Let the first term of the GP be b.

1st term = b
2nd term = br
3rd term = br²

For the AP,

3rd term = a + 2d
4th term = a + 3d
6th term = a + 5d


So b = a + 2d ①
br = a + 3d ②
br² = a + 5d ③

② - ① :
br - b = d

③ - ② :

br² - br = 2d

Sub d = br - b,

br² - br = 2br - 2b

br² - 3br + 2b = 0

r² - 3r + 2 = 0 (divide by the constant b)
(Shown)

thank you ! I need help with the other parts also :(

r² - 3r + 2 = 0

(r - 1)(r - 2) = 0

r = 1 or r = 2

(Reject r = 1 as that would imply d = b(1) - b = 0, but d is non-zero as stated in the question)

So r = 2

since r = 2,

d = b(2) - b = b

Sub b = d into ①,

d = a + 2d
a = -d

Since sum of first 4 terms of the AP = -6,

a + (a + d) + (a + 2d) + (a + 3d) = -6

4a + 6d = -6

sub a = -d,

4(-d) + 6d = -6

2d = -6

d = -3

so a = 3

Edited

mth term of G

= b r^(m-1)
= -3 (2^(m-1)) (since b = d)
= -3/2 (2^m)

mth term of A

= a + (m-1)d
= 3 + (m-1)(-3)
= 3 - 3m + 3
= 6 - 3m

AP : 3,0,-3,-6,-9,-12,-15,...
GP : -3,-6,-12,-24,-48,-96,-192,...


Difference between mth term of G and A must be more than 10000. As we can see, the GP terms are always much smaller than the AP terms when comparing term by term.

So for the difference, we want Am - Gm > 10000 as we are looking for a positive difference here.

6 - 3m - (-3/2 (2^m) ) > 10000

6 - 3m + 3/2 (2^m) > 10000

Using GC,

m > 12.707

so minimum m = 13

If mth term of A is 6 - 3m, then nth term an = 6 - 3n

So xn = (2k)^(6 - 3n)

And x(n-1)

= (2k)^(6 - 3(n-1))
= (2k)^(9 - 3n)


xn / x(n-1) = (2k)^(6 - 3n) / (2k)^(9 - 3n)

= (2k)^(6 - 3n - 9 + 3n)

= (2k)-³

= (1/2k)³

= 1/8k³


So this sequence xn is also a GP with common ratio 1/8k³

For the series to converge, |common ratio| < 1

So |(1/8k³)| < 1

(⅛) /|k³| < 1

⅛ < |k³|

|k|³ > (½)³

|k| > ½

k < -½ or k > ½