Formula Used.

a_n = a + (n–1)d

a₃ = a + (3–1)d

a₃ = a + 2d

If 3^rd term of A.P is given as 8

Then,

a + 2d = 8

we get a = 8–2d ......eq 1

a₁₀ = a + (10–1)d

a₁₀ = a + 9d

If 10^th term of A.P is given as 20 + 6^th term

Then,

a₆ = a + (6–1)d

= a + 5d

a + 9d = 20 + [a + 5d]

we get

a–a + 9d–5d = 20

4d = 20

d = = 5 ......eq 2

Putting d in eq 1 we get ;

a = 8–(2×5)

= 8–10 = –2

As a = –2 and d = 5

Then;

a₁ = a + (n–1)d = –2 + (1–1)(5) = –2

a₂ = a + (n–1)d = –2 + (2–1)(5) = –2 + (5)×1 = –2 + 5 = 3

a₃ = a + (n–1)d = –2 + (3–1)(5) = –2 + (5)×2 = –2 + 10 = 8

a₄ = a + (n–1)d = –2 + (4–1)(5) = –2 + (5)×3 = –2 + 15 = 13

a_n = a + (n–1)d = –2 + (n–1)(5) = –2–5 + 5n = –7 + 5n

∴ The A.P is –2, 3, 8, 13, ……, 5n–7