The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

Let the three terms of the GP be: $\frac{a}{r}, a, ar$

Then,

$$\frac{a}{r} + a + ar = 26$$

$$a \left(\frac{1}{r} + 1 + r\right) = 26$$

Also, sum of squares:

$$\frac{a^2}{r^2} + a^2 + a^2 r^2 = 364$$

$$a^2 \left(\frac{1}{r^2} + 1 + r^2\right) = 364$$

Now,

$$\left(\frac{1}{r} + 1 + r\right)^2 = \frac{1}{r^2} + 1 + r^2 + 2\left(\frac{1}{r} + r + 1\right)$$

$$\left(\frac{26}{a}\right)^2 = \frac{364}{a^2} + 2\left(\frac{26}{a}\right)$$

$$\frac{676}{a^2} = \frac{364}{a^2} + \frac{52}{a}$$

$$676 = 364 + 52a$$

$$52a = 312$$

$$a = 6$$

$$\text{Now, } \frac{1}{r} + 1 + r = \frac{26}{6} = \frac{13}{3}$$

$$3 + 3r + 3r^2 = 13r$$

$$3r^2 - 10r + 3 = 0$$

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

$$r = 3 \text{ or } r = \frac{1}{3}$$

Hence, the three terms are 2, 6, 18.