Question

If the sum of the first `n` terms of a sequence is `S_n = n^2-3n`, then what is the sixth term of the sequence?

Answer 1

`a_6=8`

Explanation:

`"assuming that the sum of n terms is"`

`S_n=n^2-3n`

`S_1=1-3=-2rArra_1=-2`

`S_2=4-6=-2rArra_2=S_2-a_1=-2-(-2)=0`

`S_3=9-9=0`

`rArra_3=S_3-a_2-a_1=0-0-(-2)=2`

`S_4=16-12=4`

`rArra_4=S_4-a_3-a_2-a_1=4-2-0-(-2)=4`

`S_5=25-15=10`

`rArra_5=10-4-2-0-(-2)=6`

`S_6=36-18=18`

`rArra_6=18-6-4-2-0-(-2)=8`

`"the first 6 terms are "-2,0,2,4,6,8`

Answer 2

`8`

Explanation:

Assuming you intended:

`S_n = n^2-3n`

We can substitute `n=0,1,...,6` into this formula and write `S_0, S_1,...,S_6` as:

`0, -2, -2, 0, 4, 10, 18`

and can deduce that the first six terms of the sequence are:

`-2, 0, 2, 4, 6, 8`

So the sixth term is `8`

The general term in standard form is:

`a_n = -2+2(n-1)`

Answer 3

`8.`

Explanation:

Let `S_n` denote the sum of the first `n`terms of a Seq.

`{t_i}_(i=1)^(i=n).`

`:. S_n=t_1+t_2+t_3+...+t_(n-1)+t_n,`

`=[t_1+t_2+t_3+...+t_(n-1)]+t_n,`

`rArr S_n=S_(n-1)+t_n....................(star).`

Therefore, by `(star),`

`"The Reqd. Term="t_6=S_6-S_5,`

`=(6^2-3*6)-(5^2-3*5),...[because, S_n=n^2-3n],`

`=36-18-25+15,`

`rArr t_6=8.`

N.B.: In general, `t_n=2n-4.`