In an AP: 

(i) Given a = 5, d = 3, a_n = 50, find n and S_n . 

(ii) Given a = 7, a₁₃ = 35, find d and S₁₃. 

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂. 

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀. 

(v) Given d = 5, S₉ = 75, find a and a₉ . 

(vi) Given a = 2, d = 8, S_n = 90, find n and a_n. 

(vii) Given a = 8, an = 62, S_n = 210, find n and d. 

(viii) Given a_n = 4, d = 2, S_n = –14, find n and a.

 (ix) Given a = 3, n = 8, S = 192, find d. 

(x) Given l = 28, S = 144, and there are total 9 terms. Find a.

(i) Given:

a = 5, d = 3, a_n = 50, find n and S_n.

Using the formula,

aₙ = a + (n - 1)d

50 = 5 + (n - 1)3

45 = (n - 1)3

15 = n - 1

n = 16

Using the formula,

Sₙ = n/2 [a + l]

S₁₆ = 16/2 [ 5 + 50]

= 8 × 55

= 440

Therefore,

n = 16 and S₁₆ = 440

(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.

Using the formula,

aₙ = a + (n - 1)d

a₁₃ = a + (13 - 1)d

35 = 7 + 12d

35 - 7 = 12d

d = 28/12

d = 7/3

Using the formula,

Sₙ = n/2 [a + l]

S₁₃ = 13/2 [7 + 35]

= 13/2 × 42

= 13 × 21

= 273

Therefore,

S₁₃ = 273 and d = 7/3

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.

Using the formula,

a_n = a + (n - 1)d

a₁₂ = a + (12 - 1)3

37 = a + 33

a = 4

Using the formula,

S_n = n/2 [a + l]

S₁₂ = 12/2 [4 + 37]

= 6 × 41

S₁₂ = 246

Therefore,

S₁₂ = 246 and a = 4

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀

Using the formula, 

a_n = a + (n - 1)d

a₃ = a + (3 - 1)d

15 = a + 2d   - - - - - (1)

Using the formula,

S_n = n/2 [2a + (n - 1)d]

S₁₀ = 10/2 [2a + (10 - 1)d]

125 = 5[2a + 9d]

25 = 2a + 9d  - - - - -(2)

On multiplying equation (1) by 2, we obtain,

30 = 2a + 4d  - - - - -(3)

Solving equations (2) and (3) we get,

- 5 = 5d

d = - 1

From equation (1),

15 = a + 2(- 1)

15 = a - 2

a = 17

a₁₀ = a + (10 - 1)d

a₁₀ = 17 + 9 (- 1)

a₁₀ = 17 - 9

a₁₀ = 8

Therefore,

a₁₀ = 8 and d = - 1

(v) Given d = 5, S₉ = 75, find a and a₉ . 

Using the formula,

S_n = n/2 [2a + (n - 1)d]

75 = 9/2 [2a + (9 - 1)5]

75 = 9/2(2a + 40)

25 = 3(a + 20)

25 = 3a + 60

a = - 35/3

Now we get,

a₉= a + (9 - 1) × 5

= - 35/3 + 8 × 5

= - 35/3 + 40

= (- 35 + 120)/3

= 85/3

Therefore, a₉ = 85/3 and a = - 35/3

(vi) Given a = 2, d = 8, S_n = 90, find n and a_n . 

Using the formula,

S_n = n/2 [2a + (n - 1) d]

90 = n/2 [4 + (n - 1) 8]

90 = n [2 + (n - 1)4]

90 = n [2 + 4n - 4]

90 = n [4n - 2]

90 = 4n² - 2n

4n² - 2n - 90 = 0

4n² - 20n + 18n - 90 = 0

4n (n - 5) + 18(n - 5) = 0

(n - 5)(4n + 18) = 0

Either (n - 5) = 0 or (4n + 18) = 0

n = 5 or n = - 9/2

However, n can not be negative.

Therefore, n = 5

we get,

a₅ = 2 + (5 - 1)8

a₅ = 2 + 4 × 8

a₅ = 2 + 32

a₅ = 34

Therefore, n = 5 and a₅ = 34

(vii) Given a = 8, an = 62, S_n = 210, find n and d. 

Using the formula,

S_n = n / 2 [a + l]

210 = n / 2 [8 + 62]

210 = n × 35

n = 6

Using the formula,

a_n = a + (n - 1)d

we get,

62 = 8 + (6 - 1) d

62 - 8 = 5d

54 = 5d

d = 54/5

Therefore,  n = 6 and d = 54/5

(viii) Given a_n = 4, d = 2, S_n = –14, find n and a.

Using the formula,

 a_n = a + (n - 1)d

4 = a + (n - 1)2

4 = a + 2n - 2

a = 6 - 2n  .... (1)

S_n = n/2 [a + l]

- 14 = n/2 [6 - 2n + 4]     [from(1)]

- 14 = n (5 - n)

- 14 = 5n - n²

n² - 5n - 14 = 0

n² - 7n + 2n -14 = 0

n (n - 7) + 2 (n - 7) = 0

(n - 7)(n + 2) = 0

Either n - 7 = 0 or n + 2 = 0

n = 7 or n = - 2

However, n can not be negative.

Therefore, n = 7

From equation (1), we get

a = 6 - 2n

a = 6 - 2 × 7

a = 6 - 14

a = - 8

Therefore, a = - 8 and n = 7 

 (ix) Given a = 3, n = 8, S = 192, find d.

Using the formula,

S_n = n/2 [2a + (n - 1) d]

192 = 8/2 [2 × 3 + (8 - 1) d]

192 = 4[6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

Therefore, d = 6

(x) Given l = 28, S = 144, and there are total 9 terms. Find a.

Using the formula,

S_n = n/2 [a + l]

144 = 9/2 (a + 28)

32 = a + 28

a = 4

Therefore, a = 4