**Solution:**

We know that the nth term of the AP is;

a_{n} = a + (n − 1) d

a_{4} = a + (4 − 1) d

a_{4} = a + 3d

Similarly,

a_{8} = a + 7 d

a_{6} = a + 5 d

a_{10} = a + 9 d

Given that,

a_{4 }+ a_{8} = 24

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12 …………………………………………………… **(i)**

a_{6 }+ a_{10} = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 …………………………………….. **(ii)**

On subtracting equation **(i)** from **(ii)**, we get;

2d = 22 − 12

2d = 10

d = 5

From equation **(i)**, we have;

a + 5d = 12

a + 5(5) = 12

a + 25 = 12

a = −13

a_{2} = a + d = −13 + 5 = −8

a_{3} = a_{2 }+ d = −8 + 5 = −3

Therefore, the first three terms of this AP are −13, −8, and −3.