In an A.P. (with usual notations):

(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 d = 5, S₉ = 75, find a and a₉

(iv) given a = 8, a_n = 62, S_n = 210, find n and d

(v) given a = 3, n = 8, S = 192, find d.

(i) a = 5, d = 3, a_n = 50.

By formula a_n = a + (n - 1)d

⇒ 50 = 5 + (n - 1)3
⇒ 50 = 5 + 3n - 3
⇒ 50 = 2 + 3n
⇒ 50 - 2 = 3n
⇒ 48 = 3n
⇒ n = 16.

By formula $S_n = \dfrac{n}{2}[2a + (n - 1)d]$

$S_n = \dfrac{16}{2}[2 \times 5 + (16 - 1)3] \\[1em] = 8[10 + 15 \times 3] \\[1em] = 8 \times 55 \\[1em] = 440.$

Hence, n = 16 and S_n = S₁₆ = 440.

(ii) a = 7, a₁₃ = 35

By formula a_n = a + (n - 1)d

⇒ 35 = 7 + (13 - 1)d
⇒ 35 = 7 + 12d
⇒ 35 - 7 = 12d
⇒ 28 = 12d
⇒ d = $\dfrac{28}{12}$ (Dividing by 4)

⇒ d = $\dfrac{7}{3}.$

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$S_{13} = \dfrac{13}{2}[2 \times 7 + (13 - 1)\Big(\dfrac{7}{3}\Big)] \\[1em] = \dfrac{13}{2}[14 + 12 \times \dfrac{7}{3}] \\[1em] = \dfrac{13}{2} \times [14 + 28] \\[1em] = \dfrac{13}{2} \times 42 \\[1em] = 13 \times 21 \\[1em] = 273.$

Hence, d = $\dfrac{7}{3}$ and S_n = S₁₃ = 273.

(iii) d = 5, S₉ = 75

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$S_9 = \dfrac{9}{2}[2 \times a + (9 - 1)5] \\[1em] \Rightarrow 75 = \dfrac{9}{2}[2a + 8 \times 5] \\[1em] \Rightarrow 75 = \dfrac{9}{2} \times [2a + 40] \\[1em] \Rightarrow 75 = 9(a + 20) \\[1em] \Rightarrow 75 = 9a + 180 \\[1em] \Rightarrow 9a = 75 - 180 \\[1em] \Rightarrow 9a = -105 \\[1em] \Rightarrow a = -\dfrac{105}{9} \\[1em] \Rightarrow a = -\dfrac{35}{3}$

By formula a_n = a + (n - 1)d,

$\Rightarrow a_9 = -\dfrac{35}{3} + (9 - 1)5 \\[1em] = -\dfrac{35}{3} + 40 \\[1em] = \dfrac{-35 + 120}{3} \\[1em] = \dfrac{85}{3}.$

Hence, a = $-\dfrac{35}{3} \text{ and } a_9 = \dfrac{85}{3}.$

(iv) a = 8, a_n = 62, S_n = 210.

By formula a_n = a + (n - 1)d

⇒ 62 = 8 + (n - 1)d
⇒ 62 - 8 = (n - 1)d
⇒ (n - 1)d = 54          (Eq 1)

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow 210 = \dfrac{n}{2}[2 \times 8 + (n - 1)d] \\[1em]$

Putting value of (n - 1)d from Eq 1 in above equation,

$\Rightarrow \dfrac{n}{2}[16 + 54] = 210 \\[1em] \Rightarrow \dfrac{n}{2} \times 70 = 210 \\[1em] \Rightarrow 35n = 210 \\[1em] \Rightarrow n = \dfrac{210}{35} \\[1em] \Rightarrow n = 6 \\[1em] \text{From Eq 1, } (n - 1)d = 54 \\[1em] \Rightarrow (6 - 1)d = 54 \\[1em] \Rightarrow 5d = 54 \\[1em] \therefore d = \dfrac{54}{5} \text{ and } n = 6.$

Hence, the value of n = 6 and d = $\dfrac{54}{5}.$

(v) a = 3, n = 8, S = 192.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ 192 = $\dfrac{8}{2}[2 \times 3 + (8 - 1)d]$
⇒ 192 = 4[6 + 7d]

⇒ $\dfrac{192}{4}$ = 6 + 7d
⇒ 48 = 6 + 7d
⇒ 7d = 42
⇒ d = 6.

Hence, the value of d is 6.