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Bunuel
Sequence A = {1, 5, 9, 13, ...}
Sequence B = {56, 58, 60, 62, ...}

What is the smallest positive integer n, such that the sum of 2n terms of sequence A is greater than the sum of n terms of sequence B?

A. 8
B. 9
C. 10
D. 12
E. 14


Are You Up For the Challenge: 700 Level Questions

For sequence 1, a = 1, d =4.
For seq 2, a =56, d =2
We need to find, a + (2n -1) 4 > a +(n-1)2
or, 1 + 8n - 4 > 56 + 2n -2
or, 6n -3 > 54
or, 6n > 57
or, n > 9. something
so, the smallest possible integer n can take is 10.
C is the answer
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