GO Classes 2023 | IIITH Mock Test 1 | Question: 61

Answer 1

Sum if “n” natural number Ʃn = n(n+1)/2

                                                                                          Number Of Elements    Sum of group   

The number in 1st group =(1)                                              1                                1

The number in 2nd group =(2,3,4)                                       3                           Ʃ4 -Ʃ1 = 9

The number in 3rd group =(5,6,7,8,9)                                  5                          Ʃ9 -Ʃ4 = 35

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The number in 14th group =(…………,196)                           27                      

The number in 15th group =(197 , … … … …, 225)              29                       Ʃ225 -Ʃ196 = 6119

The number in nth group =(1+ (n-1)^2,  …  …  … , n^2)    (2n-1)                   Ʃn^2 -Ʃ( n-1)^2 

 

Answer -c      6119

Comments

how do i know that 14th group is ending at 196 and 15th is starting at 197?

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put n=15 in the nth group formula you get 197, 1 less than that is  last number of the previous group as they all r consecutive numbers.

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last element of the list is $n^2$

Answer 2

nth group starts with $(n-1)$$^{2}$ $+$ $1$ and ends with $n$$^{2}$ and every group consists $2n-1$ numbers
$15th$ group starts with 196 and ends with $225$

$sum$ = $\frac{No of terms}{2}$ * $[first$ $term$+$last$ $term]$ = $\frac{29}{2} * [(196+1) + 225]$ = 6119