Omg~the G.P question made all of us(Pre-U1 Beta)pening...lll
Finally ,i solved the question ~
Therefore, i wan to share the question + solution wif everyone

Q12.
A geometric series has first term 1 and the common ratio r is positive.
The sum of the first 5 terms of a G.P. is twice the sum of the terms from the
6th to the 15th inclusive.
Show that r^5(r to the power of 5) = 1/2(√3-1)

Solution
U1+U2+U3+U4+U5 =
2[U6+U7+U8+U9+U10+U11+U12+U13+U14+U15]
a+ar+ar^2+ar^3+ar^4 =
2(a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7+ar^8+ar^9+ar^10+ar^11+ar^12+ar^13+ar^14)

a=1

(1+r+r^2+r^3+r^4) =
2(r^5+r^6+r^7+r^8+r^9+r^10+r^11+r^12+r^13+r^14)
(1+r+r^2+r^3+r^4) =
2[r^5(1+r+r^2+r^3+r^4)+r^10(1+r+r^2+r^3+r^4)]

1=2r^5+2r^10
2r^10+2r^5-1=0
Let d = r^5
Then 2d^2+2d-1=0
Using Quadratic Formula
because r is positive

therefore
d=[-2+√(2)^2-4(2)(-1)]/2(2)
=(-2+√12)/4
=[-2+(√4.√3)]/4
=(-2+2√3)/4
=(-1+√3)/2
=1/2(√3-1)

d=r^5
Thus, r^5 = 1/2(√3-1) ....shown

Done it,happy~