Let the 3 terms of the GP are: a, ar, ar²

Given, a + ar + ar² = 13 

=> a(1 + r + r² )= 13 .................1

Again, a² + (ar)² + (ar² )² = 91

=> a² + a² r² + a² r⁴ = 91

=> a² (1 + r² + r⁴ ) = 91 .............2

Square equation 1, we get

     a² (1 + r + r² )² = 169

=> a² (1 + r² + r⁴ + 2r + 2r² + 2r³ ) = 169 ..........3

Now, equation3 - equation2, we get

      2ra² (1 + r + r² ) = 169 - 91

=> 2ra*13 = 169 - 91

=> 26ar = 78

=> ar = 78/26

=> ar = 3

=> r = 3/a ...............4

From eqaution1, we get

      a{1 + 3/a + (3/a)² } = 13

=> a{1 + 3/a + 9/a² } = 13

=> a + 3 + 9/a  = 13

=> a + 9/a  = 13 - 3

=> a + 9/a  = 10

=> (a² + 9)/a  = 10

=> a² + 9  = 10a

=> a² - 10a + 9 = 0

=> (a - 9)*(a - 1) = 0

=> a = 1, 9

From eqaution 4, we get

      r = 3/1 and r = 3/9

=> r = 3 and r = 1/3

Now, the geometric terms are when a = 1 and r = 3 

    1, 1*3, 1*3² 

=  1, 3, 9

Again, the geometric terms are when a = 9 and r = 1/3 

    9, 9* (1/3), 9*(1/3)² 

=  9, 9/3, 9/9

= 9, 3, 1

So, the geometric series are: 1, 3, 9 or 9, 3, 1