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Sum of first 4 columns would cancel each other.Left would be last two columns.
Sum of series formula: S=n/2[2a+(n-1)d]
where n=no of terms
a= first term
d=constant difference between each term
So on substituting value for (3,6,9,12,15,18)
S=63
and for (4,8,12,16,20,24)
S= 84
total sum =63+84 =147
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Bunuel
What is the sum of the numbers in the grid below?


A. 146
B. 147
C. 148
D. 149
E. 150


Attachment:
2020-03-17_1502.png

for the given sequence we see that for each row we get multiple of 7 right from 7 to 42
so sum of all multiples of 7 ; from 7 to 42 is ;
42/7 * ( 7+42)/2
=> 147
IMO B
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Bunuel
What is the sum of the numbers in the grid below?


A. 146
B. 147
C. 148
D. 149
E. 150


Attachment:
2020-03-17_1502.png

Solution:

We see that the first four columns sum to a total of zero. The sum of the last two columns is:

6 x (18 + 3) / 2 + 6 x (24 + 4) / 2 = 3 x 21 + 3 x 28 = 63 = 84 = 147

Alternate Solution:

We see that if we add each row, the sum of the first four numbers is 0. Thus, we only need to add the last two numbers of each row. Therefore, the rows (counting from the top) sum to 7, 14, 21, 28, 35 and 42, respectively. These are the first 6 positive multiples of 7 and the sum of these numbers is:

6 x (7 + 42)/2 = 3 x 49 = 147

Answer: B
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