To find a1 in this sequence, we need to determine the common difference between consecutive terms.

The common difference can be found by subtracting the value of the previous term from the value of the next term.

Given that:
S3 = 35
S10 = 120

Using the formula for the sum of an arithmetic series:
S3 = (n/2)(2a1 + (n-1)d)
35 = (3/2)(2a1 + 2d)

Solving for equation for d, we'll get:
35 = 3a1 + 3d

S10 = (n/2)(2a1 + (n-1)d)
120 = (10/2)(2a1 + (10-1)d)
120 = 5(2a1 + 9d)
24 = 2a1 + 9d
12 = a1 + 4.5d

Substitute 3a1 + 3d = 35 into 12 = a1 + 4.5d we get:
12 = 35 - 3d + 4.5d
12 = 35 + 1.5d
-23 = 1.5d
d = -23 / 1.5
d = -15.33

Substitute the value of d back into the equation 3a1 + 3d = 35:
3a1 + 3(-15.33) = 35
3a1 - 46 = 35
3a1 = 35 + 46
3a1 = 81
a1 = 81 / 3
a1 = 27

Therefore, a1 = 27.