IIT JEE , AIEEE Forum - Mathematics , Algebra

perin

find the 8th and nth term of the series : 1/n + (2n+1)/n + (4n+1)/n + ............
can the 8th term be deduced from the nth term, if not, why?

nivedh_89

as it forms an A.P.,
common difference(d) = 2

term8 = a+7d...where a = 1st term......
= 1/n+14...
=(14n+1)/n

nivedh_89

nth term = a+(n-1)d

=1/n+(n-1)2
=1/n+2n-2
=1+2n^2-2n/n

Yes the 8th term can be deduced from the nth term...............................to verify,substitute n=8 in both the expressions of 8th and nth terms,both will come out to be equal.........

jatinroxx

I dont understand y ws this question even posted...

Its such a walk in the park...

abhi_go_iit

given seq. is an AP with common difference(d) =2..

first term (a)=1/n....

T8= 1/n+7d i.e.(1/n+7*2)= 14n+1/n

for Nth term TN=1/n+(N-1)d i.e 1/n+(N-1)2= 2Nn-2n+1/d

n clearly T8 can deduced from TN by putting N=8.....

rate me if u like..................

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