Subjects algebra

Arithmetic Sequence Ffa68E

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Arithmetic Sequence Ffa68E


1. **Stating the problem:** We are given an arithmetic sequence defined by terms involving $P$ and $E$, specifically the terms $P$, $E$, $1000E$, and $-P$. We need to analyze or find a relationship involving these terms. 2. **Understanding arithmetic sequences:** An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This difference is called the common difference $d$. 3. **Setting up the problem:** Given the terms $a_1 = P$, $a_2 = E$, $a_3 = 1000E$, and $a_4 = -P$, since these are consecutive terms of an arithmetic sequence, the differences between consecutive terms must be equal: $$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d$$ 4. **Writing the equations:** $$E - P = 1000E - E = -P - 1000E$$ Simplify each: $$E - P = 999E$$ $$999E = -P - 1000E$$ 5. **From the first equality:** $$E - P = 999E \implies -P = 999E - E = 998E \implies P = -998E$$ 6. **From the second equality:** $$999E = -P - 1000E$$ Substitute $P = -998E$: $$999E = -(-998E) - 1000E = 998E - 1000E = -2E$$ 7. **Solve for $E$:** $$999E = -2E \implies 999E + 2E = 0 \implies 1001E = 0 \implies E = 0$$ 8. **Find $P$ using $E=0$:** $$P = -998E = -998 \times 0 = 0$$ 9. **Conclusion:** The only way for the terms $P$, $E$, $1000E$, and $-P$ to form an arithmetic sequence is if $E=0$ and $P=0$. **Final answer:** $$P = 0, \quad E = 0$$