The formula to learn for the nth term of arithmetic
sequences is:
an = a1 + (n-1)·d
I'll work the second one. But I'll just give you the answer
to the first one, so you'll have to do it by yourself, but
you do it exactly like the second one.
The answer to the first one is
an = 8n - 59
---------------------
2. a6=-31; a14=-135
an = a1 + (n-1)·d
Substitute n=6
a6 = a1 + (6-1)d
-31 = a1 + 5d
Substitute n=14
a14 = a1 + (14-1)d
-135 = a1 + (13)d
-135 = a1 + 13d
So we have the system of equations:
-135 = a1 + 13d
-31 = a1 + 5d
Multiply the second equation through by -1
-135 = a1 + 13d
31 = -a1 - 5d
Add those two equations:
-135 = a1 + 13d
31 = -a1 - 5d
-----------------
-104 = 8d
= d
= d
Substituting that in
-31 = a1 + 5d
-31 = a1 + 5(-13)
-31 = a1 - 65
34 = a1
So
an = a1 + (n-1)·d
becomes
an = 34 + (n-1)·(-13)
an = 34 - 13(n-1)
an = 34 - 13n + 13
an = 47 - 13n
Edwin