Which is greater, D or N, and by how much?

D is the sum of the odd numbers from 1 through 99 inclusive, and N is the sum of the even numbers from 2 through 98 inclusive:
           D = 1+3+5+ ........+99 and N = 2+4+6+ ...... +98
Which is greater, D or N, and by how much?

MetroEye: Refer Progression03 for formulae.

Number of terms in odd number series = \(\dfrac{t_n+1}{2}\)= \(\dfrac{99+1}{2}\) = \(\dfrac{100}{2}\)  = 50

Number of terms in even number series = \(\dfrac{t_n}{2} = \dfrac{98}{2}\)  = 49

Sum, D = \( n^2\) = 50*50

Sum, N = n(n+1) = 49*50

=> D is greater than N.

Difference between D and N = D-N = 50*50-49*59 = 50(50-49) = 50

Hence D is greater than N by 50.

After how many stops will the train be full?

A train can hold 78 passengers. The train starts out empty and picks up 1 passenger at first stop, 2 passengers at the second stop, 3 passengers at the third stop, and so forth. After how many stops will the train be full?

MetroEye: Refer to Preogression02

The sequence in which passengers are picked up is 1,2,3.....
So when the sum of consecutive numbers reach 78, the train will be full.

i.e; 1+2+3+......+N = 78

Here S = 78

N = \(\dfrac{\sqrt{1+8S} -1}{2}\) = \(\dfrac{\sqrt{1+8*78} -1}{2}\) = \(\dfrac{\sqrt{1+624} -1}{2}\) = \(\dfrac{\sqrt{625} -1}{2}\) = \(\dfrac{25-1}{2}\) = 12

=> 1+2+3+4+5+6+7+8+9+10+11+12 = 78

So the train will be full after 12 stops.

What is the 100th number in the sequence?

If we count by 3s starting with 1, the following sequence is obtained: 1,4,7,10, ... What is the 100th number in the sequence?

MetroEye: t_n=a+(n-1)d
where t_n = n^th term of the sequence
a = first term of the sequence
n = number of terms in the sequence
d = common difference of the sequence

In the given sequence, a=1,n=100,d=3 
=> t₁₀₀=1+(100-1)3=298
So the 100^th term of the sequence is 298