Find d and write the next four terms of the following A.P.'s
(i) `0,-3,-6,-9`,……………
(ii) `1/6,1/3,1/2`…………..

Answer

To solve the problem, we need to find the common difference \(d\) for each arithmetic progression (A.P.) and then write the next four terms.

### (i) For the A.P. \(0, -3, -6, -9, \ldots\)

**Step 1: Find the common difference \(d\)**
The common difference \(d\) can be calculated by subtracting any term from the subsequent term.
Let's take the first two terms:
\[ d = -3 - 0 = -3 \]

**Step 2: Write the next four terms**
To find the next terms, we will keep adding the common difference \(d\) to the last term.
- The last term we have is \(-9\).
- Next term:
\[ -9 + (-3) = -12 \]
- Next term:
\[ -12 + (-3) = -15 \]
- Next term:
\[ -15 + (-3) = -18 \]
- Next term:
\[ -18 + (-3) = -21 \]

Thus, the next four terms are:
\(-12, -15, -18, -21\).

### (ii) For the A.P. \(\frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \ldots\)

**Step 1: Find the common difference \(d\)**
We can calculate \(d\) by subtracting the first term from the second term:
\[ d = \frac{1}{3} - \frac{1}{6} \]
To perform this subtraction, we need a common denominator:
The LCM of 3 and 6 is 6.
So,
\[ d = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} \]

**Step 2: Write the next four terms**
Now we will add the common difference \(d\) to the last term.
- The last term we have is \(\frac{1}{2}\).
- Next term:
\[ \frac{1}{2} + \frac{1}{6} \]
To add these, we need a common denominator (which is 6):
\[ \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \]
- Next term:
\[ \frac{2}{3} + \frac{1}{6} \]
Again, using a common denominator:
\[ \frac{4}{6} + \frac{1}{6} = \frac{5}{6} \]
- Next term:
\[ \frac{5}{6} + \frac{1}{6} = \frac{6}{6} = 1 \]
- Next term:
\[ 1 + \frac{1}{6} = \frac{6}{6} + \frac{1}{6} = \frac{7}{6} \]

Thus, the next four terms are:
\(\frac{2}{3}, \frac{5}{6}, 1, \frac{7}{6}\).

### Final Answers:
1. For the first A.P.: Next four terms are \(-12, -15, -18, -21\).
2. For the second A.P.: Next four terms are \(\frac{2}{3}, \frac{5}{6}, 1, \frac{7}{6}\).