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 .
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 .
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