Find the 96th term of the arithmetic sequence 1, -12, -25,...

6 answers
Question:

Find the 96th term of the arithmetic sequence 1, -12, -25,...

Answers

-220, The arithmetic sequence is an=a1+(n-1)d
a1-the first term
an-is the nth term
d-common difference between terms

-1059

Step-by-step explanation:

Katelyn, this is super easy to understand.

Anyways, to find the 96th term of the sequence, first subtract -14 and -3. You may be thinking it is -17, but it is not. -17 is only if you are subtracting -14 and 3.

(-14) - (-3) = -14 + 3 = -11.

That means the difference between each number is -11.

Now all you have to do is take the -11 and multiply it by 96, which is -1056. Then, add the -1056 to -3, which is -1059. That should be your answer.

Your school may have taught you a different way, but this is my way.

-1152

Step-by-step explanation:

F(1)=-16

F(2)=-28

F(3)=-40

Notice the sequence is decreasing by -12 for every increase in x

You can conclude the function for this sequence is going to be

F(x)=-12x

Plug 96 as X.

F(96)=-1152

1124.

Step-by-step explanation:

The common difference (d)   is -16 - (-28) = 12,  ( also -28 - (-40) = 12).

The nth term of an A.S.  is a1 + d(n - 1)  where a1 = first term, d=common difference.

so the 96th term of the given A.S. is:

-16 + 12(96 - 1)

= -16 + 12*95

= 1124.

The 96th term of the given AP is (-1048)

Step-by-step explanation:

We have, the given AP is :

-3, -14, -25....

First term is -3

Common difference is -14 - (-3)= -11

Let [tex]a_{96}[/tex] is the 96th term of the AP.

The nth term of the AP is given by :

[tex]a_n=a+(n-1)d[/tex]

Put all the values, such that,

[tex]a_{96}=(-3)+(96-1)\times (-11)\\\\a_{96}=-1048[/tex]

So, the 96th term of the given AP is (-1048).

             a₉₆ = - 1234

Step-by-step explanation:

-12 - 1 = -13   and  -25 - (-12) = -13  ⇒  d = -13

[tex]a_n=a_1+d(n-1)\\\\a_1=1\\d=-13\\n=96\\\\a_{96}=1+(-13)(96-1)\\\\a_{96}=1-13\cdot95\\\\a_{96}=1-1235\\\\a_{96}=-1234[/tex]

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