7

Arrange the number like this
1 2 3 4 5 6 7 8 9 10
1 3 6 10 15 21 28 36 45 55 (sum in ascending order)
55 54 52 49 45 40 34 27 19 10 (sum in descending order)

It clearly shows the sum till 7(28) is greater than the sum in reverse till 8(27) by 1.

Ans D.

I think backsolving this kind of question is better way to go.

Start at 6 and see which side you need to go..

If you use the formula for sum of consecutive digits: n/2*(F+L), where F and L are the first and last numbers in the sequence and n is the total number of numbers in the set, you have the following 2 equations:

A) X/2(1+X)
B) 1+(10-X/2)*(X+10)

Then you can plug in starting at C. In doing so, when I got to answer D, the two equations came to 28 (for A) and 27 (for B). I chose D b/c the sums were closer than choices C and E.

dpatwa wrote:If you use the formula for sum of consecutive digits: n/2*(F+L), where F and L are the first and last numbers in the sequence and n is the total number of numbers in the set, you have the following 2 equations:

A) X/2(1+X)
B) 1+(10-X/2)*(X+10)

Then you can plug in starting at C. In doing so, when I got to answer D, the two equations came to 28 (for A) and 27 (for B). I chose D b/c the sums were closer than choices C and E.

I know 'A', but what formula is 'B'?