Here is one way to minimally insert terminals so that A1 ⇒ A2 ⇒ … yields exactly

  24 17 16 9 1 14 12 13 4 11 10 2 3 19 7 5 8 23 25 6 22 15 18 21 20.

1.  We choose the second alternative for A2 (rule_3):
      A2 → 24 A33 A31 A37 A3 23 A3 22 A36 A33

    Expanding:
      – A33⇒17  
      – A31⇒A32 A7⇒16 9 1  
      – A37⇒A29 19⇒(14 12 13 4 11 10 2 3) 19  
      – first A3⇒A4 A5⇒7 5 8  
      – 23  
      – second A3⇒6  
      – 22  
      – A36⇒A31⇒15  
      – A33⇒20  

    That gives
      24 17 (16 9 1) (14 12 13 4 11 10 2 3 19) (7 5 8) 23 6 22 15 20

    which is missing 25 at pos 19, then needs 18 at pos 23 and 21 at pos 24.

2.  To fix:
    a)  In rule_7 (A3→6) insert a 25 before the 6 so that A3→25 6.  
        Now the second A3 yields 25 6 (pos 19–20).

    b)  In rule_3 insert “18 21” between A36 and A33 so that after A36⇒A31⇒15 it produces 18 21 before the final A33⇒20.

    No other insertions are needed.

Updated production rules:
rule_3: A2 -> 24 A33 A31 A37 A3 23 A3 22 A36 18 21 A33
rule_7: A3 -> 25 6