The sum up to n terms of an Arithmetic Series is given by $\rm S_{n} = \frac{n}{2} \cdot (2a + (n - 1) d)$, where the symbols have their usual meaning. Given, $\rm S_{n} = 315$.

The given progression $\rm 42 + 39 + 36 + \dots$ is an Arithmetic Series whose first term $\rm (a) = 42$.

The common difference of the series is $\rm t_{2} - t{1} = 39 - 42$ $\rm \Rightarrow d = -3$.

Let n number of terms of the series must be taken to make the sum 315, then by using the formula for the sum of Arithmetic Series, we get,

$\rm S_{n} = \frac{n}{2} \cdot (2a + (n - 1)d)$

$\rm or, 315 = \frac{n}{2} \cdot (2 \cdot 42 + (n - 1) (-3))$

Multiplying both sides of the equation by 2, we get,

$\rm or, 315 \cdot 2 = 2 \cdot \frac{n}{2} \cdot (84 - 3n  +3 )$

$\rm or, 630 = n (87 - 3n)$

$\rm or, 630= 87 n - 3n^{2}$

$\rm or, 3n^{2} – 87n + 630 = 0$

$\rm or, 3 (n^{2} - 29 n + 210) = 0$

$\rm or, n^{2} - 29n + 210 = 0$

The above equation is quadratic in n. We use the mid-term factorization formula to solve it.

$\rm or, n^{2} - (15 + 14)n + 210 = 0$

$\rm or, n^{2} - 15n - 14n + 210 = 0$

$\rm or, n(n  - 15) - 14 ( n - 15) = 0$

$\rm or, (n - 14) (n - 15) = 0$

Either

$\rm (n - 14) = 0$

$\rm \therefore n = 14$

Or

$\rm (n - 15) = 0$

$\rm \therefore n = 15$

Hence, the required number of terms to make the sum of the Arithmetic Series is 14 or 15.

To explain the double answer, we find the 14th and the 15th term of the Arithmetic Series.

$\rm t_{14} = a + (14 - 1)d = 42 + (13) \cdot (-3)$

$\rm \therefore t_{14} = 42 - 39 = 3$

And

$\rm t_{15} = a + (15 - 1)d = 42 + 14 \cdot (-3)$

$\rm \therefore t_{15} = 42 - 42 = 0$

Hence, the sum of the Arithmetic Series will be 315 by adding the first 14 terms. However, the fifth term is 0, so the sum would still be 315 by adding the first 15 terms.