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Please help- Four different integers form an increasing AP.One of these numbers is equal to the sum of the squares of the other three numbers.Then- find all the four numbers.

I assumed the numbers to be $a,a+d,a+2d,a+3d$ and wlog let $a+3d$ this number and according to question-

$a^2+(a+d)^2+(a+2d)^2=a+3d$

but I could not solve the above equation.

Dr. Sonnhard Graubner
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Rohan
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2 Answers2

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$$(a-d)^2+a^2+(a+d)^2=a+2d\iff3a^2+2d^2=a+2d$$

$$\iff3\left(a-\frac16\right)^2+2\left(d-\frac12\right)^2=\frac1{12}+\frac12=\frac7{12}$$

$$\iff(6a-1)^2+6(2d-1)^2=7$$

As $a,d$ are integers, $$6(2d-1)^2\le7\implies(2d-1)^2\le\dfrac76\implies d=0,1$$

Both cases $\implies (6a-1)^2=1\iff6a-1=\pm1\implies a=0$ as $a$ is an integer

lab bhattacharjee
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It might help to remember that the $n$th term of an AP is $a_1 + (n-1)d$ where $a_1$ is the first term of the AP and $d$ is the common difference. Setting $a_1^2 + (a_1 + d)^2+(a_1 +2d)^2=a_1 + 3d$ should help you get the answer.

Ari
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  • I tried to do the same thing but could not solve the equation – Rohan Oct 24 '14 at 13:22
  • Have you considered the fact that $a_1$ can be negative but $d$ must be positive? – Ari Oct 24 '14 at 13:35
  • Yes I thought about it ,but still I'm not able to draw out any conclusions – Rohan Oct 24 '14 at 13:43
  • I found a solution. As ugly as it may be, multiply out the parts of the equation and then apply the quadratic formula. From there it is a simple matter of finding which positive, integral, $d$ result in a positive discriminant. – Ari Oct 25 '14 at 19:40
  • It worked ! Thanks Ari! – Rohan Oct 26 '14 at 10:07