1. **State the problem:** We need to create a sequence of multiples of 2 starting from 2 (i.e., 2, 4, 6, ...) such that the sum of these numbers equals 15000. 2. **Formula used:** The sum of the first $n$ multiples of 2 is given by the arithmetic series formula: $$ S_n = 2 + 4 + 6 + \cdots + 2n = 2(1 + 2 + 3 + \cdots + n) $$ Using the formula for the sum of the first $n$ natural numbers: $$ 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2} $$ So, $$ S_n = 2 \times \frac{n(n+1)}{2} = n(n+1) $$ 3. **Set up the equation:** We want the sum to be 15000, so: $$ n(n+1) = 15000 $$ 4. **Solve the quadratic equation:** $$ n^2 + n - 15000 = 0 $$ Using the quadratic formula: $$ n = \frac{-1 \pm \sqrt{1 + 4 \times 15000}}{2} = \frac{-1 \pm \sqrt{60001}}{2} $$ 5. **Calculate the discriminant:** $$ \sqrt{60001} \approx 244.95 $$ 6. **Find the positive root:** $$ n = \frac{-1 + 244.95}{2} = \frac{243.95}{2} = 121.975 $$ Since $n$ must be an integer, we check $n=121$ and $n=122$. 7. **Check sums for $n=121$ and $n=122$:** - For $n=121$: $$ S_{121} = 121 \times 122 = 14762 $$ - For $n=122$: $$ S_{122} = 122 \times 123 = 15006 $$ 8. **Conclusion:** The sum 15000 is not exactly achievable with an integer number of terms in this sequence. The closest sums are 14762 (for 121 terms) and 15006 (for 122 terms). 9. **Final answer:** The sequence of the first 122 multiples of 2 sums to 15006, which is the closest to 15000. **Sequence:** 2, 4, 6, ..., 244 **Sum:** 15006