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If S n denotes the sum to n terms of the series 1 = n = 9 1+22+333+......+999+...... , then for n = >2 (9 times) a). S n -S n-1 =1\9(10 n -n 2 +n) b).S c). 9(S n -S n-1 )=n(10 n -1) d).S 3 = 356

If Sn denotes the sum to n terms of the series 1<=n<=9


1+22+333+......+999+...... , then for n=>2


                    (9 times)


 


a). Sn-Sn-1=1\9(10n-n2+n)                           b).S


c). 9(Sn-Sn-1)=n(10n-1)                               d).S3= 356

Grade:11

1 Answers

AJIT AskiitiansExpert-IITD
68 Points
10 years ago

Dear swapnil ,

there is a method of solving objective questions called solving by elimination. although we can try about solving for Sn -Sn-1 but in case of objective exams where speed matters it is better to use the former one.

so here , Sn -Sn-1 = an  , i.e. nth term.

a) apply n=2 , we get a2 not equal to 22 so the option is eliminated.

b) not clearly mentioned by you.

c)apply two a2 is 22 but it you shud check also for n =9 where it gives wrong answer so option is eliminated.

d) manually see it is right .

so option d is correct.

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.


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