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Sequences and Triangles question

Here is the question:

Screenshot-20190909210349-1416x171.png

I would expect the answer to use a pythagoras calculation with a, a+d and a+2d, and I would expect the consecutive terms to be in descending order when they are reciprocals, so x is the biggest but its reciprocal would be the smallest.

I first tried the cosine rule using cos 90 as zero but the algebra becomes huge.

Is there a method I am missing?

:confused:
Original post by Gmart
Here is the question:

Screenshot-20190909210349-1416x171.png

I would expect the answer to use a pythagoras calculation with a, a+d and a+2d, and I would expect the consecutive terms to be in descending order when they are reciprocals, so x is the biggest but its reciprocal would be the smallest.

I first tried the cosine rule using cos 90 as zero but the algebra becomes huge.

Is there a method I am missing?

:confused:


Yes, Pythagoras.

I'd start with the AP. Your terms are consecutive, so work out a relationship between x,y,z.

Hint:

Spoiler



Then for the right-angled triangle.

Work out the squares of each side, then show that the sum of two of them equals the third - using the relationship you derived form the AP. Actual working takes very few lines. Seeing what to do requires a bit of thought though.
Reply 2
Thanks, I'll give it a go :smile:
Reply 3
Should I put everything in terms of a & d?
Original post by Gmart
Should I put everything in terms of a & d?


The parameters of the AP are not required at all. See the hint in my first post.
Reply 5
Surely any relationship between x, y & z is going to be concerning d?
Original post by Gmart
Surely any relationship between x, y & z is going to be concerning d?


No. If you use the a,d of the AP, you can eliminate them to get a relationship in x,y,z only. Or, follow my hint, and avoid a,d entirely.
Reply 7
Original post by ghostwalker
No. If you use the a,d of the AP, you can eliminate them to get a relationship in x,y,z only. Or, follow my hint, and avoid a,d entirely.

OK!! :eek:
Reply 8
Original post by ghostwalker
No. If you use the a,d of the AP, you can eliminate them to get a relationship in x,y,z only. Or, follow my hint, and avoid a,d entirely.

OK!! :eek:
Reply 9
Here's my work so far:

1x,z+x2xz,1z \dfrac{1}{x}, \dfrac{z+x}{2xz}, \dfrac{1}{z}

Putting this into pythagoras:

(x+z)2=y2+(xy+z)2 (x+z)^2 = y^2 + (x-y+z)^2
Reply 10
Makes y2=0y^2=0

How can that be!?

Arrggh!
Reply 11
I'm gonna sleep on it - thanks for your help! :cry:
Original post by Gmart
Here's my work so far:

1x,z+x2xz,1z \dfrac{1}{x}, \dfrac{z+x}{2xz}, \dfrac{1}{z}



Well you've eliminated y from your AP, rather than getting a relationship between the three variables.

1y=....\dfrac{1}{y}=.... then multiply up to get it into a useful format.


Putting this into pythagoras:

(x+z)2=y2+(xy+z)2(x+z)^2 = y^2 + (x-y+z)^2


Do read the rest of my first post.

You need to show that's the case, not just state it. And you have a sign error there.

Also, it's not immediately obvious, unless I'm missing something, which is the hypotenuse and which are the two other sides, so putting it straight into Pythagoras is a little premature.

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