66Assuming m≠n, the extremum is attained for the index m+n2 if that's an integer which will be when both are either even or odd. If one of them is even and one odd then the extremum is taken twice at m+n2 &ndash 12 and m+n2+12
Denote by Sk the sum to k terms of an AP.
If for some two indices m and n it is known that Sm=Sn, for what index i, does Si attain a real extremum.
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8 Answers
6666The solution just involves quadratics.
Since for the indices m and n, we have Sm = Sn so
\dfrac{m}{2}\left(2a+(m-1)d\right)=\dfrac{n}{2}\left(2a+(n-1)d\right)
which gives
\dfrac{1}{2}-\dfrac{a}{d}=\dfrac{m+n}{2} ------ (1)
assuming m≠n.
Next,
S_k = \dfrac{k}{2}\left(2a+(k-1)d\right)
which can finally be written as
S_k = \dfrac{d}{2}\left(k-\left(\dfrac{1}{2}-\dfrac{a}{d}\right)\right)^2 -\dfrac{d}{2}\left(\dfrac{1}{2}-\dfrac{a}{d}\right)^2
which gives, using (1),
S_k = \dfrac{d}{2}\left(k-\dfrac{m+n}{2}\right)\right)^2 -\dfrac{d}{2}\left(\dfrac{m+n}{2}\right)^2
From where its not difficult to get the conclusion I already stated.
341The motivation for the problem is the fact that the expression for the sum to n terms of an AP is a quadratic in n.
We know that for a quadratic f(x), points y and z such that f(y) = f(z), y≠z, lie symmetrically about the extremum point xo and so x_0 = \frac{y+z}{2}
Hence if m and n are of the same parity, S_{\frac{m+n}{2}} is our real extremum.
Otherwise, S_{\frac{m+n-1}{2}} = S_{\frac{m+n+1}{2}} will be the real extremum.
Cool, na?
62I couldnt stop posting a solution to this one..
The thing is summations works very much like integration..
The sum of a few terms is zero in between, which directly means that the area of the curve is zero... (I know this is not absolutely correct.. but am trying to give a perspective of my own)
SO what it says is that plot of the terms of the AP will be a straight line cutting the x axis exactly at the mid point of m and n...
So this would mean that the area will be maximum or minimum at the point where it cuts the x axis...
As a "unanswered point" , which i expect answers from users other than kaymant sir, bipin and ofcourse prophet sir is that
If that is the case, then why is Kaymant sir's answer not simply (m+n)/2 ??
341because its an index and so it has to be a natural number. So when m and n are not integers \frac{m+n+1}{2} and \frac{m+n-1}{2} are the integers that are closest to m+n/2 and since they lie symmetrically on either side of this point the sums at these 2 points will be equal.
There are quite a few AP summation problems in the textbooks that are very simply solved by directly assuming a quadratic of the form an^2+bn
49KAYMANT SIR: ""The solution just involves quadratics.
Since for the indices m and n, we have Sm = Sn so
which gives
------ (1)
assuming m≠n.
Next,
which can finally be written as
which gives, using (1),
From where its not difficult to get the conclusion I already stated.""
the equations cannot be seen...[2][2][2][2][2]
66Is it? But I can see it perfectly. Anyway, here are the equations without the latex.
The solution just involves quadratics.
Since for the indices m and n, we have Sm = Sn so
m2(2a+(m-1) d) = n2 (2a + (n-1)d)
which gives
12 &ndash ad = m+n2 ----- (1)
assuming m≠n.
Next,
Sk = k2 (2a +(k-1)d)
which can finally be written as
Sk = d2 (k &ndash (1/2 &ndash a/d))2 &ndash d2 (1/2 &ndash a/d)2
which gives, using (1),
Sk = d2 ( k &ndash m+n2)2 &ndash d2 (m+n2)2
From where its not difficult to get the conclusion I already stated.