1. **State the problem:** We need to compute the sum of the sequence: 3 + 6 + 9 + ... + 117 + 120 + 117 + ... + 9 + 6 + 3. 2. **Analyze the sequence:** This sequence increases by 3 up to 120, then decreases by 3 back down to 3. It is symmetric around the maximum term 120. 3. **Identify the arithmetic progression (AP) going up:** The terms increasing are 3, 6, 9, ..., 120. 4. **Find the number of terms in the increasing part:** The first term $a_1 = 3$, common difference $d = 3$, last term $a_n = 120$. Use the formula for the $n$th term of an AP: $$a_n = a_1 + (n-1)d$$ $$120 = 3 + (n-1)3$$ $$120 - 3 = 3(n-1)$$ $$117 = 3(n-1)$$ $$n-1 = 39$$ $$n = 40$$ So, there are 40 terms increasing from 3 to 120. 5. **Sum of the increasing part:** Use the sum formula for AP: $$S_n = \frac{n}{2}(a_1 + a_n)$$ $$S_{40} = \frac{40}{2}(3 + 120) = 20 \times 123 = 2460$$ 6. **Sum of the decreasing part:** The decreasing part is 117 + 114 + ... + 6 + 3, which is the same as the increasing part except the last term 120 is excluded. Number of terms in decreasing part is 39 (since 120 is not repeated twice). Sum of decreasing part: $$S_{39} = \frac{39}{2}(3 + 117) = \frac{39}{2} \times 120 = 39 \times 60 = 2340$$ 7. **Total sum:** Add the sums of increasing and decreasing parts: $$2460 + 2340 = 4800$$ **Final answer:** $$\boxed{4800}$$