1. **Stating the problem:** We have a sequence of patterns made from tiles, where the number of tiles increases by the same amount each time. We want to find the number of tiles in the 4th pattern. 2. **Given data:** - Pattern 1 has 4 tiles. - Pattern 2 has 7 tiles. - Pattern 3 has 11 tiles. 3. **Find the common difference:** The difference between Pattern 2 and Pattern 1 is $7 - 4 = 3$ tiles. The difference between Pattern 3 and Pattern 2 is $11 - 7 = 4$ tiles. Since the difference is not constant, the sequence is not arithmetic based on these numbers alone. 4. **Check the given formula:** The problem states the $n^{th}$ term of a sequence is given by: $$a_n = 6(3 - n)$$ 5. **Calculate the first four terms using the formula:** - For $n=1$: $$a_1 = 6(3 - 1) = 6 \times 2 = 12$$ - For $n=2$: $$a_2 = 6(3 - 2) = 6 \times 1 = 6$$ - For $n=3$: $$a_3 = 6(3 - 3) = 6 \times 0 = 0$$ - For $n=4$: $$a_4 = 6(3 - 4) = 6 \times (-1) = -6$$ 6. **Interpretation:** The formula $a_n = 6(3 - n)$ does not match the tile counts given for the patterns (4, 7, 11). It produces 12, 6, 0, -6 which are not the tile counts. 7. **Find the actual pattern for tile counts:** The tile counts are 4, 7, 11 for patterns 1, 2, 3. Calculate the differences: - $7 - 4 = 3$ - $11 - 7 = 4$ The differences are not constant, so let's check the second difference: - $4 - 3 = 1$ This suggests a quadratic pattern. 8. **Find the quadratic formula:** Assume $a_n = An^2 + Bn + C$ Use the known terms: - $a_1 = A(1)^2 + B(1) + C = A + B + C = 4$ - $a_2 = 4A + 2B + C = 7$ - $a_3 = 9A + 3B + C = 11$ 9. **Solve the system:** From $a_1$: $A + B + C = 4$ From $a_2$: $4A + 2B + C = 7$ From $a_3$: $9A + 3B + C = 11$ Subtract $a_1$ from $a_2$: $$ (4A + 2B + C) - (A + B + C) = 7 - 4 \Rightarrow 3A + B = 3 $$ Subtract $a_2$ from $a_3$: $$ (9A + 3B + C) - (4A + 2B + C) = 11 - 7 \Rightarrow 5A + B = 4 $$ Subtract the two equations: $$ (5A + B) - (3A + B) = 4 - 3 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2} $$ Plug $A = \frac{1}{2}$ into $3A + B = 3$: $$ 3 \times \frac{1}{2} + B = 3 \Rightarrow \frac{3}{2} + B = 3 \Rightarrow B = 3 - \frac{3}{2} = \frac{3}{2} $$ Plug $A$ and $B$ into $A + B + C = 4$: $$ \frac{1}{2} + \frac{3}{2} + C = 4 \Rightarrow 2 + C = 4 \Rightarrow C = 2 $$ 10. **Final formula:** $$ a_n = \frac{1}{2}n^2 + \frac{3}{2}n + 2 $$ 11. **Calculate the 4th term:** $$ a_4 = \frac{1}{2} \times 16 + \frac{3}{2} \times 4 + 2 = 8 + 6 + 2 = 16 $$ **Answer:** The 4th pattern has **16 tiles**. 12. **First four terms of the sequence:** - $a_1 = 4$ - $a_2 = 7$ - $a_3 = 11$ - $a_4 = 16$