Question

Consider a strictly increasing sequence 6, 7, 8, 9, x, y, 40, 50, 60, 70 (where $x, y \in Z$). If the number of ordered pairs of $(x, y)$ such that no five consecutive terms of the sequence forms an arithmetic progression is $K$, then the sum of digits of $K$ is

Answer 1

18

Answer 2

Let the given sequence be $a_1, a_2, \ldots, a_{10}$. The sequence is 6, 7, 8, 9, x, y, 40, 50, 60, 70. The sequence is strictly increasing, so $a_1 < a_2 < \ldots < a_{10}$. From the given values, we have $9 < x < y < 40$, where $x, y$ are integers. The possible integer values for $x$ range from 10 to 38. For a fixed $x$, $y$ can take integer values from $x+1$ to 39. The total number of ordered pairs $(x, y)$ satisfying $9 < x < y < 40$ is the number of ways to choose 2 distinct integers from the set $\{10, 11, \ldots, 39\}$, which has $39 - 10 + 1 = 30$ integers. The number of such pairs is $\binom{30}{2} = \frac{30 \times 29}{2} = 15 \times 29 = 435$.

No five consecutive terms of the sequence form an arithmetic progression (AP). The possible sets of five consecutive terms are:

1. (6, 7, 8, 9, x): This is an AP if the common difference is 1, which requires $x = 9+1 = 10$.
2. (7, 8, 9, x, y): This is an AP if the common difference is 1, which requires $x = 9+1 = 10$ and $y = 10+1 = 11$.
3. (8, 9, x, y, 40): If this is an AP, the common difference must be $9-8=1$. Then $x=9+1=10$, $y=10+1=11$, and $40=11+1=12$. This is false, so this sequence can never be an AP.
4. (9, x, y, 40, 50): If this is an AP, the common difference must be $50-40=10$. Then $40=y+10 \implies y=30$, $y=x+10 \implies x=20$, and $x=9+10 \implies 9=20+10=30$. This is false, so this sequence can never be an AP.
5. (x, y, 40, 50, 60): This is an AP if the common difference is $50-40=10$ (from the last three terms). This requires $y = 40-10 = 30$ and $x = 30-10 = 20$. So, if $(x, y) = (20, 30)$, this sequence is an AP.
6. (y, 40, 50, 60, 70): This is an AP if the common difference is $50-40=10$ (from the last four terms). This requires $y = 40-10 = 30$. So, if $y=30$, this sequence is an AP.

The pairs $(x, y)$ that cause a five-term AP to form are:

• $x=10$ (from case 1). These are pairs $(10, y)$ where $10 < y < 40$, i.e., $11 \le y \le 39$. Number of such pairs is $39 - 11 + 1 = 29$.
• $(x, y) = (10, 11)$ (from case 2). This pair is included in the set where $x=10$.
• $(x, y) = (20, 30)$ (from case 5). This pair satisfies $9 < 20 < 30 < 40$.
• $y=30$ (from case 6). These are pairs $(x, 30)$ where $9 < x < 30$, i.e., $10 \le x \le 29$. Number of such pairs is $29 - 10 + 1 = 20$.

We need to find the number of pairs $(x, y)$ from the total 435 pairs such that none of these conditions are met. Let $A$ be the set of pairs where $x=10$. $|A|=29$. Let $B$ be the set of pairs where $(x, y)=(10, 11)$. $|B|=1$. Let $C$ be the set of pairs where $(x, y)=(20, 30)$. $|C|=1$. Let $D$ be the set of pairs where $y=30$. $|D|=20$.

The condition "no five consecutive terms form an AP" means that the pair $(x, y)$ must not be in $A \cup B \cup C \cup D$. We need to find $|A \cup B \cup C \cup D|$.

$B = \{(10, 11)\}$. Since $10=10$, $B \subseteq A$. So $A \cup B = A$. $A \cup B \cup C \cup D = A \cup C \cup D$.

We calculate $|A \cup C \cup D| = |A| + |C| + |D| - (|A \cap C| + |A \cap D| + |C \cap D|) + |A \cap C \cap D|$.

• $A: x=10$. Pairs are $(10, y)$ for $11 \le y \le 39$.

• $C: (x, y) = (20, 30)$.

• $D: y=30$. Pairs are $(x, 30)$ for $10 \le x \le 29$.

• $A \cap C$: Pairs where $x=10$ and $(x, y)=(20, 30)$. Impossible, $10 \neq 20$. $|A \cap C|=0$.

• $A \cap D$: Pairs where $x=10$ and $y=30$. This is the pair $(10, 30)$. $|A \cap D|=1$.

• $C \cap D$: Pairs where $(x, y)=(20, 30)$ and $y=30$. This is the pair $(20, 30)$. $|C \cap D|=1$.

• $A \cap C \cap D$: Pairs where $x=10$, $(x, y)=(20, 30)$, and $y=30$. Impossible. $|A \cap C \cap D|=0$.

$|A \cup C \cup D| = |A| + |C| + |D| - (|A \cap D| + |C \cap D|) + 0$ $|A \cup C \cup D| = 29 + 1 + 20 - (1 + 1) = 50 - 2 = 48$.

The number of pairs $(x, y)$ to be excluded is 48. The total number of possible pairs is 435. The number of allowed pairs, $K$, is $435 - 48 = 387$.

The question asks for the sum of digits of $K$. $K = 387$. Sum of digits of $K = 3 + 8 + 7 = 18$.

The final answer is $\boxed{18}$.