(i) 2, 4, 8, 16, . . .

(ii) 5 7 2, , 3, ,2 2 . . .

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

(iv) – 10, – 6, – 2, 2, . . .

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, . . .

(vi) 0.2, 0.22, 0.222, 0.2222, . . .

(vii) 0, – 4, – 8, –12, . . .

(viii) -1/2, -1/2, -1/2, -1/2, . . .

(ix) 1, 3, 9, 27, . . .

(x) a, 2a, 3a, 4a, . . .

(xi) a, a^{2}, a^{3}, a^{4}, . . .

(xii) √2, √8, √18 , √32, . . .

(xiii) √3, √6, √9 , √12 , . . .

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2}, . . .

(xv) 1^{2}, 5^{2}, 7^{2}, 73, . . .

**Solution:**

**(i) 2, 4, 8, 16 …**

Here, the common difference is;

a_{2} – a_{1} = 4 – 2 = 2

a_{3} – a_{2} = 8 – 4 = 4

a_{4} – a_{3} = 16 – 8 = 8

Since a_{n} _{+1} – a_{n }or the common difference is not the same every time.

Therefore, the given series does not form an A.P.

**(ii) 2, 5/2, 3, 7/2 ….**

Here,

a_{2} – a_{1} = 5/2 – 2 = 1/2

a_{3} – a_{2} = 3 – 5/2 = 1/2

a_{4} – a_{3} = 7/2 – 3 = 1/2

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = 1/2, and the given series are in A.P.

The next three terms are;

a_{5} = 7/2 + 1/2 = 4

a_{6} = 4 + 1/2 = 9/2

a_{7} = 9/2 + 1/2 = 5

**(iii) -1.2, – 3.2, -5.2, -7.2 …
**

Here,

a_{2} – a_{1} = (-3.2) – (-1.2) = -2

a_{3} – a_{2} = (-5.2) – (-3.2) = -2

a_{4} – a_{3} = (-7.2) – (-5.2) = -2

Since a_{n} _{+1} – a_{n} or common difference is the same every time.

Therefore, d = -2, and the given series are in A.P.

Hence, the next three terms are;

a_{5} = – 7.2 – 2 = -9.2

a_{6} = – 9.2 – 2 = – 11.2

a_{7} = – 11.2 – 2 = – 13.2

**(iv) -10, – 6, – 2, 2 …
**

Here, the terms and their difference are;

a_{2} – a_{1} = (-6) – (-10) = 4

a_{3} – a_{2} = (-2) – (-6) = 4

a_{4} – a_{3} = 2 – (-2) = 4

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = 4 and the given numbers are in A.P.

Hence, the next three terms are;

a_{5} = 2 + 4 = 6

a_{6} = 6 + 4 = 10

a_{7} = 10 + 4 = 14

**(v) Given, 3, 3 + √2, 3 + 2√2, 3 + 3√2
**

Here,

a_{2} – a_{1} = 3 + √2 – 3 = √2

a_{3} – a_{2} = (3 + 2√2) – (3 + √2) = √2

a_{4} – a_{3} = (3 + 3√2) – (3 + 2√2) = √2

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = √2, and the given series forms an A.P.

Hence, the next three terms are;

a_{5} = (3 + √2) + √2 = 3 + 4√2

a_{6} = (3 + 4√2) + √2 = 3 + 5√2

a_{7} = (3 + 5√2) + √2 = 3 + 6√2

**(vi) 0.2, 0.22, 0.222, 0.2222 ….
**

Here,

a_{2} – a_{1} = 0.22 – 0.2 = 0.02

a_{3} – a_{2} = 0.222 – 0.22 = 0.002

a_{4} – a_{3} = 0.2222 – 0.222 = 0.0002

Since a_{n} _{+1} – a_{n} or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

**(vii) 0, -4, -8, -12 …
**

Here,

a_{2} – a_{1} = (-4) – 0 = -4

a_{3} – a_{2} = (-8) – (-4) = -4

a_{4} – a_{3} = (-12) – (-8) = -4

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = -4, and the given series forms an A.P.

Hence, the next three terms are;

a_{5} = -12 – 4 = -16

a_{6} = -16 – 4 = -20

a_{7} = -20 – 4 = -24

**(viii) -1/2, -1/2, -1/2, -1/2 ….
**

Here,

a_{2} – a_{1} = (-1/2) – (-1/2) = 0

a_{3} – a_{2} = (-1/2) – (-1/2) = 0

a_{4} – a_{3} = (-1/2) – (-1/2) = 0

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = 0, and the given series forms an A.P.

Hence, the next three terms are;

a_{5} = (-1/2) – 0 = -1/2

a_{6} = (-1/2) – 0 = -1/2

a_{7} = (-1/2) – 0 = -1/2

**(ix) 1, 3, 9, 27 …
**

Here,

a_{2} – a_{1} = 3 – 1 = 2

a_{3} – a_{2} = 9 – 3 = 6

a_{4} – a_{3} = 27 – 9 = 18

Since a_{n} _{+1} – a_{n} or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

**(x) a, 2 a, 3 a, 4 a …**

Here,

a_{2} – a_{1} = 2a – a = a

a_{3} – a_{2} = 3a – 2a = a

a_{4} – a_{3} = 4a – 3a = a

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = a, and the given series forms an A.P.

Hence, the next three terms are;

a_{5} = 4a + a = 5a

a_{6} = 5a + a = 6a

a_{7} = 6a + a = 7a

**(xi) a, a ^{2}, a^{3}, a^{4} …**

Here,

a_{2} – a_{1} = a^{2 }– a = a( a – 1)

a_{3} – a_{2} = a^{3 }– a^{2 }= a^{2}( a – 1)

a_{4} – a_{3} = a^{4} – a^{3 }= a^{3}( a – 1)

Since a_{n} _{+1} – a_{n} or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

**(xii) √2, √8, √18, √32 …
**

Here,

a_{2} – a_{1} = √8 – √2 = 2√2 – √2 = √2

a_{3} – a_{2} = √18 – √8 = 3√2 – 2√2 = √2

a_{4} – a_{3} = 4√2 – 3√2 = √2

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = √2, and the given series forms an A.P.

Hence, the next three terms are;

a_{5} = √32 + √2 = 4√2 + √2 = 5√2 = √50

a_{6} = 5√2 + √2 = 6√2 = √72

a_{7} = 6√2 + √2 = 7√2 = √98

**(xiii) √3, √6, √9, √12 …**

Here,

a_{2} – a_{1} = √6 – √3 = √3 × √2 – √3 = √3(√2 – 1)

a_{3} – a_{2} = √9 – √6 = 3 – √6 = √3(√3 – √2)

a_{4} – a_{3} = √12 – √9 = 2√3 – √3 × √3 = √3(2 – √3)

Since a_{n} _{+1} – a_{n} or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

**(xiv) 1 ^{2}, 3^{2}, 5^{2}, 7^{2} …**

Or, 1, 9, 25, 49 …..

Here,

a_{2} − a_{1} = 9 − 1 = 8

a_{3} − a_{2 }= 25 − 9 = 16

a_{4} − a_{3} = 49 − 25 = 24

Since a_{n} _{+1} – a_{n} or the common difference is not the same every time.

Therefore, the given series doesn’t form an A.P.

**(xv) 1 ^{2}, 5^{2}, 7^{2}, 73 …**

Or 1, 25, 49, 73 …

Here,

a_{2} − a_{1} = 25 − 1 = 24

a_{3} − a_{2 }= 49 − 25 = 24

a_{4} − a_{3} = 73 − 49 = 24

Since a_{n} _{+1} – a_{n} or the common difference is the same every time.

Therefore, d = 24, and the given series forms an A.P.

Hence, the next three terms are;

a_{5} = 73 + 24 = 97

a_{6} = 97 + 24 = 121

a_{7 }= 121 + 24 = 145