Fwd: Sir kindly solve this question

[Varun Gupta]

Ques 29

S1= { 1 }

S2= {2,3}

S3= {4,5,6}

S1 contains 1 nos.  

S2 contains 2 nos.

S3 contains 3 nos.

S40 will contain 40 elements and the Last element of S40 will be (40*41)/2 ...i.e ans is 820.

Ques.30

S50 last element will be 50*51 /2 =  1275

S50 contains 50 elements so the first element will be 1226

Diff. 1275-1226 = 49 

---------- Forwarded message ----------
From: Ankush Asthana <ankush...@gmail.com>
Date: Mon, Jan 16, 2012 at 12:56 AM
Subject: Sir kindly solve this question
To: "gupta....@gmail.com" <gupta....@gmail.com>





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[Varun Gupta]

---------- Forwarded message ----------
From: Varun Gupta <gupta....@gmail.com>
Date: Tue, Jan 17, 2012 at 10:11 PM
Subject: Re: Sir kindly solve this question
To: Ankush Asthana <ankush...@gmail.com>


To count sum of n nos. the formula is n*n+1 / 2.

Try to observation that the Last element of every set also gives the count of the nos. before it and itself also .

Now elements in Set 1 -  1

Set 2 - 2 elements , Set 3 - 3 elements ...................Set 40 - 40 elements .

Last element of set 40 will be equal to total no. of elements in First 40 sets ..

B/R , Varun

On Tue, Jan 17, 2012 at 1:30 AM, Ankush Asthana <ankush...@gmail.com> wrote:

Thank u sir but I didn't understand why the last no. Will be 40*41/2

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