In an arithmetic sequence whose 4th term is 14 and whose 11th term is 70, the first term and common difference are
A. a = -18, d = 8 B. a = -7, d = 7 C. a = -10, d = 8 D. a = -14, d = 7
Correct Answer: C
Explanation
To find the first term (\(a\)) and common difference (\(d\)) of the arithmetic sequence, we use the information provided about the 4th term and the 11th term. The formulas for the \(n\)-th term in an arithmetic sequence are:
\[ a_n = a + (n - 1) \times d \]
Given: - The 4th term (\(a_4\)) is 14. - The 11th term (\(a_{11}\)) is 70.
Step-by-Step Solution:
1. Set up the equations using the given terms:
For the 4th term: \[ a + (4 - 1) \times d = 14 \] \[ a + 3d = 14 \]
For the 11th term: \[ a + (11 - 1) \times d = 70 \] \[ a + 10d = 70 \]
2. Solve these equations simultaneously:
- Subtract the first equation from the second to eliminate \(a\): \[ (a + 10d) - (a + 3d) = 70 - 14 \] \[ 7d = 56 \] \[ d = \frac{56}{7} = 8 \]
- Substitute \(d = 8\) into the first equation: \[ a + 3 \times 8 = 14 \] \[ a + 24 = 14 \] \[ a = 14 - 24 = -10 \]
The first term is \( a = -10 \) and the common difference is \( d = 8 \).
