In a sequence the sum of the nth terms is #3n^2+5n#.If its mth term is 164, then what is the value of m?

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Answer 1 · 2017-11-15T16:29:39

# m=27.#

Let #S_n and T_n# denote the sum of the first #n# terms and the

#n^(th)# term of the sequence in question, respectively.

Then, #S_n=T_1+T_2+...+T_(n-1)+T_n.#

#:. S_n={T_1+T_2+...+T_(n-1)}+T_n.#

But, #T_1+T_2+...+T_(n-1)=S_(n-1).#

#rArr S_n=S_(n-1)+T_n............(ast).#

Since, #S_n=3n^2+5n, :. S_(n-1)=3(n-1)^2+5(n-1).#

Therefore, by #(ast),# we have,

#T_n=S_n-S_(n-1),#

#=3n^2+5n-{3(n-1)^2+5(n-1)},#

#=3n^2+5n-(3n^2-6n+3+5n-5),#

# :. T_n=6n+2.#

Now, given that #T_m=164 rArr 6m+2=164.#

#:. m=(164-2)/6=27.#