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If sum of n termsof a sequende is Sn the...

If sum of n termsof a sequende is `S_n` then its nth term `t_n=S_n-S_(n-1)`. This relation is vale for all `ngt-1` provided `S_0= 0.` But if `S_!=0`, then the relation is valid ony for `nge2` and in hat cast `t_1` can be obtained by the relation`t_1=S_1.` Also if nth term of a sequence `t_1=S_n-S_(n-1)` then sum of n term of the sequence can be obtained by putting `n=1,2,3,.n` and adding them. Thus `sum_(n=1)^n t_n=S_n-S_0.` if `S_0=0, then sum_(n=1)^n t_n=S_n.` On the basis of above information answer thefollowing questions: If the sum of n terms of a sequence is `10n^2+7n` then the sequence is (A) an A.P. having common difference 20 (B) an A.P. having common difference 7 (C) an A.P. having common difference 27 (D) not an A.P.

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If sum of n terms of a sequence is S_n then its nth term t_n=S_n-S_(n-1) . This relation is valid for all ngt-1 provided S_0= 0. But if S_1=0 , then the relation is valid only for nge2 and in hat cast t_1 can be obtained by the relation t_1=S_1. Also if nth term of a sequence t_1=S_n-S_(n-1) then sum of n term of the sequence can be obtained by putting n=1,2,3,.n and adding them. Thus sum_(n=1)^n t_n=S_n-S_0. if S_0=0, then sum_(n=1)^n t_n=S_n. On the basis of above information answer the following questions: If the sum of n terms of a sequence is 10n^2+7n then the sequence is (A) an A.P. having common difference 20 (B) an A.P. having common difference 7 (C) an A.P. having common difference 27 (D) not an A.P.

If sum of n termsof a sequende is S_n then its nth term t_n=S_n-S_(n-1) . This relation is vale for all ngt-1 provided S_0= 0. But if S_!=0 , then the relation is valid ony for nge2 and in hat cast t_1 can be obtained by the relation t_1=S_1. Also if nth term of a sequence t_1=S_n-S_(n-1) then sum of n term of the sequence can be obtained by putting n=1,2,3,.n and adding them. Thus sum_(n=1)^n t_n=S_n-S_0. if S_0=0, then sum_(n=1)^n t_n=S_n. On the basis of above information answer thefollowing questions:If nth term of a sequence is n/(1+n^2+n^4) then the sum of its first n terms is (A) (n^2+n)/(1+n+n^2) (B) (n^2-n)/(1+n+n^2) (C) (n^2+n)/(1-n+n^2) (D) (n^2+n)/(2(1+n+n^2)

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