There are 4 terms of an increasing arithmetic progression (AP) where one term is equal to the sum of the squares of the other three.
Solution 1
Using your values for the AP and guessing that the last term is the sum of the squares of the rest, I get $$a+3d=(a-3d)^2+(a-d)^2+(a+d)^2\\ =3a^2-6ad+11d^2\\ a=\frac 16(6d+1\pm \sqrt{(6d+1)^2-44d^2})\\ =\frac 16(6d+1\pm\sqrt{1+24d-8d^2})$$ which is nicely real as long as $d$ is not too big. I don't find a unique solution.
Solution 2
Start with any arithmetic progression $a,a+d,a+2d,a+3d$. Select any term as the one that will equal the sum of the squares of the other three: say, $a+d$. Now solve for $k$: $$k(a+d)=k^2(a^2 + (a+2d)^2+(a+3d)^2)$$ In other words, take $$k=\frac{a+d}{a^2 + (a+2d)^2+(a+3d)^2}$$ Now, by construction, the arithmetic progression $$ka,k(a+d),k(a+2d),k(a+3d)$$ satisfies your conditions.
It should be clear to you, therefore, that there are many solutions to your problem. Perhaps you forgot to tell us that the terms must be integers?
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Siddharth Garg
Updated on August 01, 2022Comments
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Siddharth Garg over 1 year
If there are 4 terms of an increasing arithmetic progression (AP) where one term is equal to the sum of the squares of the other three, find the terms.
I have tried solving the question by taking the terms as $a-3d$, $a-d$, $a+d$ & $a+3d$. I am getting $a=-0.5$ but $d$ is coming out to be a complex number, which shouldn't happen.
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Evargalo over 6 yearsI don't know how you found $a=-0.5$, but this is not correct.
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Hypergeometricx over 6 years@Evargalo - Nice nickname - clever 8-letter compression :)
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Evargalo over 6 years@hypergeometric thank you. And congratulations, not every mathematician can guess its etymology.
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Siddharth Garg over 6 yearsEdited the question.
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TonyK over 6 yearsUnder these new rules, the only solutions are $(0,0,0,0)$ and $(\frac13,\frac13,\frac13,\frac13)$.
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Siddharth Garg over 6 yearsShouldn't all the terms of the A.P. be non-negative as they are equal to sum of squares which is always positive?
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TonyK over 6 years@SiddharthGarg: Your comment seems to imply that each term of the AP is equal to the sum of the squares of the other three terms. Please let us know if this is so, before we all waste even more of our time on it.
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Evargalo over 6 yearsBy the way, that would make for a richer problem... but probably without any solution ?
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Siddharth Garg over 6 yearsYes, TonyK. That is exactly what the question is. I have edited the question too to make it more clear.