## Quant Question Of The Day : 15

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# Numbers

Amit has the list of first 50 even numbers i.e { 2 , 4, 6 , ….., 100} and Sumit has the list of first 50 odd numbers i.e {1, 3, 5, …….., 99} . Gautam creates a list by adding the square of each number in Amit’s list to the square of the corresponding number in Sumit’s list. Varsha creates a list by taking twice the product of corresponding numbers in Amit’s list and Sumit’s list. If the positive difference between the sum of the numbers in Gautam’s list and the sum of the numbers in Varsha’s list is K, Find the value of K^{2}.

### Solve our previous ‘questions of the day’

1.Quant Question Of The Day : 14

2. Quant Question Of The Day : 13

3. Quant Question Of The Day : 12

4. Quant Question Of The Day : 11

2500 ?

2500 ?

2500.

1²+2²+3²+………….+100²

Which is

1+4+25+………

Subtract

4+24+60

Which gives 1 (50times)

i.e k

Hence k=50

K²= 2500

Well done Shrey 🙂

2500

Though I am able to solve this question but I have a doubt as the series 2{(1*2)+(3*4)+(5*6)+……(99*100)}

How to find the sum for such series. Here the difference in both the series is coming out 1 for each term but in a different set of question how to find the sum of such series.

Hello Abhishek ,

Nth term of the series 1 × 2 + 3 × 4 + 5 × 6 +……+99 × 100 : tn = ( 2n -1) ( 2n )

Sigma 4n² – 2n from 1 to 50 =

4[n • (n +1)•(2n +1) ]/6 – 2 [ n ( n +1)/2 ]

4 [ 50 × 51 × 101 ]/6 – 2 (50 × 51/2 = 169150 🙂

Got it. Thanks a lot Raman sir.

K=50.

So, k²= 2500.

Perfect !

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2500