the third and fourth terms of an arithmetic sequence are -3 and 0.?

Question

how do i found the 1000th term

Answer 1

Since you're dealing with an AP(arithmetric sequence), let the nth term be Tn

Generally, Tn=a+(n-1)d

Where a=first term; n=number of terms; d=common difference

From the first point, 3rd term=-3 and from the second point, 4th term=0

This means, 

T3=a+(3-1)d=-3

a+2d=-3..................1

Also, T4=a+(4-1)d=0

a+3d=0

a=-3d.......................2

Substitute a as -3d into equation 1;

-3d+2d=-3

-d=-3

d=3

Also, from equation 2, a=-3(3)=-9

Hence, a=-9, d=3

Tn=-9+(n-1)3

Tn=-9+3n-3=3n-12

Tn=3n-12

Finally, when n=1000

T1000=3(1000)-12

          =3000-12

          =2988

Hence, 1000th term is 2988  

Answer 2

-9, -6, -3, 0, ...

-9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, ...

a_n = 3 (n - 4) 

The 1000th term is 2988

Answer 3

I think of the difference as a "step."

-3 to 0 means it increases in steps of 3

You have the 4th term, which is 0.

You want the 1000th term which is 996 steps away.

So just add 996 step to the 4th term.

0 + 3*996 = 2988 is the 1000th term

Answer 4

 The third and fourth terms of an arithmetic sequence are -3 and 0.

 -9, -6, -3, 0, ...

 a_n = 3 (n - 4) 

 The 1000th term is 3(996) = 2988.

Answer 5

Common difference is 3

Hence, 2nd term is -6 and 1st term is -9

nth term is -9 + 3(n - 1)

i.e. 3n - 12

so, 1000th term is 3(1000) - 12 => 2988

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Answer 6

For an arithmetic sequence:

a₁

a₂ = a₁ + d ← where d is the common difference, i.e.: 5

a₃ = a₂ + d = a₁ + 2d = - 3

a₄ = a₃ + d = a₁ + 3d = 0

You can obtain a system with 2 equations:

(1) : a₁ + 2d = - 3

(2) : a₁ + 3d = 0

Then you calculate (1) - (2):

(a₁ + 2d) - (a₁ + 3d) = - 3 - 0

a₁ + 2d - a₁ - 3d = - 3

d = 3

Recall a₃ = a₁ + 2d = - 3

a₁ + 2d = - 3

a₁ = - 3 - 2d → we've just seen that: d = 3

a₁ = - 9

Recall:

a₁

a₂ = a₁ + d

a₃ = a₁ + 2d

a₄ = a₁ + 3d

…and you can generalize writing:

a(n) = a₁ + (n - 1).d → for the 1000th term, n = 1000

a₁₀₀₀ = - 9 + (1000 - 1).3

a₁₀₀₀ = - 9 + (999 * 3)

a₁₀₀₀ = 2988

Answer 7

nth term of an Arithmetic Sequence is given by -

T(n)  = First term (suppose-a) + ( n-1 ) (common difference ( say d )

Hence ---

T(3) = a + ( 3 - 1 ) d   =  - 3

=> a + 2 d  =  - 3  .................... (1)   Similarly --

T(4) =  a + ( 4 - 1 ) d  =  0

=> a + 3 d  =  0  ............. (2)

from (1) and (2) we get 

a  =  - 9  and d  =  3

Hence  T(1000)  =  - 9 + ( 1000 - 1 ) 3  =  3000 - 12  =  2988.......Answer

Answer 8

Difference is 3.

2nd tem is -6

1st term is -9

n-t term is a + (n-1)d

1000th term = -9 + (1000-1)3 = 2988

Answer 9

if -3 and 0 are third or fourth terms,it mean it is increasing by 3, so if we start from 0 we have 996 terms left to get to 1000 terms.

so the 1000th term is the term x how much it is increasing

we have 996 terms if we start from 0

996 x 3

=2988