Question 13 (2 points) In an arithmetic sequence $t_{8}=-49$ and $t_{15}=-84$, find the value of $t_{1}$. A
Solution
Step 1 ：In an arithmetic sequence, the difference between any two successive terms is constant. This difference is often called the 'common difference'. We can find the common difference by subtracting \(t_{8}\) from \(t_{15}\) and then dividing by the difference in their positions (15 - 8 = 7).
Step 2 ：Given that \(t_{8} = -49\) and \(t_{15} = -84\), we can calculate the common difference as follows: \(\frac{t_{15} - t_{8}}{15 - 8} = \frac{-84 - (-49)}{7} = -5.0\).
Step 3 ：Once we have the common difference, we can find \(t_{1}\) by subtracting 7 times the common difference from \(t_{8}\).
Step 4 ：Calculating \(t_{1}\) gives us: \(t_{1} = t_{8} - 7 \times (-5.0) = -49 - 7 \times (-5.0) = -14.0\).
Step 5 ：Final Answer: The value of \(t_{1}\) is \(\boxed{-14}\).
