Hall and Knight book for Higher Algebra gave a good intepretation of the quadratic equation
$$dn^2 + (2a - d)n - 2s = 0$$ where $a$ is the first term , $d$ common difference and $s$ sum of all the terms, in an AP.
As per it if we have say $a, d$ and $s$ and try to find the value of $n$, we get two roots of the above quadratic equation where possibility is one is positive while other negative.
If say the roots are $l$ and $m$, and $m$ is negative then if we add backwards from the first $l$-th term of the AP till $|m|$ number of terms the sum is also equal to $s$.
They seems to have given a proof and I can understand it is in relation to the fact that $lm$ is the product of the roots of the equation but I still find the proof little difficult to understand
Can u please send a proof easy for me to understand
Last two steps where I drew arrows
The Internet Archive has the Fourth Edition (1891) of this (venerable old) text; I found it helpful to read the preceding section in order to find what "this interpretation" referred to. The problem under discussion is finding the number of terms in an arithmetic series when the sum, first term, and constant difference are given. The "average of first and last terms" sum formula leads us to $$ s \ \ = \ \ n· \left( \ \frac{a \ + \ [a + (n - 1)·d]}{2} \ \right) \ \ \Rightarrow \ \ dn^2 \ + \ (2a - d)·n \ - 2s \ \ = \ \ 0 \ \ , \ \ \ \mathbf{[ \ 1 \ ] } $$ for which there are two solutions for $ \ n \ \ . \ $ The authors are showing that the second solution, which contemporary authors would generally discard as "spurious", does have an interpretation if we count from the "final" to the "initial term" by taking the constant difference as $ \ (-d) \ \ . $ [I am recapitulating the set-up for this since it may help in understanding the author's manipulations in their proof.]
So saying, their "reversed summation" starts with the term $ \ a + (n_1 - 1)·d \ $ and thus ends with $ \ [ \ a + (n_1 - 1)·d \ ] \ + \ (n_2 - 1)·(-d) \ = \ a + (n_1 - n_2)·d \ \ . \ $ We'll call the sum of this arithmetic series $ \ \mathcal{S} \ \ , $ making no assumption about its possible relation to $ \ s \ \ . \ $ The sum formula used above here produces $$ \mathcal{S} \ \ = \ \ n_2 · \left( \ \frac{ [a + (n_1 - 1)·d] \ + \ a + (n_1 - n_2)·d }{2} \ \right) \ \ = \ \ n_2 · \left[ \ \frac{ 2a + (2n_1 - n_2 - 1)·d }{2} \ \right] $$ [which now gets us to the point you asked about] $$ = \ \ \frac12 · ( \ 2a·n_2 \ + \ 2·d·n_1·n_2 \ - \ d· n_2^2 \ - \ d·n_2 \ ) \ \ , $$ which can be re-arranged as $$ = \ \ \frac12 · ( \ 2·d·n_1·n_2 \ \ \mathbf{-} \ \ [ \ d· n_2^2 \ - \ 2a·n_2 + \ d·n_2 \ ] \ ) \ \ \ \mathbf{[ \ 2 \ ]} $$ [and we now spot that there are typoes in the penultimate line of their argument!]. (Who do we complain to about a 130-year-old textbook?)
Referring to the quadratic equation in $ \ \mathbf{[1]} \ $ above, we see that since $ \ -n_2 \ $ satisfies $ \ dn_2^2 \ - \ 2an_2 \ + \ dn_2 \ = \ 2s \ \ , \ $ the term in square brackets in $ \ \mathbf{[2]} \ $ is equal to $ \ 2s \ \ . \ $ The authors are also applying the Viete relation which states that the product of the roots of the quadtatic equation is equal to the ratio of the constant term to the leading coefficient; hence, $ \ n_1·(-n_2) \ = \ \frac{-2s}{d} \ \Rightarrow \ d·n_1·n_2 \ = \ 2s \ \ . \ $ Consequently, $$ \mathcal{S} \ \ = \ \ \ \ \frac12 · ( \ 2·2s \ \ \mathbf{-} \ \ [ \ 2s \ ] \ ) \ \ = \ \ s \ \ , $$ thereby verifying the authors' claim.
Another reason that the "negative difference" interpretation may not be discussed much can be seen in the following example. For the arithmetic series with initial term $ \ a \ = \ -3 \ , \ $ common difference $ \ d \ = \ 7 \ , \ $ and sum $ \ s \ = \ 2755 \ \ , \ $ our quadratic equation is $$ \ \ 7n^2 + (2·[-3] - 7)n - 2·2755 \ = \ 7n^2 - 13n - 5510 \ = \ (n - 29)·(7n + 190) \ = \ 0 \ \ . \ $$ In the "forward direction", the $ \ n_1 \ = \ 29 \ $ terms runs from $ \ -3 \ $ to $ \ 193 \ \ , \ $ confirming the sum $ \ s \ = \ 29·\frac{190}{2} \ = \ 2755 \ \ . \ $ As for the "backward" sum, while $ \ -n_2 \ = \ -\frac{190}{7} \ \ , \ $ doesn't "make sense" as the number of terms in a series, it does formally produce in the series with $ \ a \ = \ 193 \ \ , \ \ d \ = \ -7 \ \ $ a "final term" of $ 193 \ + \ \left( \ \frac{190}{7} - 1 \ \right)·(-7) \ = \ 10 \ \ $ and the sum $$ \mathcal{S} \ \ = \ \ \frac{190}{7}·\left(\frac{193 \ + \ 10}{2} \ \right) \ \ = \ \ 2755 \ \ . $$ So $ \ n_2 \ $ is not necessarily an integer and does not lead to an arithmetic series in the usual sense: it becomes understandable why this second solution is usually neglected.