### Instruction: Find the eighth term of the sequence $1440, 1716, 1848, \ldots$, whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.### Response: I notice that the sequence is formed by multiplying the corresponding terms of two arithmetic sequences, which means that each term is of the form $a_n b_n$, where $a_n$ and $b_n$ are the $n$th terms of the two arithmetic sequences.
I also notice that the first term is $1440 = 120 \times 12$, which suggests that $a_1 = 120$ and $b_1 = 12$.
To check this hypothesis, I can see if the second term also matches this pattern.
The second term is $1716 = 143 \times 12$, which is consistent with $a_1 = 120$ and $b_1 = 12$, since $a_2 = a_1 + d_a = 120 + 23 = 143$ and $b_2 = b_1 + d_b = 12 + 0 = 12$, where $d_a$ and $d_b$ are the common differences of the two arithmetic sequences.
To further check this hypothesis, I can see if the third term also matches this pattern.
The third term is $1848 = 154 \times 12$, which is again consistent with $a_1 = 120$ and $b_1 = 12$, since $a_3 = a_2 + d_a = 143 + 11 = 154$ and $b_3 = b_2 + d_b = 12 + 0 = 12$.
So far, I have found that $a_1 = 120$, $b_1 = 12$, $d_a = 23$, and $d_b = 0$.
To find the eighth term of the sequence, I need to find $a_8 b_8$.
To find $a_8$, I can use the formula for the $n$th term of an arithmetic sequence: $a_n = a_1 + (n - 1) d_a$.
Plugging in $n = 8$, I get $a_8 = 120 + (8 - 1) 23 = 120 + 161 = 281$.
To find $b_8$, I can use the same formula, but with $b_1$ and $d_b$.
Plugging in $n = 8$, I get $b_8 = 12 + (8 - 1) 0 = 12 + 0 = 12$.
Therefore, the eighth term of the sequence is $a_8 b_8 = 281 \times 12 = 3372$.