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by rajeshvellanki » Mon Apr 23, 2007 11:43 am
okay here you go:

The only number which satisfies the given condition is : 101

and a(n)=a(1)+(n-1)d

199=100+(n-1)d

==> n=100 terms

Probability of selecting 1 number out of 100 numbers is 1/100 hence D

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by Cybermusings » Wed Apr 25, 2007 6:06 am
Since we have to pick a number between 100 and 199 both inclusive...it is imperative that the hundreds digit is always 1....for the hundreds and last digit always to be 1 more than the middle digit...the last digit has to be 1 and the middle digit 0

There is only 1 such number i.e. 101

So probability = 1/100
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