Question

Consider the arithmetic sequence where the 15th term is 0 and the 40th term is -50.

Answer 1

Answer:

$a_n=28-2(n-1)$

Step-by-step explanation:

We want to find the formula for this problem in the explicit form of:

$a_n=a_1+d(n-1)$, where $a_n$ is the nth term, $a_1$ is the first term, and d is the common difference

Here, we can say that the "first term" is the 15th term, which is 0, so instead of a_1, we have a_15. The nth term is the 40th term, which is -50, so instead of a_n, we have a_40:

$a_n=a_1+d(n-1)$

$a_{40}=a_{15}+d(40-15)$

$-50=0+d(25)$

d = -2

Now, we need to find the first term:

$a_{15}=a_1+d(15-1)$

$0=a_1+(-2)(14)$

a_1 = 28

Finally, our equation is: $a_n=28-2(n-1)$

Hope this helps!