Let a is the first term and d is the common difference of the AP

Given, a₃ = 15

=> a + 2d = 15

=> a = 15 - 2d ..............1

Again given

      S₁₀ = 125

=> (10/2)*{2a + (10 - 1)d} = 125

=> 5(2a + 9d) = 125

=> 2a + 9d = 125/5

=> 2a + 9d = 25

From equation 1, we get

      2(15 - 2d) + 9d = 25

=> 30 - 4d + 9d = 25

=> 30 + 5d = 25

=> 5d = 25 - 30

=> 5d = -5

=> d = -5/5

=> d = -1

From equation 1,

     a = 15 - 2(-1)

=> a = 15 + 2

=> a = 17

Now, a₁₀ = a + 9d

=> a₁₀ = 17 + 9*(-1)

=> a₁₀ = 17 - 9

=> a₁₀ = 8

So, the value of d is -1 and a₁₀ is 8