INTERESTING AP !!!!!

Algebra Level pending

If a , b , c , d a,b,c,d are distinct integers in AP such that d = a 2 + b 2 + c 2 d=a^2+b^2+c^2 ,

then a + b + c + d a+b+c+d is :

1 0 2 3

Shivansh Mittal by shivansh mittal , 22, India

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2 solutions

Manish Dash
Jul 18, 2015

The given numbers can be represented as b-k, b b+k, b+ 2k where k is the common difference.

So we we can write:

b + 2 k = ( b − k ) 2 + b 2 + ( b + k ) 2 b+2k=(b-k{ ) }^{ 2 }+{ b }^{ 2 }+(b+k{ ) }^{ 2 }

=> b + 2 k = 3 b 2 + 2 k 2 b+2k={ 3b }^{ 2 }+2{ k }^{ 2 }

=> b ≤ 3 b 2 b\le 3{ b }^{ 2 } -----(i) and 2 k ≤ 2 k 2 2k\le 2{ k }^{ 2 } ----------(ii)

Solving eq(ii), we get k=0,1

k=0 is rejected since the numbers are distinct. ∴ \therefore k=1.

Solving eq(i), we get b=0, 1/3

Since the numbers are integers hence b ≠ 1 3 b\neq \frac { 1 }{ 3 } ∴ \therefore b=0.

∴ \therefore The numbers a,b,c,d are -1, 0, 1, 2 respectively.

∴ a + b + c + d = 2 \therefore \boxed {a+b+c+d=2}

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Moderator note:

I disagree that we can reach the conclusions of (i) and (ii).

What we have is an inequality of specific values, as opposed to an inequality of variables. For example, if we have 1 + 4 ≤ 2 + 3 1 + 4 \leq 2 + 3 , does that imply that 1 ≤ 2 1 \leq 2 and 4 ≤ 3 4 \leq 3 ?

Even in the event of inequality of variables, we cannot do such a direct comparison. For example, we know that AM-GM tells us a 2 + b 2 ≥ 2 a b a^2 + b^2 \geq 2 ab . We cannot compare the term a b ab and conclude that 0 ≥ 2 a b 0 \geq 2ab hence 0 ≥ 2 0 \geq 2 .

Since it is an AP with integers with four terms, a+d = b+c . So a+b+c+d=2(a+d). So the sum can only be even. It can only be 0 or 2. But with sum 0, four integers CAN NOT be in AP. So if the options given contain an answer, it must be 2 ~~~~~~ \boxed{\color{#D61F06}{ \large 2} } .Since the four integer terms sum up to 2, some terms have to be negative. Thus the condition given is not necessary. And even without the given condition, not only we can know their sum (=2), I think there is only one unique solution -1, 0, 1, 2.

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sorry for the inconvenience caused!! the previous post was not put by me and i believe that it was hacked...but now its recovered!! @shivansh mittal sorry for the inconvenience caused!!

A Former Brilliant Member - 6 years, 3 months ago

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