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December 18th, 2014, 02:25 AM  #1 
Member Joined: Dec 2014 From: India Posts: 61 Thanks: 0  addition of geometric progression
how the addition of geometric series gives 2 and product gives 40

December 18th, 2014, 05:37 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Short answer: The 2 and the 40 are determined from the coefficient of the $\displaystyle x^3$ term and the constant term in the quartic. Long answer: If the roots of the quartic are $\displaystyle a3d, ad, a+d, a+3d$ then by multiplying (xroot) brackets together we should get the original quartic: i.e. $\displaystyle (x  (a3d))(x  (ad))(x  (a+d))(x  (a+3d)) = x^4 2x^3  21x^2 + 22x^3 + 40$ If we multiply out this whole lefthand side, we can collect up all of the $\displaystyle x^3$terms, $\displaystyle x^2$terms and $\displaystyle x$terms to equate the coefficients with $\displaystyle a$ and $\displaystyle d$. I'm not going to multiply out the whole thing, but I know that part of it is $\displaystyle (x  (a3d))(x  (ad))(x  (a+d))(x  (a+3d)) = x^4 + ((a3d)(ad)(a+d)(a+3d))x^3 + ... + ((a3d))((ad))((a+d))((a+3d))$ The coefficient in front of the $\displaystyle x^3$ term is $\displaystyle ((a3d)(ad)(a+d)(a+3d)) = 4a$ If we compare this with the actual one, which is 2, we know that $\displaystyle a$ must be 1/2. Note that as a consequence of multiplying out the brackets we have ended up with a coefficient that is calculated by summing up all of the roots and placing a minus sign in front. Also, the last term, the constant, is $\displaystyle ((a3d))((ad))((a+d))((a+3d))$ $\displaystyle = (a3d)(ad)(a+d)(a+3d)$ $\displaystyle = (a3d)(a+3d) \cdot (ad)(a+d)$ $\displaystyle = \left(a^2  9d^2\right)\left(a^2d^2\right)$ Note that as a consequence of multiplying out the brackets we have ended up with a constant term that is the product of our roots. Comparing this with the actual constant term, which is 40, gives $\displaystyle \left(a^2  9d^2\right)\left(a^2d^2\right) = 40$. We can also substitute the a=1/2 that we know, so $\displaystyle \left(\left(\frac{1}{2}\right)^2  9d^2\right)\left(\left(\frac{1}{2}\right)^2d^2\right) = 40$ $\displaystyle \left(\frac{1}{4}  9d^2\right)\left(\frac{1}{4}d^2\right) = 40$ $\displaystyle \left(1  36d^2\right)\left(14d^2\right) = 640$ $\displaystyle 1  40d^2 + 144d^4 = 640$ $\displaystyle 144d^4  40d^2  639 = 0$ Solving this gives $\displaystyle d = \pm 3/2$, as explained in the textbook. Now we know $\displaystyle a$ and $\displaystyle d$, we know all the solutions. Last edited by Benit13; December 18th, 2014 at 05:40 AM. 
December 18th, 2014, 06:00 PM  #3  
Member Joined: Dec 2014 From: India Posts: 61 Thanks: 0  Quote:
Thank you now i understood  
December 19th, 2014, 01:21 AM  #4 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
No problem 

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addition, geometric, progression 
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