**Solution:**

Let the first term of an A.P = a

and the common difference of the given A.P = d

As we know that

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+ 9d

Sum of 4th and 8th terms of an A.P = 24 ( given )

a_{4} +a_{8} = 24

a + 3d + a + 7d = 24

2a + 10 d = 24

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

Sum of 6 th and 10 th term of an A.P = 44 ( given )

a_{6} +a_{10} = 44

a + 5d +a+ 9d = 44

2a + 14 =44

**a + 7d = 22 .....................(ii)**

Solving (i) & (ii)

a +7 d = 22

a + 5d = 12

- - -

2d = 10

d = 5

From equation (i) ,

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

**So, the first three terms are -13 ,-8,-3**