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Title: progressions
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Posted On: Sep 07, 2010 01:52 PM
 

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

   
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swapnil sudhir
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Sep 11, 2010 01:39 PM

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.

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