Arithmetic Progression formula proof

In summary, the conversation discusses the proof for deriving the formula for the sum of an arithmetic series. It involves using two different ways of summing the series and then adding them together to obtain the formula n(2a+(n-1)d)/2. The conversation also includes a demonstration of the proof using the two different ways of summing the series.
  • #1
Kartik.
55
1
The proof says that -
Let,
Sn= a+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)----->1
Sn= (a+(n-2)d)+(a+(n-1)d)+...+a+(a+d)+(a+2d)------>2

Now if we have to add such things(1 and 2) how would we do that?
 
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  • #2
Kartik. said:
The proof says that -
Let,
Sn= a+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)----->1
Sn= (a+(n-2)d)+(a+(n-1)d)+...+a+(a+d)+(a+2d)------>2

Now if we have to add such things(1 and 2) how would we do that?


Since there doesn't seem to be any explicit order in line (2) it is hard to know what you want...I'd rather do the following:

(1) == [itex]S_n=(a)+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)[/itex]

(2) == [itex]S_n=(a+(n-1)d)+(a+(n-2)d)+...+(a+d)+(a)[/itex] ...These two are the same sum in inverse order.

Now sum both lines (1)+(2) in the same order (from left to right):

[itex]2S_n=(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d) = n(2a+(n-1)d)[/itex] , and from here you obtain the well known formula.

DonAntonio
 
  • #3
I am not sure what you are trying to do. It looks like one method of deriving the formula for the sum of an arithmetic series but, if so, you have the second sum wrong. You should be just "swapping ends" of the first series. You should have
a+ (a+ d)+ (a+ 2d)+ ...+ (a+ (n-2)d)+ (a+ (n-1)d and
(a+ (n-1)d+ (a+ (n-2)d)+ ...+ (a+ d)+ a

Now add "vertically". The two first numbers in each sum are a and a+ (n-1)d. They add to 2a+ (n-1)d, of course. The next term in each sum is a+d and a+ (n-2)d. They add to 2a+ (1+ n- 2)d= 2a+ (n- 1)d again. The next to last terms are a+ (n- 2)d and a+ d. The add to 2a+ (n-2+ 1)d= 2a+ (n-1)d once more. Finally, the last terms are a+ (n-1)d and a which add to, yet again, 2a+ (n-1)d.

Basically, as move to the right in the first sum, we are always adding "d". Since the second sum is just the first sum reverse, we add "d" as we move right which means that moving left we are subtracting "d". Because, adding vertically, we will have added d to one term and subtracted d from the other, the sums are always the same, 2a+ (n- 1)d. There are n terms so the sum of the two is n(2a+ (n-1)d. Since the two are the same, the sum of either one is half that" n(2a+ (n-1)d)/2.
 
  • #4
HallsofIvy said:
I am not sure what you are trying to do. It looks like one method of deriving the formula for the sum of an arithmetic series but, if so, you have the second sum wrong. You should be just "swapping ends" of the first series. You should have
a+ (a+ d)+ (a+ 2d)+ ...+ (a+ (n-2)d)+ (a+ (n-1)d and
(a+ (n-1)d+ (a+ (n-2)d)+ ...+ (a+ d)+ a

Now add "vertically". The two first numbers in each sum are a and a+ (n-1)d. They add to 2a+ (n-1)d, of course. The next term in each sum is a+d and a+ (n-2)d. They add to 2a+ (1+ n- 2)d= 2a+ (n- 1)d again. The next to last terms are a+ (n- 2)d and a+ d. The add to 2a+ (n-2+ 1)d= 2a+ (n-1)d once more. Finally, the last terms are a+ (n-1)d and a which add to, yet again, 2a+ (n-1)d.

Basically, as move to the right in the first sum, we are always adding "d". Since the second sum is just the first sum reverse, we add "d" as we move right which means that moving left we are subtracting "d". Because, adding vertically, we will have added d to one term and subtracted d from the other, the sums are always the same, 2a+ (n- 1)d. There are n terms so the sum of the two is n(2a+ (n-1)d. Since the two are the same, the sum of either one is half that" n(2a+ (n-1)d)/2.


Thanks!..I was trying to do the thing you just demonstrated
 
  • #5
DonAntonio said:
Since there doesn't seem to be any explicit order in line (2) it is hard to know what you want...I'd rather do the following:

(1) == [itex]S_n=(a)+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)[/itex]

(2) == [itex]S_n=(a+(n-1)d)+(a+(n-2)d)+...+(a+d)+(a)[/itex] ...These two are the same sum in inverse order.

Now sum both lines (1)+(2) in the same order (from left to right):

[itex]2S_n=(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d) = n(2a+(n-1)d)[/itex] , and from here you obtain the well known formula.

DonAntonio

Thanks ! :D
 

1. What is the formula for finding the nth term in an arithmetic progression?

The formula for finding the nth term in an arithmetic progression is a + (n-1)d, where a is the first term in the progression and d is the common difference between consecutive terms.

2. How do you prove the formula for finding the nth term in an arithmetic progression?

The formula for finding the nth term in an arithmetic progression can be proven using mathematical induction. This involves showing that the formula holds true for the first term, and then assuming it holds true for the (n-1)th term and using this to prove it for the nth term.

3. Can the formula for arithmetic progression be used to find the sum of a series?

Yes, the formula for arithmetic progression can be used to find the sum of a series. The formula for the sum of an arithmetic series is n/2*(a+l), where n is the number of terms, a is the first term, and l is the last term.

4. How do you use the formula for arithmetic progression to solve real-life problems?

The formula for arithmetic progression can be used to solve real-life problems involving linear relationships, such as calculating the growth rate of investments or predicting future values based on past data. It can also be used in various fields of science, such as physics and chemistry, to model and predict patterns and trends.

5. Is the formula for arithmetic progression applicable to all types of progressions?

No, the formula for arithmetic progression is only applicable to arithmetic progressions, which are sequences where each term is obtained by adding a constant value to the previous term. It cannot be used for other types of progressions, such as geometric or harmonic progressions.

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