The sum of the first three terms of a G.P. is 13 and sum of their squares is 91. Determine the G.P.

Let the first term of G.P be a

⇒ second term = ar and third term = ar².

(where, r is the common ratio)

∵ sum of three terms is 13

⇒ a(1 + r + r²) = 13r……..(1)

Also, sum of their squares is 91.

⇒ a² (1 + r² + r⁴) = 91r² ………(2)

Now, Squaring (1) dividing by (2)

⇒ 7( 1 + r² + r) = 13( 1 + r² - r)

⇒ (7 + 7r² + 7r) = ( 13 + 13r² - 13 r)

⇒ 6r² - 20r + 6 = 0

⇒ 2(3r² - 10r + 3) = 0

⇒ 3r² - 10r + 3 = 0

⇒ 3r² - 9r-r + 3 = 0

⇒ (3r – 1)(r – 3) = 0

Substituting r in equation (1), we get-

⇒a(1 + 3 + 9) = 13 × 3

And,

⇒13a = 13

And

⇒ a = 1 and a = 9.

Now, G.P is-

a, ar, ar²,….

⇒ If r = 3 and a = 1 then,

⇒1, 1×3, 1×3², ……

= 1, 3 , 9 are the first three terms.

And, If and a = 9 then,

= 9, 3, 1 are the first three terms.

⇒ 1,3,9,… … or 9,3,1,… … is the G.P.