Bee or b?

As $a,b,c$ is a GP we have that $b = ar$ and $c = ar^{2}$ for some real number $r \ne 0$ . The given equations can then be written as

$a(1 + r + r^{2}) = 26$ , (A), and $a^{2}(1 + r^{2} + r^{4}) = 364$ , (B).

Squaring equation (A) yields $a^{2}(1 + 2r + 3r^{2} + 2r^{3} + r^{4}) = 676$ , and when we subtract equation (B) from this result we end up with

$a^{2}(2r + 2r^{2} + 2r^{3}) = 312 \Longrightarrow a^{2}r(1 + r + r^{2}) = 156$ .

Dividing this last result by equation (A) yields $b = ar = \dfrac{156}{26} = \boxed{6}$ .

Note that a,b and c are in Geometric Progression

$a$ = first term of GP

$ar$ = second term of GP

$ar^2$ = third term of GP

$a(r^2+r+1)=26$

$a^2(r^4+r^2+1)=364$

$r^4+r^2+1=(r^2+r+1)(r^2-r+1)$

$a(r^2+r+1)*a(r^2-r+1)=364$

$26a(r^2-r+1)=364$

$a(r^2-r+1)=14$ ------>subtract this from the first equation

$2ar=12$

$ar=6=b$

a,b and c are in G.P.

Therefore, b=√ac→b²=ac

a²+b²+c²=364

a²+ac+c²=364

a²+2ac+c²-ac=364

(a+c)²-ac=364

By first eq, a+c is 26-b

(26-b)²-b²=364

Solving for b gives 6.