Solution:

We all know what the AP’s nth phrase is;

a_n = a+(n−1)d

a₄ = a+(4−1)d

a₄ = a+3d

We can write in the same way.,

a₈ = a+7d

a₆ = a+5d

a₁₀ = a+9d

Given that,

a₄+a₈ = 24

a+3d+a+7d = 24

2a+10d = 24

a+5d = 12 …………………………………………………… (i)

a₆+a₁₀ = 44

a +5d+a+9d = 44

2a+14d = 44

a+7d = 22 …………………………………….. (ii)

When we subtract equation (i)from equation (ii), we get,

2d = 22 − 12

2d = 10

d = 5

From equation (i), we get,

a+5d = 12

a+5(5) = 12

a+25 = 12

a = −13

a₂ = a+d = − 13+5 = −8

a₃ = a₂+d = − 8+5 = −3

As a result, this A.P.’s first three terms are 13, 8, and 3.