Given: the sum of the 4^th and the 8^th terms of an A.P. is 24, and the sum of the 6^th and 10^th term is 34

⇒ a₄ + a₈ = 24 and a₆ + a₁₀ = 34

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

When n = 4:

∴ a₄ = a + (4 – 1)d

⇒ a₄ = a + 3d

When n = 6:

∴ a₆ = a + (6 – 1)d

⇒ a₆ = a + 5d

When n = 8:

∴ a₈ = a + (8 – 1)d

⇒ a₈ = a + 7d

When n = 10:

∴ a₁₀ = a + (10 – 1)d

⇒ a₁₀ = a + 9d

According to question:

a₄ + a₈ = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ 2(a + 5d) = 24

⇒ a + 5d = 12………(i)

a₆ + a₁₀ = 34

⇒ a + 5d + a + 9d = 34

⇒ 2a + 14d = 34

⇒ 2(a + 7d) = 34

⇒ a + 7d = 17………(ii)

Subtracting equation (i) from (ii):

a + 7d – (a + 5d) = 17 – 12

⇒ a + 7d – a – 5d = 5

⇒ 2d = 5

Put the value of d in equation (i):

Hence, the value of a and d areand respectively