In an AP :

given S₁₀ = 125, a₃ = 15 find d and a₁₀

d is the common difference of the series

a is the first term of the series

a₃ = 15(Given in the Question)

 ∴S₁₀ =10/2  [2×a + (10-1)×d]

Solving (i) and (ii) we get:-

     2a + 9d= 25
     a   + 2d= 15  (Multiplying the Equation by 2 and changing the sign)

⇒ 2a + 9d = 25
    2a + 4d = 30
     -      -       -
--------------------
⇒   5d = (-5)
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∴ d= (-1)

 After putting value of d in any of the above equation we get a=17

∴ a ₁₀ = 17 + (10-1)(-1)
          = 8