i)

let first term of the GP be a and common ratio be r.

let the common difference for the AP be d.


a1 = a
a3 = ar² = a + d ①
-a2 = -ar = a + 2d ②


① x 2 - ② :

2ar² - (-ar) = 2a + 2d - (a + 2d)

2ar² + ar = a
2ar² + ar - a = 0
a(2r² + r - 1) = 0
a(2r -1)(r + 1) = 0

a = 0 or r = ½ or r = -1

(a = 0 is rejected as the terms of the GP are non zero.
r = -1 is rejected as question says series is convergent , so |r| < 1 )

so common ratio = ½




ii)

Summation on the left :
a4 + a6 + a8 + ...+ a18 + a20 + a22

There are 10 terms in this summation.

Summation on the right :

a1 + a3 + a5 + a7 + a9 + a11 + ...

The common ratios for both summations are r².


First term on the left summation is a4 = ar³
First term on the right is a1= a


So S10 for left side = (S∞)² for right side.

ar³(1 - (r²)^10)/(1 - r²) = (a/(1 - r²))²

ar³(1 - r^20)/(1 - r²) = a²/(1 - r²)²

ar³(1 - r^20)(1 - r²) = a²

ar³(1 - r^20)(1 - r²) - a² = 0

a( r³(1 - r^20)(1 - r²) - a) = 0

a = 0 or a = r³(1 - r^20)(1 - r²)
(N.A , as explained above in i)


so sub r = ½,

a = (½)³(1 - (½)^20)(1 - (½)²)

a = (⅛)(¾)(1 - (½)^20)

a = 3/32 (1 - (½)^20)


a = 3/32 (1 - 1/(2^20) )

a = 3/32 (2^20 - 1)/(2^20)

a = 3(2^20 - 1)/(2^5 × 2^20)

a = 3(2^20 - 1)/(2^25)

so b = 3/(2^25)

Thank u @J

Welcome.

I think for the final answer you could leave it as 3/33554432 if you want, as 2^25 = 33554432