Arithmetical progressions-2

In my previous blog,I had introduced the subject.
I referred to a progression with the first term 'a' =32 and common difference 'd'=5.
Here is how one can find subsequent terms-
For the second term you have to add the value 'd=5' once to 'a=32' giving you 32+5=37.
For the third term you have to add twice the same value giving you 32+10 =42.
For the fourth term you have to add thrice giving you 32+15=47 and so on.
Hence for the 16th term you have to add the value 15 times to give you 32+15*5=32+75=107
For the 45th term you have to similarly add the value 44 times to give you 32+44*5=32+220=252.
Now for finding the sums-
If the total no of terms is odd ,there will always be a middle term ,double the value of which will equal the total  of the value of the first term and the value of the last term.The total of the value of the second term and the penultimate term will be the same amount..This process continues throughout.
When there are 45 terms in all ,the middle term will be the 23 rd term viz 32+22*5=32+110=142.Double this value is 284.
The total value of the first  and 45th terms is  (32+252)=284..Similarly the total of the values of  the second term (37) and 44th term (247) is (37+247)=284..Similarly you get the same result considering the third term(42) and 43rd term (242) viz (42+242)=284 and so on..
The total of all the 45 terms will hence be 45 times the value of the middle term viz 45*142=6380.
You can check the correctness of this method by first  considering 3 terms then 5 terms,then 7 terms and so on
When the total no of terms is even there will be two middle terms and hence you have to take the mean value of those two terms.
When there are 16 terms,the 8th term(67) and the 9th term(72) will be the middle terms and hence you have to take their mean and multiply by 16,to get the total of all the 16 terms.You will have 16*(67+72)/2=8*139=1112 .

Arithmetical Progressions

I wish to familiarise the topic Arithmetical Progressions
Any sequence of numbers starting with a particular number and progressively increasing with equal additions can be stated to be forming such progressions.
The start number can be called the first term-generally denoted by 'a'
The additions in equal values can be called the 'common difference-generally denoted by 'd'
'a' can be anything like 5,    100      ,3/7    ,4.05,    6^2   etc
'd' can also be similarly anything-can also have negative values.
A few examples shown below-
100,103,106,109,112......
(4.5)   .(5.2)    (5.9)    (6.6)....
7/45......9/45    11/45...13/45....15/45    17/45
1000,992,984,976,968...  etc
Given the first term  like 32 and common difference like 5,we have methods to find out the 16th or 45th etc terms.Also the sums of the various terms upto the 16th term or 45th term.
I will explain the methods in my next blog.

Remainders

I am posing a question for you to solve.
I want you to find the smallest number which gives a remainder 2 when divided by 3,a remainder 3 when divided by 5 and a remainder 5 when divided by 7., 
How can you find more such numbers?

Clock face-Times..Rejoinder.

No one has yet given me a reply.The solution is based on the following facts-
After 12 noon,the hands first merge a little after 01-05
Next a little after 02-10...Next a little after 03-15 etcetc.
In fact,they will be together again several times and for the 11th time they will be at 12 midnight.
Thus in 12 hours,11 intervals occur
Each interval will be {1 +(1/11)}hours or 1 hour and {5+(5/11)} minutes.
The times of merger are thus {5+(5/11)} minutes after 1 PM...{10+(10/11)}minutes after 2PM,{16+(4/11}minutes after 3PM..,{21+(9/11)}minutes after 4PM  etc etc .

Forming fractions

Suppose we have 2 fractions like 2/7 and 4/9.
We may like to find more fractions with values between these two.There is a method for this.In general if a/b and c/d are 2 fractions,then the fraction (a+c)/(b+d) will have values between those two.You can thus find as many fractions as you want between 2/7 and 4/9.
I have shown some of them below.
Let A=2/7 and B=4/9.The first fraction in between C=(2+4)/(7+9)=6/16=3/8
Next D between 2/7 and 3/8=5/15=1/3.Next E between 3/8 and 4/9=7/17and so on.The following is a more detailed list-
2/7,5/17,3/10,4/13,1/3,5/14,4/11,7/19,3/8,5/13,2/5,9/22,7/17,18/43,11/26,3/7,4/9.You can expand the list to whatever length you wish.

Clock face-times when the hour hand and minute hand merge

This is a problem for you.
Both the hands on a clock face merge at 12noon or midnight.
After 12noon,they again merge a little later than  01-05 and again a little later than 02-10.
Can you find the actual times in each case?
Also subsequent stages.

Number 153

I am now introducing a new black hole number 153.
Take any number divisible by 3 like 423.
Find the cubes of digits in 423 and add.You get 99.
Repeat the same process with 99.You get 1458.
Repeat the processes again.
1458 gives you 702...702 gives you 351...351 gives you 153..and finally 153 gives you again 153.
The same result will be arrived at with any number divisible by 3
Second example   63.The repeat processes give you 243,99 and again as in previous case 153.
Third example 168.The results are 729,378,882,1032,36 and again 153 like for 63
You can consider even larger numbers divisible by 3 also.More repetitions may be required.