\
144 = a \times \frac{1}{8} \\
a = 144 \times 8 = 1152
\]
Thus, the first term \( a = 1152 \).

#### (iii) Sum of the first six terms:
The sum of the first \(n\) terms of a geometric progression is:
\[
S_n = \frac{a(1 - r^n)}{1 - r}
\]
For the first six terms:
\[
S_6 = \frac{1152(1 - \left(\frac{1}{2}\right)^6)}{1 - \frac{1}{2}} \\
S_6 = \frac{1152(1 - \frac{1}{64})}{\frac{1}{2}} \\
S_6 = \frac{1152 \times \frac{63}{64}}{\frac{1}{2}} \\
S_6 = 1152 \times \frac{63}{32} \\
S_6 = 2268
\]
So, the sum of the first six terms is **2268**.

---

Do you want further details or clarifications? Here are some related questions:

1. What is the formula for the sum of an arithmetic sequence?
2. How would the approach change if the integers between 1 and 1,000 were divisible by another number?
3. How can the general form of a geometric sequence be used to find any term?
4. What are the applications of geometric progressions in real-world problems?
5. Can you derive the sum of an infinite geometric progression?

**Tip:** When solving geometric progression problems, dividing terms helps eliminate unknowns quickly, leading to an easier path to find the ratio.

How many integers are there between 1 and 1,000 which are divisible by 17? The 4th and 7th terms of a G.P. are 144 and 18 respectively. Find: (i) The common ratio, (ii) The first term, (iii) The sum of the first six terms.

This problem involves finding how many integers between 1 and 1,000 are divisible by 17, and solving a geometric progression where the 4th and 7th terms are given as 144 and 18 respectively. The solution includes calculating the common ratio, the first term, and the sum of the first six terms.

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