Let 3, 6, 9, 12, ... upto 78 terms and 5, 9, 13, 17, ... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ______.

We have two series: Series 1 is 3, 6, 9, 12, ..., up to 78 terms (AP with $$a = 3, d = 3$$), whose last term is $$3 + 77 \times 3 = 234$$. Series 2 is 5, 9, 13, 17, ..., up to 59 terms (AP with $$a = 5, d = 4$$), whose last term is $$5 + 58 \times 4 = 237$$.

The terms of Series 1 are given by $$3n$$ for $$n = 1, 2, ..., 78$$, and those of Series 2 by $$5 + 4(m-1) = 4m + 1$$ for $$m = 1, 2, ..., 59$$. For common terms, $$3n = 4m + 1$$, which implies $$3n \equiv 1 \pmod{4} \Rightarrow n \equiv 3 \pmod{4}$$ (since $$3 \times 3 = 9 \equiv 1 \pmod{4}$$). Thus $$n = 3, 7, 11, 15, ...$$ giving the common terms $$9, 21, 33, 45, ...$$, an AP with first term 9 and common difference 12.

These common terms must be at most $$234$$ (from Series 1) and $$237$$ (from Series 2). In general they are $$9 + 12(k-1) = 12k - 3$$ for $$k = 1, 2, 3, ...$$, so $$12k - 3 \leq 234 \Rightarrow k \leq 19.75$$, hence $$k \leq 19$$. Also, $$12(19) - 3 = 225 \leq 237$$, confirming that there are 19 common terms.

Finally, the sum of these common terms is $$S = \frac{19}{2}(9 + 225) = \frac{19}{2} \times 234 = 19 \times 117 = 2223$$. Therefore, the sum of the common terms is 2223.