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Question Number 210629 by peter frank last updated on 14/Aug/24

Answered by Rasheed.Sindhi last updated on 14/Aug/24

$$\{\begin{array}{l}x=a+(a+d)+(a+2d)+,...+(a+(m-1)d)\\ y=(a+md)+(a+(m+1)d)+...+(a+(2m-1)d)\\ z=(a+2md)+(a+(2m+1)d)+...+(a+(3m-1)d)\end{array}$$$$\{\begin{array}{l}x-ma=\left(d\right)+\left(2d\right)+,...+\left((m-1)d\right)\\ y-ma=\left(md\right)+\left((m+1)d\right)+...+\left((2m-1)d\right)\\ z-ma=\left(2md\right)+\left((2m+1)d\right)+...+\left((3m-1)d\right)\end{array}$$$$\{\begin{array}{l}\frac{x-ma}{d}=\left(1\right)+\left(2\right)+,...+\left((m-1)\right)\\ \frac{y-ma}{d}=\left(m\right)+\left((m+1)\right)+...+\left((2m-1)\right)\\ \frac{z-ma}{d}=\left(2m\right)+\left((2m+1)\right)+...+\left((3m-1)\right)\end{array}$$$$...$$

Answered by mr W last updated on 14/Aug/24

$$saythedifferencefromthe{(m+1)}^{th}$$$$andthefirsttermist,then$$$$eachterminthesecondgroupis$$$$bytlargerthanthecorresponding$$$$terminthefirstgroup.thesumof$$$$thesecondgroupisbymtlargerthan$$$$thesumoffirstgroup,etc.$$$$a_1,a_2,...,a_m\leftarrow thefirstmterms,theirsumisx$$$$b_1,b_2,...,b_m\leftarrow thenextmterms,theirsumisy$$$$c_1,c_2,...,c_m\leftarrow thelastmterms,theirsumisz$$$$b_1=a_1+t,b_2=a_2+t,...,b_m=a_m+t$$$$c_1=b_1+t,c_2=b_2+t,...,c_m=b_m+t$$$$y=x+mt$$$$z=y+mt$$$$\Rightarrow x+z=2y$$$$\Rightarrow x^2+z^2+2xz=4y^2$$$$\Rightarrow x^2+z^2-2xz=4y^2-4xz$$$$\Rightarrow {(x-z)}^2=4(y^2-xz)✓$$

Commented by mm1342 last updated on 14/Aug/24

$$veryexcellent\underset{―}{\underbrace{⋚}}$$

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