Let a be the first term and r be the common ratio of the G.P. According to question, a = 1, $a_3 + a_5$ = 90 ⇒ $ar^2 + ar^4$ = 90 ⇒ $1r^2 + 1r^4$ = 90 ⇒ $r^4 + r^2$ – 90 = 0 ⇒ ($r^2$ – 9) ($r^2$ + 10) = 0 ⇒ $r^2$ = 9, – 10 but $r^2$ ≠ –10 ∴ $r^2$ = 9 ⇒ r = ± 3. Hence the common ratio of G.P. is ± 3.