Sn = (2a1 + (n-1)d) < 1000

Sn = [2*7 + (n-1)12] < 1000
     
Multiply both sides by 2

      n[2*7 + (n-1)12] < 2000

      n[14 + 12n - 12] < 2000             
      
            n[2 + 12n] < 2000

             2n + 12n² < 2000

      12n² + 2n - 2000 < 0

Divide through by 2

        6n² + n - 1000 < 0    

Use the quadratic formula to determine the zeros
of 

f(n) = 6n² + n - 1000

They are approximately 12.82688011 and -12.99354678

We know n cannot be a negative number.  We think therefore
the answer will be the greatest integer less than 12.82688011
which is 12.

To prove it we substitute n = 12

Sn = [2*7 + (n-1)12]

S12 = [14 + (12-1)12]

S12 = 6[14 + 11×12]

S12 = 6[14 + 132]

S12 = 6[146]

S12 = 876

Then we substitute n = 13 to see if that runs over 1000.

To prove it we substitute n = 12

Sn = [2*7 + (n-1)12]

S13 = [14 + (13-1)12]

S13 = [14 + 12×12]

S13 = [14 + 144]

S13 = [158]

S13 = 1027

This runs over 1000, so we now have proved that

the answer is 12.

---------------

For the second part of the problem, it is done exactly the same
way, just use 2000 instead of 1000.

The answer is  18.

Edwin