Question

In an arithmetic sequence {an}, a1 +a11 = 2019. Find a6. Help. 40 Pts

Answer 1

1009.5

Explanation:

We know a term is made up of:

1. The first term added

2. The common difference added (The amount between two terms.)

So, we can say the first term is the variable x, and d for common difference. a11 then equals x+10d.

The question asks for a6, which is x+5d. It also gives that a1+a11=2019, so x+x+10d=2019. 2x+10d=2019, or 2(x+5d).

Do you notice something now? a6 is x+5d, and a1+a11=2(x+5d). a1+a11=2(a6)! That just means a6=2019/2=1009.5

I hope you understand! Thank you!

Answer 2

Answer: 1009.5

Work Shown:

a1 = unknown = x

a2 = x+d, where d is some constant

a3 = x+2d

a4 = x+3d

a5 = x+4d

a6 = x+5d

and so on until we reach

a11 = x+10d

Add a1 and a11 to get

a1+a11 = x+x+10d = 2x+10d = 2(x+5d)

Note how we get 2(x+5d) and above shows a6 = x+5d

So we see that a1+a11 is exactly double that of a6

This means a6 must be half that of a1+a11

If a1+a11 = 2019, then a6 = (a1+a11)/2 = 2019/2 = 1009.5