1. The problem states that $x$, 17, $3x - y^2 - 2$, and $3x + y^2 - 30$ are four consecutive terms of an increasing arithmetic sequence. 2. In an arithmetic sequence, the difference between consecutive terms is constant. Let this common difference be $d$. 3. Using the terms, we write the equations for the common difference: $$17 - x = d$$ $$3x - y^2 - 2 - 17 = d$$ $$3x + y^2 - 30 - (3x - y^2 - 2) = d$$ 4. Simplify the differences: $$17 - x = d$$ $$3x - y^2 - 19 = d$$ $$3x + y^2 - 30 - 3x + y^2 + 2 = d$$ $$2y^2 - 28 = d$$ 5. Since all equal $d$, set them equal to each other: $$17 - x = 3x - y^2 - 19$$ $$3x - y^2 - 19 = 2y^2 - 28$$ 6. Solve the first equation: $$17 - x = 3x - y^2 - 19$$ $$17 + 19 = 3x - y^2 + x$$ $$36 = 4x - y^2$$ $$4x - y^2 = 36$$ 7. Solve the second equation: $$3x - y^2 - 19 = 2y^2 - 28$$ $$3x - y^2 - 19 - 2y^2 + 28 = 0$$ $$3x - 3y^2 + 9 = 0$$ $$3x - 3y^2 = -9$$ $$x - y^2 = -3$$ 8. From step 7, express $x$: $$x = y^2 - 3$$ 9. Substitute $x$ into step 6: $$4(y^2 - 3) - y^2 = 36$$ $$4y^2 - 12 - y^2 = 36$$ $$3y^2 - 12 = 36$$ $$3y^2 = 48$$ $$y^2 = 16$$ 10. Substitute $y^2 = 16$ back to find $x$: $$x = 16 - 3 = 13$$ 11. Now find the four terms: $$x = 13$$ $$17$$ $$3x - y^2 - 2 = 3(13) - 16 - 2 = 39 - 18 = 21$$ $$3x + y^2 - 30 = 39 + 16 - 30 = 25$$ 12. Verify the common difference: $$17 - 13 = 4$$ $$21 - 17 = 4$$ $$25 - 21 = 4$$ 13. Sum the four terms: $$13 + 17 + 21 + 25 = 76$$ 14. Check divisibility of 76 by the given options: - Divisible by 5? No (ends with 6) - Divisible by 2? Yes (76 is even) - Divisible by 3? No (7+6=13, not multiple of 3) - Divisible by 11? No (7-6=1, not multiple of 11) - Divisible by 7? No (7*10=70, 7*11=77) Final answer: The sum is divisible by 2.