The sum of the 4^th term and the 8^th term of an A.P. is 24 and the sum of the 6^th and 10^th term is 44. Find the first three terms of the A.P.

Given,

⇒ a₄ + a₈ = 24

⇒ a + (4 - 1)d + a + (8 - 1)d = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24 ………(i)

Also,

⇒ a₆ + a₁₀ = 44

⇒ a + (6 - 1)d + a + (10 - 1)d = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44 ………(ii)

Subtracting (i) from (ii) we get,

⇒ 2a + 14d - (2a + 10d) = 44 - 24

⇒ 4d = 20

⇒ d = 5.

Substituting value of d in (i) we get,

⇒ 2a + 10(5) = 24

⇒ 2a = 24 - 50

⇒ 2a = -26

⇒ a = -13

First three terms of A.P. = a, a + d, a + 2d.

= -13, -13 + 5, -13 + 2(5)

Hence, the first three terms of A.P. are -13, -8, -3.