If the sum of three consecutive numbers in a geometric progression is 26 and the sum of their squares is 364, then the product of those numbers is

Let the terms of the GP are $$\dfrac{a}{r},a, ar$$

Where a is the first term and r is the common ratio.

As per the condition, $$\dfrac{a}{r}+a+ar=26------------(i)$$

$$\dfrac{a}{r}(1+r+r^2)=26$$

Now, squaring both side,

$$\dfrac{a^2}{r^2}(1+r+r^2)^2=676$$

As per the the second condition,

$$(\dfrac{a}{r})^2+a^2+(ar)^2=364$$

$$\dfrac{a^2}{r^2}(1-r+r^2)(1+a^2+a^2r^2)=364-------------(ii)$$

From equestion (i) and (ii)

$$\Rightarrow \dfrac{1-r+r^2}{1+r+r^2}=\dfrac{364}{676}$$

$$\Rightarrow \dfrac{1-r+r^2}{1+r+r^2}=\dfrac{7}{13}$$

$$\Rightarrow 13 - 13r + 13r^2 = 7 + 7r + 7r^2$$

$$\Rightarrow (3r-1)(r-3) =0$$

So,$$r=3$$ and $$r=\dfrac{1}{3}$$

Now, substituting the value of r=3,

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

$$a(1+3+9)=26$$

$$a(13)=26$$

$$a=\dfrac{26}{13}=2$$

Hence, the product of these numbers $$=a\times \dfrac{a}{r}\times ar =a^3=6^3=216$$