Given series: 5, 8, 11, ...

We need to find the least value of n for which the sum of n terms is not less than 670.

Let's find the formula for the nth term of the series first.

The series is an Arithmetic Progression (AP) with the first term (a) = 5 and common difference (d) = 3.

The nth term of an AP is given by the formula: an = a + (n-1)d

Substituting the values, we get: an = 5 + (n-1)3 = 3n + 2

Now let's find the sum of n terms of the series using the formula for the sum of an AP.

The sum of n terms of an AP is given by the formula: Sn = n/2[2a + (n-1)d]

Substituting the values, we get: Sn = n/2[2(5) + (n-1)3] = n/2[3n + 7]

We need to find the least value of n for which Sn is not less than 670.

So, n/2[3n + 7] ≥ 670

Multiplying both sides by 2, we get: 3n² + 7n ≥ 1340

Rearranging, we get: 3n² + 7n - 1340 ≥ 0

Now we need to solve this quadratic inequality to find the values of n that satisfy it.

We can factorize the quadratic expression as follows: (3n - 35)(n + 38) ≥ 0

This gives us two critical points: n = 35/3 and n = -38

We need to test the intervals (-∞, -38), (-38, 35/3), and (35/3, ∞) to see which values of n satisfy the inequality.

Testing the interval (-∞, -38), we can take n = -40. Substituting in the inequality, we get: 3(-40)² + 7(-40) - 1340 = -4580 < 0,="" which="" means="" this="" interval="" does="" not="" satisfy="" the="" />

Testing the interval (-38, 35/3), we can take n = 7. Substituting in the inequality, we get: 3(7)² + 7(7) - 1340 = -41 < 0,="" which="" means="" this="" interval="" does="" not="" satisfy="" the="" />

Testing the interval (35/3, ∞), we can take n = 20. Substituting in the inequality, we get: 3(20)² + 7(20) - 1340 = 190 ≥ 0, which means this interval satisfies the inequality.

Therefore, the least value of n for which the sum of the series is not less than 670 is 20.

Hence, the correct option is (a) 20.