arithmetic progession question

Question

if $\sqrt{a-x}, \sqrt x, \sqrt{a+x}$ are in AP provided $a>x$ and $a,x$ are positive integers then what is the least possible value of $x$?

Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem — JKnecht, Mar 02 '16 at 14:17

GoodDeeds · Answer 1 · 2016-03-02T14:24:32.153

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$$\sqrt{a-x},\sqrt{x},\sqrt{a+x}$$ are in an AP. By the definition of an arithmetic progression, $$\sqrt{a+x}-\sqrt{x}=\sqrt{x}-\sqrt{a-x}$$ Thus $$2\sqrt{x}=\sqrt{a+x}+\sqrt{a-x}$$ Squaring both sides, $$4x=a+x+a-x+2\sqrt{a^2-x^2}$$ $$2x-a=\sqrt{a^2-x^2}$$ Squaring both sides, $$4x^2+a^2-4ax=a^2-x^2$$ $$5x^2-4ax=0$$ $$x(5x-4a)=0$$ Since $x\ne0$, $$x=\frac{4a}{5}$$ Since $a,x$ are positive integers, for the least value, $a$ must be the smallest positive integer divisible by $5$. Thus, $a=5$.

Correspondingly, the smallest value of $x$ is $x=4$.

edited Mar 02 '16 at 14:24

answered Mar 02 '16 at 14:21

GoodDeeds

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So the smallest value for x is when a = 5. Then x = 4. – peter.petrov Mar 02 '16 at 14:22
@peter.petrov Thank you, added it to the answer. – GoodDeeds Mar 02 '16 at 14:24
What i hav done is.......2√x=√a-x+√a+x..squaring both the sides..4x=a-x+a+x+2(√a-x)(√a+x)....=>4x=2(a+(√a)^2-(√x)^2..=> 2x=2a-x..=> 3x=2a...=>x=2a/3....therefore the leadt value of x would be 2......... – user319310 Mar 02 '16 at 15:11
@user319310 But $\sqrt{(a+x)(a-x)}=\sqrt{a^2-x^2}\ne a-x$. Or did I interpret your question wrongly? – GoodDeeds Mar 02 '16 at 15:15

score 0 · Answer 2 · answered Mar 02 '16 at 15:27

GoodDeeds's answer is arguably the "right" way to go about solving this problem, but it's possibly worth noting that it could be tackled by means of brute-force search. That is, the conditions require that $a-x\lt x\lt a$, which can be re-expressed as

$$x\lt a\lt2x$$

Now we can rule out $x=1$ immediately, since there is no integer $a$ satisfying $1\lt a\lt2$.

If $x=2$, the inequality $2\lt a\lt4$ is satisfied only by $a=3$, but $1,\sqrt2,\sqrt5$ is not an arithmetic progression. (Strictly speaking, we need to show that $\sqrt5-\sqrt2\not=\sqrt2-1$, but the non-equality seems clear enough.)

Likewise, if $x=3$, the inequality $3\lt a\lt6$ is satisfied only by $a=4$ and $5$, but $1,\sqrt3,\sqrt7$ and $\sqrt2,\sqrt3,\sqrt8$ are not arithmetic progressions.

It's only at $x=4$ that we find a value of $a$, namely $a=5$, for which $\sqrt{a-4},2,\sqrt{a+4}$ is an arithmetic progression.

arithmetic progession question

2 Answers2