How to use arithmetric progession formula in these question ?

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1 Answer
Jan 4, 2018

#6^(th)# term to #18^(th)# term are #-12,-15,-18,-21,-24,-27,-30,-33,-36,-39,-42,-45,-48#

Explanation:

We have #S_n=(9n)/2-(3n^2)/2#. Let the #n^(th)# term of A.P. be written as #a_n#.

Now as #S_1=a_1#, but #S_1=9/2-3/2=6/2=3#, hence #a_1=3#

and as #S_2=(9xx2)/2-(3xx2^2)/2=9-6=3#, but #S_2=a_1+a_2#

hence #a_2=S_2-a_1=3-3=0#

Hence common difference is #d=0-3=-3#

Let us check as #S_n=n/2(2a_1+(n-1)d)#

= #n/2(2xx3+(n-1)xx(-3))#

= #n/2(6-3n+3)=n/2(9-3n)=(9n)/2-(3n^2)/2#

and #a_n=3+(n-1)(-3)=6-3n#

Hence #a_6=6-3xx6=-12#

and the sequence from #a_6# to #a_18# is

#-12,-15,-18,-21,-24,-27,-30,-33,-36,-39,-42,-45,-48#