I have assumed $x=a-d,y=a$ and $z=a+d$ by which I get $a=7$ and numbers could be $1,7,13$ or $13,7,1$ or $7,7,7$, but the total no of solutions set is $13$.
I am not getting it. Please help.
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1What about $(6,7,8)$, $(5,7,9)$, $(4,7,10)$, $(3,7,11)$, $(2,7,12)$? – Ángel Mario Gallegos Apr 03 '18 at 16:53
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gap between $x$ and $y$ and $y$ and $z$ does not have to be the same. $2, 7, 12$ or $3, 5, 13$ will also work. – Vasili Apr 03 '18 at 16:53
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Please use MathJax. – Shaun Apr 03 '18 at 16:53
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1The problem statement says they are in an arithmetic progression. – Joshua Wang Jan 19 '21 at 04:06
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Since $z-y=y-x$. $x+z=2y$ and hence $3y=21$. So, $y=7$.
$x+z=2y=14$. So, $z=14-x$.
For any value of $x$, $x,7,14-x$ are in A.P.
If a natural number means a positive integer, them $x$ can take the values $1,2,3,\dots,13$.
CY Aries
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@Vasya $(x,y,z)$ is an ordered triple does not mean that the three numbers are in ascending order. The ordered triples $(1,2,3)$ and $(1,3,2)$ are not the same. But they are the same if they are regarded as unordered triples. – CY Aries Apr 03 '18 at 17:05