Question

Identify the 31st term of an arithmetic sequence where a1 = 26 and a22 = –226.

Answer 1

Well first find d.
a1 = 26
a22 = –226
a22=a1+21d
Substitute!
-226=26+21d
21d=-252
d= -12
Now lets find the 31st term.
a31=a1+30d
a31=26+30 * (-12)
a31=26 + -360
a31= -334

Answer 2

how did you get the d part

Answer 3

\[t_{n}=a+(n-1)d\]

Answer 4

what?.. i don't get it

Answer 5

\[t_{n}\] is the nth term.In this case I put it as 22nd term.d is the common diff. which was found.And a is a1

Answer 6

so its always lower like if it was 17 it would be a1+16?

Answer 7

a1+16d

Answer 8

okay let me see if i get this one by myself

Answer 9

a1 = –12 and a27 = 66.
a1=-12
a27=66
a27=a1+26d?