Question

What is the 35th term of the arithmetic sequence where a1 = 13 and a17 = -83?

Answer 1

i thought a taught you :P

Answer 2

ar^0+ar^1+ar^2...

Answer 3

i am getting -3115

Answer 4

a1=a=13
since a17=-83 then the formula is

ar^16 it is 16 because they start at one but our formulas exponents start at 0
-83=13*r^16

sfr do you know how to solve for r?

Answer 5

Yes and i understood what you meant by it but i do not have the formula to plug the #'s in :"(

Answer 6

r = e^(ln(b/a)/n)

if you wanna solve for r :) in

a r^n=b :)

Answer 7

a=13 and a17=-83. => a+16d=-83
=> d=-6
now put this in the formula for the sum
S=35/2[2*13+34*-6]=-3115

Answer 8

-200
-197
-194
-191

Answer 9

\[An=a1+(n-1)d \]

Answer 10

and then

Answer 11

after you find d

Answer 12

you plug it all into the other equation right?

Answer 13

a1 = 13

are you sure that is right?

Answer 14

ooh sorry i am wrong i read that is geometric sequence not arithmetic lol *blush* hold on

Answer 15

nooooo lol

Answer 16

ignore all i said and lets start over :P I am so sorry

Answer 17

tis all good but im strugglin with this topic lol

Answer 18

-83 = 13 + +(17, -1) d
solve for d :)

Answer 19

d=-6 so use first formula here it is the same formula i use to find d

http://en.wikipedia.org/wiki/Arithmetic_progression

Answer 20

a_n=a_1+(n-1)*d n is ith element an is ith term d is difference and a1 is first term

Answer 21

13+(35-1)*-6=-191