This progression is so radical
a − x , x , a + x are first three terms of an arithmetic progression with integral terms where a > x > 0 . Find the least composite value of a .
The answer is 20.
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1 solution
by mistake i wrote 10
The method of removing the square root(s)...
BruteForce method... simply love it :)
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Another way,
2 x = a + x + a − x
2 x = a + x − a − x 2 x
a + x − a − x = x
a + x + a − x = 2 x
2 a + x = 3 x
4 ( a + x ) = 9 x
4 a = 5 x
Why it is tagged with Calculus? And why is this level 4, I think it should be level 3. Nice problem btw. I like it when I write the terms out in terms of a :
a − 5 4 a , 5 4 a , a + 5 4 a
⇒ 5 a , 2 5 a , 3 5 a
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Sequences and series come under calculus , though in schools of india it is taught in algebra
Why not x=4 and a=5?
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Hey, 5 is not a composite number...it's a prime...
what about a=5 and x=4
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For a=5 and x=4 all the conditions are satisfied a>x>0 and the terms of A.P also remain integer . In first attempt I tried this
5 is a prime..... so will be excluded.
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I got at that very moment I gave a reason see the top comment
Sorry I got it , it is asked about least composite value of a.
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Then how did you get this question right ? Also wouldn't x=8 be an answer ?
Ok understood .
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At x=8 the AP won't consist of integers.
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@Shubhendra Singh – Yeah, just now I got it . I was primarily focused on finding the values of x and a .
Thanks
Even I got my first try wrong for the same reason ^_^
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Pls answer my question , Ok I understood
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@Arpit Agarwal – For 8 the terms will not remain integers and since a least composite number is asked(which is the trickest part of the question) 5 is excluded
Same method. Great solution.
Because the sucesion is with integrals terms
Why can't a=10 and x=8????????
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because in question it is stated that ap consist of integral terms
Using the property of AP we get
2 x = a + x + a − x
Whole square and rearrange to get
4 x 2 + a 2 − 4 a x = a 2 − x 2
This gives 5 x = 4 a
By this Least composite value of a = 2 0 when x = 1 6
So the answer is 2 0