The sum of the 4^th term and the 8^th terms of an A.P. is 24 and the sum of 6^th and the 10^th terms of the same A.P. is 34. Find the first three terms of the A.P.

According to question,

a₄ + a₈ = 24 …….(i)

a₆ + a₁₀ = 34 …….(ii)

Solving (i) we get,

⇒ a + (4 - 1)d + a + (8 - 1)d = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 ………(iii)

Solving (ii) we get,

⇒ a + (6 - 1)d + a + (10 - 1)d = 34

⇒ a + 5d + a + 9d = 34

⇒ 2a + 14d = 34

⇒ a + 7d = 17 ………(iv)

Subtracting (iii) from (iv) we get,

⇒ a + 7d - (a + 5d) = 17 - 12

⇒ 2d = 5

⇒ d = $\dfrac{5}{2}$.

Substituting d in (iv) we get,

⇒ a + $7 \times \dfrac{5}{2}$ = 17

⇒ a + $\dfrac{35}{2}$ = 17

⇒ a = $17 - \dfrac{35}{2} = \dfrac{34 - 35}{2} = -\dfrac{1}{2}$.

A.P. = a, (a + d), (a + 2d) ………

$= -\dfrac{1}{2}, -\dfrac{1}{2} + \dfrac{5}{2}, -\dfrac{1}{2} + 2 \times \dfrac{5}{2} ……… \\[1em] = -\dfrac{1}{2}, \dfrac{-1 + 5}{2}, -\dfrac{1}{2} + 5………. \\[1em] = -\dfrac{1}{2}, 2, \dfrac{9}{2} ……..$

Hence, first three terms of the A.P. = $-\dfrac{1}{2}, 2, \dfrac{9}{2}$