(i) 2 + 5 + 8 + … + 182

First term, a = a₁ = 2

Common difference, d = a₂ – a₁ = 5 – 2 = 3

a_n term of given AP is 182

a_n = a + (n - 1)d

182 = 2 + (n - 1)3

182 = 2 + 3n – 3

182 = 3n – 1

3n = 182 + 1

n = 183/3

= 61

Now,

By using the formula,

S = n/2 (a + l)

= 61/2 (2 + 182)

= 61/2 (184)

= 61 (92)

= 5612

∴ The sum of the series is 5612

(ii) 101 + 99 + 97 + … + 47

First term, a = a₁ = 101

Common difference, d = a₂ – a₁ = 99 – 101 = -2

a_n term of given AP is 47

a_n = a + (n - 1)d

47 = 101 + (n - 1)(-2)

47 = 101 – 2n + 2

2n = 103 – 47

2n = 56

n = 56/2 = 28

Then,

S = n/2 (a + l)

= 28/2 (101 + 47)

= 28/2 (148)

= 14 (148)

= 2072

∴ The sum of the series is 2072

(iii) (a – b)² + (a² + b²) + (a + b)² + s…. + [(a + b)² + 6ab]

First term, a = a₁ = (a - b)²

Common difference, d = a₂ – a₁ = (a² + b²) – (a – b)² = 2ab

a_n term of given AP is [(a + b)² + 6ab]

a_n = a + (n - 1) d

[(a + b)² + 6ab] = (a - b)² + (n - 1)2ab

a² + b² + 2ab + 6ab = a² + b² – 2ab + 2abn – 2ab

a² + b² + 8ab – a² – b² + 2ab + 2ab = 2abn

12ab = 2abn

n = 12ab / 2ab

= 6

Then,

S = n/2 (a + l)

= 6/2 ((a - b)² + [(a + b)² + 6ab])

= 3 (a² + b² – 2ab + a² + b² + 2ab + 6ab)

= 3 (2a² + 2b² + 6ab)

= 3 × 2 (a² + b² + 3ab)

= 6 (a² + b² + 3ab)

∴ The sum of the series is 6 (a² + b² + 3ab)