Help: Sequence and Series 2

The sixth term of an arithmetic progression is 2, and its common difference is greater than 1. Show that the value of the common difference of the progression so that the product of the first, fourth and fifth therm is greatest is 85\frac{8}{5}.

I did it with the concept of Maxima and Minima. Can someone suggest any other methods, something that is purely Algebra

Note by Akhilesh Prasad
3 years, 5 months ago

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Try to use weighted AM-GM, but keep in mind that a1,a4 a_1, a_4 are both negative, so their weights must be negative as well.
Take cases in which a5 a_5 is positive or negative, and use the fact that a6=2 a_6 = 2 for the weighted AM part.

Ameya Daigavane - 3 years, 5 months ago

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Umm....I sure do know what is AM-GM inequality but i dont know what the weighted AM and GM mean

Akhilesh Prasad - 3 years, 5 months ago

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Actually it's not the actual weighted AM-GM inequality I'm telling you to use, it's this -

ka1+la4+ma53(klma1a4a5)13 \frac{ka_1 + la_4 + ma_5}{3} \geq (klm \cdot a_1 a_4 a_5)^{\frac{1}{3}}

Choose k,l,mk, l, m such that,
k,l0 k, l \leq 0 and ka1+la4+ma5=a6ka_1 + la_4 + ma_5 = a_6 .
Use the formula for the general term of an AP to find k,l,mk, l, m.

The weighted AM-GM inequality (for two variables, here) is:

ma+nbm+n(ambn)1m+n \frac{ma + nb}{m + n} \geq (a^m \cdot b^n)^{\frac{1}{m + n}}

m,nm, n are called the weights of variables a,ba, b respectively.
It's easy to derive from the general AM-GM inequality.

Ameya Daigavane - 3 years, 5 months ago

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@Ameya Daigavane I am having a really hard time figuring how to find l,m,nl,m,n.

Now, what I did was started with ka1+la4+ma5=a6ka_1 + la_4 + ma_5 = a_6 .

Then, by pondering on it a little I approached it as follows:-

k(a)+l(a+3d)+m(a+4d)=a+5dk\left( a \right) +l(a+3d)+m(a+4d)=a+5d

On comparing the coefficients I got the following two equations

k+l+m=1k+l+m=1

k+3l+4m=5k+3l+4m=5

Then as per the rule of cross multiplication, we get,

k43=l41=m31k=l3=m2\displaystyle \frac { k }{ 4-3 } =\frac { -l }{ 4-1 } =\frac { m }{ 3-1 } \quad \quad \Longrightarrow \quad \quad k=\frac { -l }{ 3 } =\frac { m }{ 2 }

Using this result when i substitute the values back into equation k+l+m=1k+l+m=1 I got 1=01=0

Then from ka1+la4+ma53(klma1a4a5)13 \frac{ka_1 + la_4 + ma_5}{3} \geq (klm \cdot a_1 a_4 a_5)^{\frac{1}{3}} ,

We see that (a1a4a5)max=ka1+la4+ma53klm3\displaystyle{ \left( { a }_{ 1 }{ a }_{ 4 }{ a }_{ 5 } \right) }_{ max }={ \frac { ka_{ 1 }+la_{ 4 }+ma_{ 5 } }{ 3klm } }^{ 3 }.

Iam not able to find the value of klmklm. If you can suggest some method.

Akhilesh Prasad - 3 years, 5 months ago

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@Akhilesh Prasad It should be 3l+4m=5 3l + 4m = 5 , not k+3l+4m=5k + 3l + 4m = 5 , for the coefficient of d d . You're on the right track!

Ameya Daigavane - 3 years, 5 months ago

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@Ameya Daigavane But doesn't this imply that the signs of kk and ll will be different, as we had assumed that k,l0k,l\le 0.

Akhilesh Prasad - 3 years, 5 months ago

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@Akhilesh Prasad No, I don't think so, let me try.

Ameya Daigavane - 3 years, 5 months ago

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@Ameya Daigavane I am still not able to do it.

Akhilesh Prasad - 3 years, 5 months ago

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Can you gimme some pointers as to where to learn it from and its applications

Akhilesh Prasad - 3 years, 5 months ago

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@Ameya Daigavane And one more thing, what does r<saras\displaystyle \sum _{ r<s }^{ }{ { a }_{ r }{ a }_{ s } } imply. Does it imply s=1nr=1naras\displaystyle \sum _{ s=1 }^{ n }{ \sum _{ r=1 }^{ n }{ { a }_{ r }{ a }_{ s } } }

Akhilesh Prasad - 3 years, 5 months ago

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No it does not, r r goes from 1 1 to s1 s - 1 , not till n n . It should be r<saras=s=1nr=1s1aras \sum_{r < s}a_ra_s = \sum_{s = 1}^n \sum_{r = 1}^{s - 1} a_ra_s assuming the sum is over all pairs.

Ameya Daigavane - 3 years, 5 months ago

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Thanks a lot, you just solved another doubt of mine

Akhilesh Prasad - 3 years, 5 months ago

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@Calvin Lin

Akhilesh Prasad - 3 years, 5 months ago

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@Siddhartha Srivastava

Akhilesh Prasad - 3 years, 5 months ago

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@Ameya Daigavane

Akhilesh Prasad - 3 years, 5 months ago

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As am on mobile right now so couldn't post another post for this one so would you please consider it and clear my doubt here only.

A straight line is drawn through the centre of a square ABCDABCD intersecting side ABAB at NN so that AN:NB=1:2AN:NB=1:2. On this line take an arbitrary point MM lying inside the square. Prove that the distances from the point MM to the sides AB,AD,BC,CDAB, AD, BC, CD of the square taken in that order, form an A.P.A.P.

I did it by considering any vertex as origin then considering sides common to it as axes and then calculating the distance of the point from the sides and then find out the common difference.

I would appreciate it if someone would tell me how to do a rigorous proof which would teach my some interesting new things given that I am a noob at coordinate geometry.

@Ameya Daigavane,@Rishabh Cool,@Calvin Lin,@Sharky Kesa

Akhilesh Prasad - 3 years, 5 months ago

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We have, d1+d4=d2+d3=3a d_1 + d_4 = d_2 + d_3 = 3a d2d1=d4d3\Rightarrow d_2 - d_1 = d_4 - d_3

If we show d1+d3=2d2d_1 + d_3 = 2d_2 , we are done.
Note that, MNYONX \triangle MNY \sim \triangle ONX

So, d13a2=d2aa2\frac{d_1}{\frac{3a}{2}} =\frac{d_2 - a}{\frac{a}{2}}

d1=3d23a=2d2d3 \Rightarrow d_1 = 3d_2 - 3a = 2d_2 - d_3

as d3=3ad2 d_3 = 3a - d_2 , and we're finished.

Ameya Daigavane - 3 years, 5 months ago

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I did it the exact same way, just with one little deviation which was I selected the length of the sides to be aa instead of 3a3a, it made things just a little harder to visualize. And did you think up anything about the question in the main post, or could you suggest me some text to read it from.

Akhilesh Prasad - 3 years, 5 months ago

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Hmm... Quick enough. Good solution.

Deeparaj Bhat - 3 years, 5 months ago

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@Deeparaj Bhat Can you please look into the question in the original post, and provide me some other way to solve than the one that I did as stated in the note, or you could help me figure out where I am going wrong in the method suggested by @Ameya Daigavane.

Akhilesh Prasad - 3 years, 5 months ago

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One more thing you gotta tell me, how do you draw and post these diagrams on Brilliant.

Akhilesh Prasad - 3 years, 5 months ago

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@Akhilesh Prasad I used Asymptote to create the diagram.

Ameya Daigavane - 3 years, 5 months ago

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@Rishabh Cool

Akhilesh Prasad - 3 years, 5 months ago

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@Deeparaj Bhat

Akhilesh Prasad - 3 years, 5 months ago

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Here's a better way. Let the length of the square be 2a2a and take the origin as the centre of the square. Also, let the coordinate axes be parallel to the sides of the square.

I think you can proceed from here.

Deeparaj Bhat - 3 years, 5 months ago

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I don't think co-ordinate geometry is the best way here.

Ameya Daigavane - 3 years, 5 months ago

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I guess it is. I was able to do it with that in under 5 minutes.

Deeparaj Bhat - 3 years, 5 months ago

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@Deeparaj Bhat Just similarity of the triangles should be enough, it's pretty fast. I'll upload a diagram.

Ameya Daigavane - 3 years, 5 months ago

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