Which of the following are A.P.'s ? If they form an A.P., find the common difference 'd' and write three more terms :
`(i) -10, -6, -2, 2, ..... " " (ii) 3, 3+sqrt(2), 3+2sqrt(2), 3+3sqrt(2), ...`
`(iii) 0, -4, -8, -12, .... " " (iv) a, 2a, 3a, 4a, .....`
(v)`sqrt(3), sqrt(6), sqrt(9), sqrt(12), .....`

Answer

To determine whether the given sequences are in Arithmetic Progression (A.P.), we need to check if the difference between consecutive terms is constant. If they form an A.P., we will also find the common difference 'd' and write three more terms.

### Solution:

**(i) Sequence: -10, -6, -2, 2, ...**

1. Calculate the differences:
- Difference between -10 and -6:
\[ -6 - (-10) = -6 + 10 = 4 \]
- Difference between -6 and -2:
\[ -2 - (-6) = -2 + 6 = 4 \]
- Difference between -2 and 2:
\[ 2 - (-2) = 2 + 2 = 4 \]

2. Since the difference is constant (d = 4), the sequence is an A.P.

3. Common difference \(d = 4\).

4. Next three terms:
- After 2:
\[ 2 + 4 = 6 \]
- After 6:
\[ 6 + 4 = 10 \]
- After 10:
\[ 10 + 4 = 14 \]

5. Thus, the next three terms are 6, 10, and 14.

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**(ii) Sequence: 3, 3 + √2, 3 + 2√2, 3 + 3√2, ...**

1. Calculate the differences:
- Difference between 3 and (3 + √2):
\[ (3 + √2) - 3 = √2 \]
- Difference between (3 + √2) and (3 + 2√2):
\[ (3 + 2√2) - (3 + √2) = 2√2 - √2 = √2 \]
- Difference between (3 + 2√2) and (3 + 3√2):
\[ (3 + 3√2) - (3 + 2√2) = 3√2 - 2√2 = √2 \]

2. Since the difference is constant (d = √2), the sequence is an A.P.

3. Common difference \(d = √2\).

4. Next three terms:
- After (3 + 3√2):
\[ (3 + 3√2) + √2 = 3 + 4√2 \]
- After (3 + 4√2):
\[ (3 + 4√2) + √2 = 3 + 5√2 \]
- After (3 + 5√2):
\[ (3 + 5√2) + √2 = 3 + 6√2 \]

5. Thus, the next three terms are \(3 + 4√2\), \(3 + 5√2\), and \(3 + 6√2\).

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**(iii) Sequence: 0, -4, -8, -12, ...**

1. Calculate the differences:
- Difference between 0 and -4:
\[ -4 - 0 = -4 \]
- Difference between -4 and -8:
\[ -8 - (-4) = -8 + 4 = -4 \]
- Difference between -8 and -12:
\[ -12 - (-8) = -12 + 8 = -4 \]

2. Since the difference is constant (d = -4), the sequence is an A.P.

3. Common difference \(d = -4\).

4. Next three terms:
- After -12:
\[ -12 - 4 = -16 \]
- After -16:
\[ -16 - 4 = -20 \]
- After -20:
\[ -20 - 4 = -24 \]

5. Thus, the next three terms are -16, -20, and -24.

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**(iv) Sequence: a, 2a, 3a, 4a, ...**

1. Calculate the differences:
- Difference between 2a and a:
\[ 2a - a = a \]
- Difference between 3a and 2a:
\[ 3a - 2a = a \]
- Difference between 4a and 3a:
\[ 4a - 3a = a \]

2. Since the difference is constant (d = a), the sequence is an A.P.

3. Common difference \(d = a\).

4. Next three terms:
- After 4a:
\[ 4a + a = 5a \]
- After 5a:
\[ 5a + a = 6a \]
- After 6a:
\[ 6a + a = 7a \]

5. Thus, the next three terms are 5a, 6a, and 7a.

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**(v) Sequence: √3, √6, √9, √12, ...**

1. Calculate the differences:
- Difference between √6 and √3:
\[ √6 - √3 \]
- Difference between √9 and √6:
\[ √9 - √6 = 3 - √6 \]
- Difference between √12 and √9:
\[ √12 - √9 = 2√3 - 3 \]

2. The differences are not constant, thus the sequence is **not an A.P.**.

### Summary of Results:
1. (i) A.P., d = 4, next terms: 6, 10, 14
2. (ii) A.P., d = √2, next terms: \(3 + 4√2\), \(3 + 5√2\), \(3 + 6√2\)
3. (iii) A.P., d = -4, next terms: -16, -20, -24
4. (iv) A.P., d = a, next terms: 5a, 6a, 7a
5. (v) Not an A.P.

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