Find the first term, the common difference, and give a recursive formula for the arithmetic sequence.
-6th term is -10; 15th term is -46
A) $a _ { 1 } = 10 , d = 4 , a _ { n } = a _ { n - 1 } + 4$
B) $a _ { 1 } = - 10 , d = - 4 , a _ { n } = a _ { n - 1 } - 4$
C) $a _ { 1 } = - 30 , d = 4 , a _ { n } = a _ { n - 1 } + 4$
D) $a _ { 1 } = 10 , d = - 4 , a _ { n } = a _ { n - 1 } - 4$

Question

Quizplus · Accepted Answer

The difference between the 6th and 15th terms is 9 terms. The difference in value is -46 - (-10) = -36. The common difference $d = \frac{-36}{9} = -4$. To find the first term $a_1$, use the formula for the nth term of an arithmetic sequence: $a_n = a_1 + (n-1)d$. For the 6th term, $-10 = a_1 + 5(-4)$, solving for $a_1$ gives $a_1 = 10$. The recursive formula for an arithmetic sequence with a common difference of -4 is $a_n = a_{n-1} - 4$.

Find the First Term, the Common Difference, and Give a Recursive