Arithmetic Progression formula proof

[Kartik.]

The proof says that -
Let,
S_n= a+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)----->1
S_n= (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?

 

[DonAntonio]

The proof says that -
Let,
S_n= a+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)----->1
S_n= (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) == $S_n=(a)+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)$

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

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

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

DonAntonio

 

[HallsofIvy]

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.

 

[Kartik.]

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

 

[Kartik.]

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) == $S_n=(a)+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)$

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

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

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

DonAntonio

Thanks ! :D