Question

Find an equation for the nth term of the arithmetic sequence.

a19 = -58, a21 = -164

Answer 1

find common difference
\[a _{n}=a _{1}+(n-1)d\]
\[a _{21}=a _{1}+(21-1)d,\]
\[a _{1}+20d=-164\]
similarly
\[a _{1}+18d=-58\]
subtract and find d
then a1
and finally write an=?

Answer 2

An=896-53(n-1) or is it (n+1)

Answer 3

\(a_{19} = a_1 + ((19-1) \times -53)\)
\(a_{19} = a_1 + (18 \times -53)\)
\(-58 = a_1 -954\)
\(896 = a_1\)

\(a_n = 896 + (n-1)*-53\)
\(a_n = 896 - 53n + 53\)

\(\bf{a_n = 949 + 53n}\)

Answer 4

That can't be right

Answer 5

why not @Kabase1234 ?

Answer 6

My answer choices are either an=896-53(n-1) or
An=896-53(n+1)

Answer 7

I meant
\(949 - 53n = a_n\)

Answer 8

I simply simplified it.

Answer 9

this would be your answer :-
an=896-53(n-1)

Answer 10

That's what I thought

Answer 11

I simplified it from:
\(a_n = 896 + (n-1) * -53\)
which is the same thing as
\(a_n = 895-53(n-1)\)

Answer 12

\(a_n = 896 - 53(n-1)\)

Answer 13

Thank you so much! Can you help with my last one??

Answer 14

Find an equation for the nth term of the arithmetic sequence.

-3, -5, -7, -9, ...

Answer 15

The second one not the first

Answer 16

common difference d is -5-(-3)

a1 = -3

plug these into the formula
an = a1 + (n - 1)d

Answer 17

Use the main formula:
\(a_n = a_1 + (n-1)d\)

\(a_1\) = -3
\(d\) = -2

Substitute it back in:
\(a_n = -3 + (n-1)*-2\)
\(a_ n = -3 -2n + 2\)
\(\bf{a_n = -1 - 2n}\)

Answer 18

-3+-2(n) or is it (n-1)

Answer 19

@calculusxy

Answer 20

calculusxy has given you the correct method and answer.

-3 + (n - 1)*-2
= -3 +n*-2 -1*-2
= -3 - 2n + 2
= -1 - 2n

Answer 21

That can't be the answer though

Answer 22

Choices:
an = -3 - 2
an = -3 + -2(n)
an = -3 + -2(n + 1)
an = -3 + -2(n - 1)

Answer 23

ah ok
what calculusxy has done is the simplify the formula

an = a1 + (n - 1)d

a1 = -3 and d = -2

an = -3 + (n - 1)-2

which can be written as

an = -3 + -2(n - 1)

Answer 24

so that is choice d