1. **Stating the problem:** We have an arithmetic sequence representing diameters of stacked trays starting at 18 cm and ending at 34 cm. The sequence is $5, 8, 11, \ldots, 131$ (though this seems unrelated to the trays). We want to find the difference between diameters for the tray whose diameter is 26 cm. 2. **Understanding the arithmetic sequence:** The diameters form an arithmetic sequence with first term $a_1 = 18$ cm and last term $a_n = 34$ cm. The common difference is $d$ (unknown). 3. **Formula for the $n$-th term of an arithmetic sequence:** $$a_n = a_1 + (n-1)d$$ 4. **Find the position $n$ of the tray with diameter 26 cm:** Set $a_n = 26$: $$26 = 18 + (n-1)d$$ $$26 - 18 = (n-1)d$$ $$8 = (n-1)d$$ 5. **Find the total number of trays $n_{total}$ using the last term 34 cm:** $$34 = 18 + (n_{total}-1)d$$ $$16 = (n_{total}-1)d$$ 6. **Divide the two equations to eliminate $d$:** $$\frac{8}{16} = \frac{n-1}{n_{total}-1}$$ $$\frac{1}{2} = \frac{n-1}{n_{total}-1}$$ 7. **Interpretation:** The tray with diameter 26 cm is exactly halfway in position between the first and last trays. 8. **Find the difference between diameters for the tray with diameter 26 cm:** The difference between diameters is the common difference $d$. 9. **Calculate $d$ using the total number of trays:** Since the problem does not give $n_{total}$ explicitly, but the height is 7 cm (possibly number of trays), assume $n_{total} = 7$. 10. **Calculate $d$:** $$d = \frac{34 - 18}{7 - 1} = \frac{16}{6} = \frac{8}{3} \approx 2.67 \text{ cm}$$ 11. **Calculate $n$ for diameter 26 cm:** $$8 = (n-1) \times \frac{8}{3}$$ $$n-1 = \frac{8}{(8/3)} = 3$$ $$n = 4$$ 12. **Difference between diameters for the tray with diameter 26 cm is the common difference:** $$d = \frac{8}{3} \approx 2.67 \text{ cm}$$ **Final answer:** The difference between diameters for the tray with diameter 26 cm is approximately $2.67$ cm.