Let the number of coins in first pile be x.
Then second pile contains x-2
Fourth pile contains 2(x-2)
And third pile contains 2(x-2) -1
Also total coins = 25
So,
x + (x-2) + 2(x-2) + 2(x-2)-1 = 25
or 6x -11 = 25
or x = 6
So the first pile contains 6 coins
second pile contains 4 coins
third pile contains 7 coins
fourth pile contains 8 coins