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?

3 Answers
Jim G.
Aug 4, 2017

#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#

George C.
Aug 5, 2017

#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)#

Ratnaker Mehta
Aug 5, 2017

# 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.#

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