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.

What is the total number of chimes the clock will strike?

A chime clock strikes 1 chime at one o' clock, 2 chimes at two o' clock, 3 chimes at three o' clock, and so forth. What is the total number of chimes the clock will strike in twelve-hour period?

MetroEye: Progression02

The total number of chimes = 1+2+3+.......+12
here n = 12

Sum = \(\dfrac{12 * (12+1)}{2}\) = 6 * 13 = 78

The total number of chimes the clock will strike in twelve-hour period = 78

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

Sum of the counting numbers

Find the sum of the counting numbers  from 1 to 25 inclusive. In other words, if S=1+2+3+4....+24+25, find the value of S.

MetroEye: The sum of first n integers/counting numbers is S= n(n+1)/2

Here n = 25
So Sum S= 25(25+1)/2 = 325

Sum of the counting numbers from 1 to 25 is 325.