Given: 6^thterm of an A.P is 12 and 8^th terms of an A.P. is 22

⇒ a₆ = 12 and a₈ = 22

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 6:

∴ a₆ = a + (6 – 1)d

⇒ a₆ = a + 5d

Similarly, When n = 8:

∴ a₈ = a + (8 – 1)d

⇒ a₈ = a + 7d

According to question:

a₆ = 12 and a₈ = 22

⇒ a + 5d = 12 ………………(i)

And a + 7d = 22…………..(ii)

Subtracting equation (i) from (ii):

a + 7d – (a + 5d) = 22 – 12

⇒ a + 7d – a – 5d = 10

⇒ 2d = 10

⇒ d = 5

Put the value of d in equation (i):

a + 5(5) = 12

⇒ a + 25 = 12

⇒ a = 12 – 25

⇒ a = -13

As, a_n = a + (n – 1)d

a₂ = a + (2 – 1)d

⇒ a₂ = a + d

Now put the value of a = 9 and d = 2 in a_n and a₂

⇒ a_n = a + (n – 1)d

⇒ a_n = -13 + (n – 1)5

⇒ a_n = -13 + 5n – 5

⇒ a_n = -18 + 5n

a₂ = a + d

⇒ a₂ = -13 + 5

⇒ a₂ = -8

Hence, 2^thterm and n^th of the given A.P. are -8 and 5n – 18 respectively