[নোটঃ এই আর্টিকেলটি (3) Number Properties বিভাগের অধীনে 〈4〉Consecutive Number চ্যাপ্টারের অন্তর্গত, যা 〈3.4.a〉চিহ্ন দিয়ে প্রকাশ করা হয়েছে]
〈3.4.a〉ধারাবাহিক সংখ্যা, তাদের গড় ও মধ্যমা নির্ণয় এবং অন্যান্য
• Consecutive multiplies কেবলমাত্র integer দ্বারা গঠিত হয়।
• The arithmetic mean (average) and median and equal to the element in the set can be found by figuring out the median on middle number.
• The mean and median to the set are equal to the average of the first and last terms.
• The sum of the elements in the set equals the arithmetic mean (average) number in the set times the number of item in the set.
• Number of items = (Last-First) +increment+[latex]1[latex].
• The average of an ODD number of consecutive integers will always be an integer ([latex]2[latex])
• The average of an ever number of consecutive integers will never be an integer ([latex]2.5[latex])
• [latex]n[latex] সংখ্যক Consecutive integer এর গুনফল অবশ্যই [latex]n[latex] দ্বারা বিভাজ্য হবে।
Consecutive বা ধারাবাহিক Integers
Consecutive integers are integers that follow one after another from a given starting point, without skipping any integers.
For example, [latex]4, 5, 6[latex], and [latex]7[latex] are consecutive integers, but [latex]4, 6, 7[latex], and [latex]9[latex] ate not. There are many other types of consecutive patterns. For example:
Consecutive Even Integers: [latex]8, 10, 12, 14[latex] ([latex]8, 10, 14[latex], and 16 is incorrect, as it skips [latex]12[latex])
Consecutive Primes: [latex]11 13, 17, 19[latex] ([latex]11, 13, 15[latex], and [latex]17[latex] is wrong, as [latex]15[latex] is not prime)
Evenly Spaced Sets
These are sequences of numbers whose values go up or down by the same amount (the increment) from one item in the sequence to the next.
For instance, the set {[latex]4, 7, 10, 13, 16[latex]} is evenly spaced because each value increases by [latex]3[latex] over the previous value.
Sets of consecutive multiples are special cases of evenly spaced sets: all of the values in the set are multiples of the increment.
For example, {[latex]12, 16, 20, 24[latex]} is a set of consecutive multiples because the values increase from one to the next by [latex]4[latex], and each element is a multiple of [latex]4[latex].
মনে রাখতে হবে consecutive multiples কেবলমাত্র integers দ্বারাই গঠিত হতে পারে।
Sets of consecutive integers are special cases of consecutive multiples: all of the values in the set increase by [latex]1[latex], and all integers are multiples of [latex]1[latex]. For example, {[latex]l2, 13, 14, 15, 16[latex]} is a set of consecutive integers because the values increase from one to the next by [latex]1[latex], and each element is an integer.
Properties of Evenly Spaced Sets
The following properties apply to all evenly spaced sets.
(1)The arithmetic mean (average) and median are equal to each other. In other words, the average of the elements in the set can be found by figuring out the median, or “middle number.”
What is the arithmetic mean of [latex]4, 8, 12, 16[latex], and [latex]20[latex]?
In this example we have [latex]5[latex] consecutive multiples of four. The median is the 3rd largest, or [latex]12[latex]. Since this is an evenly spaced set, the arithmetic mean (average) is also [latex]12[latex].
What is the arithmetic mean of [latex]4, 8, 12, 16, 20[latex], and [latex]24[latex]?
In this example we have [latex]6[latex] consecutive multiples of four. The median is the arithmetic mean (average) of the 3rd largest and 4th largest, or the average of [latex]12[latex] and [latex]16[latex]. Thus the median is [latex]14[latex]. Since this is an evenly spaced set, the average is also [latex]14[latex].
(2) The mean and median of the set are equal to the average of the FIRST and LAST terms.
What is the arithmetic mean of [latex]4, 8, 12, 16[latex], and [latex]20[latex]?
In this example, [latex]20[latex] is the largest (last) number and [latex]4[latex] is the smallest (first). The arithmetic mean and median are therefore equal to ([latex]20+4[latex])±[latex]2=12[latex].
What is the arithmetic mean of [latex]4, 8, 12, 16, 20[latex], and [latex]24[latex]?
In this example, [latex]24[latex] is the largest (last) number and [latex]4[latex] is the smallest (first). The arithmetic mean and median are therefore equal to ([latex]20+4[latex])±[latex]2=14[latex].
Thus for all evenly spaced sets, just remember: the average equals (First+Last) ±[latex]2[latex].
(3) The sum of the elements in the set equals the arithmetic mean (average) number in the set times the number of items in the set.
This property applies to all sets, but it takes on special significance in the case of evenly spaced sets because the “average” is not only the arithmetic mean, but also the median.
What is the sum of [latex]4, 8, 12, 16[latex], and [latex]20[latex]?
Counting Integers- Add One Before You Are Done
এ ধরণের গণনা সংক্রান্ত ভুল প্রায়শ:ই হয়ে থাকেঃ [latex]0[latex] থেকে [latex]10[latex] পর্যন্ত কতট integer আছে? দশটি? ভুল। আসলে আছে ১১ টি।
It is easy to forget that you have to include extremes. In this case, both extremes (the numbers [latex]0[latex] and [latex]10[latex]) must be counted.
How many integers are there from [latex]14[latex] to [latex]765[latex], inclusive?
[latex]765-14[latex], plus [latex]1[latex], yields [latex]752[latex].
How many multiples of [latex]7[latex] are there between [latex]100[latex] and [latex]150[latex]?
Here it may be easiest to list the multiples: [latex]105, 112, 119, 126, 133, 140, 147[latex]. Count the number of terms to get the answer: [latex]7[latex]. Alternatively, we could note that [latex]105[latex] is the first number, [latex]147[latex] is the last number, and [latex]7[latex] is the increment:
Number of terms = (Last – First) + Increment + [latex]1=(147-105)+7+1=6+1=7[latex].
The Sum of Consecutive Integers
Consider this problem:
What is the sum of all the integers from [latex]20[latex] to [latex]100[latex], inclusive?
Using the rules for evenly spaced sets mentioned before, we can use shortcuts:
(1) Average the first and last term to find the precise “middle” of the set: [latex]100+20=120[latex] and [latex]120[latex] divide by 2= 60.
(2) Count the number of terms: [latex]100-20=80[latex], plus [latex]1[latex] yields [latex]81[latex].
(3) Multiply the “middle” number by the number of terms to find the sum: [latex]60*81 4,860[latex].
There are a couple of general facts to note about sums and averages of evenly spaced sets (especially sets of consecutive integers):
• The average of an odd number of consecutive integers ([latex]1, 2, 3, 4, 5[latex]) will always be an integer ([latex]3[latex]). This is because the “middle number” will be a single integer.
• On the other hand, the average of an even number of consecutive integers ([latex]1, 2, 3, 4[latex]) will never be an integer ([latex]2.5[latex]), because there is no true “middle number.”
• This is because consecutive integers alternate between EVEN and ODD numbers. Therefore, the “middle number” for an even number of consecutive integers is the AVERAGE of two consecutive integers, which is never an integer.
Products of Consecutive Integers and Divisibility
মনে রাখবেনঃ
[latex]n[latex] সংখ্যক Consecutive Integers এর গুণফল অবশ্যই [latex]n[latex] দ্বারা বিভাজ্য হবে।
একই ভাবে, আপনি যে কোন তিনটি Consecutive Integers নিন এবং তাদের গুণফল বের করলে দেখতে পাবেন যে সংখ্যাটি ৩ দিয়ে বিভাজ্য। আবার ১৭ টি Consecutive Integers এর গুণফল ১৭ দিয়ে বিভাজ্য? এর কারণ কি? কারণ হলো, যে কোন তিনটি সংখ্যা নিলে তার মধ্যে অন্তত একটি পড়বে ৩ এর multiple। একই কথা সত্য ১৭ এবং n এর জন্যে।
বিষয়টা আরেকটু জটিল করা যাক। যে কোন তিনটি ধারাবাহিক সংখ্যার গুণফল কেন ৬ দিয়ে বিভাজ্য হবে? কারণ, তিনটি ধারাবাহিক সংখ্যা নিলে তার মধ্যে একটি জোড় সংখ্যা পড়বেই, যার একটি ফ্যাক্টর হলো ২। আবার ৩ এর মালটিপল ও পাওয়া যাবে। কাজেই, ওই গুণফলকে আলাদা ভাবে ৩ দিয়ে ও ২ দিয়ে ভাগ করা যাবে। সুতরাং, সংখ্যাটি ৬ দিয়ে বিভাজ্য।
Please remember this rule:
The product of [latex]k[latex] consecutive integers is always divisible by k factorial ([latex]k[latex]!).
You can use prime boxes to keep track of factors of consecutive integers. Consider the following problem:
If [latex]x[latex] is an even integer, is [latex]x(x+1)(x+2)[latex] divisible by [latex]4[latex]?
এখানে তিনটি ধারাবাহিক সংখ্যার গুণফল প্রকাশ করা হয়েছে, যার প্রথমট জোড় সংখ্যা। তাহলে তৃতীয়টাও জোড় হবে। দুইটি জোড় সংখ্যাকে আলাদা ভাবে দুইটি ২ দিয়ে ভাগ করা যাবে। সুতরাং, সম্পূর্ণ গুণফলটিকে অবশ্যই ৪ দিয়ে ভাগ করা যাবে।
