The third term of an arithmetic sequence is $5$ and the eighth term is $-20$. What is the product of the 4th and 2015th terms?

0

Step-by-step explanation:

Let n be the first term of this sequence and d the difference between two consecutive terms.

Following the arithmetic sequence formula,

The equation for the 3rd term:

n + 2d = 5

and the equation for the 8th term:

n + 7d = -20

We should start by finding d by subtracting the second equation from the first.

n + 2d - (n + 7d) = 5 - (-20)

-5d = 25

d = -5

We can then find the 1st term by plugging this number into the first equation.

n + 2 * -5 = 5

n - 10 = 5

n = 15

Now, using once again the arithmetic sequence formula, find the equation for the fourth term.

n + (4 - 1)*d

Plug in the values we found previously and solve:

15 + (4 - 1)*-5

= 15 + 3*-5

= 15 + (-15)

= 0

The 4th term is 0.

Remember that this problem is asking for the product of the 4th term and the 2015th term, and anything times zero equals to zero, so we don't even need to solve for the 2015th term!

Therefore, the answer to this problem is 0.