Question 14 (2 points) If $t_{6}=23$ and $t_{11}=38$ in an arithmetic sequence, find an expression for $t_{n}$.

Step 1 ：Given that $t_{6}=23$ and $t_{11}=38$ in an arithmetic sequence, we can find the common difference by subtracting the value of the sixth term from the value of the eleventh term and then dividing by the difference in their positions.

Step 2 ：Let's denote the common difference as $d$, the sixth term as $t_{6}$, the eleventh term as $t_{11}$, the position of the sixth term as $n_{6}$, and the position of the eleventh term as $n_{11}$. So we have $t_{6}=23$, $t_{11}=38$, $n_{6}=6$, and $n_{11}=11$.

Step 3 ：We can calculate the common difference $d$ using the formula $d = \frac{t_{11} - t_{6}}{n_{11} - n_{6}}$. Substituting the given values, we get $d = \frac{38 - 23}{11 - 6} = 3.0$.

Step 4 ：Once we have the common difference, we can find the first term $a$ of the sequence by rearranging the formula for the nth term of an arithmetic sequence, $t_{n} = a + (n-1)d$, to $a = t_{n} - (n-1)d$. Substituting the values for $t_{6}$, $n_{6}$, and $d$, we get $a = 23 - (6-1)*3.0 = 8.0$.

Step 5 ：Finally, we can find an expression for $t_{n}$ by substituting the values of $a$ and $d$ into the formula $t_{n} = a + (n-1)d$. This gives us $t_{n} = 8.0 + (n-1)*3.0$.

Step 6 ：Final Answer: The expression for $t_{n}$ in the arithmetic sequence is \(\boxed{8.0 + (n-1)*3.0}\).