Question

Find the
14th term of a sequence where the first term is 3
and the common difference is -10

.

Answer 1

To find the \( n \)-th term of an arithmetic sequence, you can use the formula:

\[
a_n = a_1 + (n - 1) \cdot d
\]

where:
- \( a_n \) is the \( n \)-th term,
- \( a_1 \) is the first term,
- \( d \) is the common difference,
- \( n \) is the term number.

In this case:
- \( a_1 = 3 \)
- \( d = -10 \)
- \( n = 14 \)

Now, plug these values into the formula:

\[
a_{14} = 3 + (14 - 1) \cdot (-10)
\]

Calculating inside the parentheses:

\[
a_{14} = 3 + 13 \cdot (-10)
\]
\[
a_{14} = 3 - 130
\]
\[
a_{14} = -127
\]

Therefore, the 14th term of the sequence is \(\boxed{-127}\).