Hey Everyone - I have been lurking around the forum and wanted to share my notes might be of some help
Part 1: Tacking those questions which ask you to find the consecutive integers in a set / consecutive multiples in a set
Consecutive IntegersNo. of consecutive integers between x and y, including only ONE of them numbers = y - x
No. of consecutive integers between x and y, inclusive of BOTH the numbers = y - x + 1
No. of consecutive integers between x and y, EXCLUDING BOTH the numbers = y - x -1
Example:
1. Number of consecutive integers between 100 and 900, including only ONE of the numbers = 900 - 100 = 800
2. Number of consecutive integers between 100 and 900, inclusive of BOTH the numbers = 900 - 100 + 1 = 801
3. Number of consecutive integers between 100 and 900, Excluding both the numbers = 900 - 100 -1 = 799
Consecutive MultiplesSteps to find the consecutive multiples of any given number between x and y
Step 1: Find the highest divisible number between x and y
Step 2: Find the lowest divisible number between x and y
Use the below formula
\(\frac{(Highest Divisible number - Lowest divisible number) }{ Given number} + 1 \)
Example
Find the consecutive multiples between 6 and 77 that are divisible by 5
Step 1: Highest number divisible by 5 within the range is 75
Step 2: Lowest number divisible by 5 within the range is 10
Applying the formula we get \(\frac{(75 - 10)}{5} + 1\)
Simplifying we get it as 14
Part 2: Finding the Sum / Average of the multiples
We know that
Average = \(\frac{Sum of the terms}{#s terms}\)
Finding sum of evenly spaced setsEvenly space sets are ones in which the gap between each successive element in the set is equal or in other terms they are in an AP
For EVENLY SPACED sets Average can be defined as
\(Average = \frac{(Last term + First term)}{2}\)
Now using the #s of terms formula from part 1 and the above formula we can compute the sum of the terms
Example:
Find the sum of all multiples of 3 between 1 and 100?
Part 1: Find the #s multiples = \(\frac{(99-3)}{3} + 1 \) = 33
Note that the set would be 3,6,9....99 (This is evenly spaced with a common difference of 3)
Part 2: Find the average of the set = \(\frac{(99+3)}{2}\) = 51
Now the sum of all terms in the set = Average * #s of terms = 51*33 = 1683
Hope it helps