Are $\gamma$ ou the smartest #3 $\gamma$

Since $a$ , $x$ , $y$ , $z$ , $b$ are in arithmetic progression, and $x + y + z = 15$ , it follows that $y = 15/3 = 5$ and $a + b = 2y = 10$ .

$(y-d) + y + (y+d) = 15$ $y=5$

$(y-2d) , (y-d) , y , (y+d) , (y+2d)$ ; $a+b = 2y$ $a+b = 2(5) = 10$

let x= a+d, y=a+2d, z=a+3d, b=a+4d

x+y+z=15 implies 3a +6d=15 which gives a+2d=5 or a=5-2d.

$\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ = $\frac{5}{3}$

$\frac{1}{a+d}$ + $\frac{1}{a+2d}$ + $\frac{1}{a+3d}$ = $\frac{5}{3}$

substituting a=5-2d

$\frac{1}{5-d}$ + $\frac{1}{5}$ + $\frac{1}{5+d}$ = $\frac{5}{3}$

on solving we get d= $\sqrt{10}$ or - $\sqrt{10}$

that gives a=5-2 $\sqrt{10}$ or 5+2 $\sqrt{10}$

that gives b=a+4d= 5+2 $\sqrt{10}$ or 5-2 $\sqrt{10}$ .

so a+b = 10.

a + x + y + z + b = 15 + a + b

Sn = ((a1+an)n)/2 ----> ((a+b)5)/2 = 15+a+b ---> 5a + 5b = 30 + 2a + 2b ---> a + b = 10

if a,x,y,z,b are in AP then,

a+x+y+z+b = 5(a+b)/2 2a+2b+30 = 5a+5b On re-arranging. 3(a+b) = 30 a+b = 30/3 = 10

X, y, z are part of an arithmetic progression, thus x+z=2y. x+y+z=3y=15. A+x+y+z+b=5 y=5 5=25, A+b+x+y+z=a+b+15=25... a+b=25-15=10 and I didn't even use the second piece of info,

x-a=y-x=z-y=b-z(a,x,y,z,b are in AP) solving : x+z=a+b=2y therefore y=3 and a+b=10

ooo