Three numbers a;b; 9; ... are in geometric sequence. If 1 is subtracted from the first term then
the sequence becomes Arithmetic. Determine the values of a and b.

If you can also solve the other circled question

if a b 9 are in GP

b÷a = 9 ÷ b

b^2 = 9a

if a - 1 b 9 are in AP

2b = a+8

b = a+8÷2

substitute

b is equal to a + 8 by 2 in b square equal to 9 a

(a+8÷2)^2 = 9a

a^2+16a+64 = 36a

a^2 -20a +64 =0

(a-16) (a-4) =0

a= 16 or a=4

when a=16 b= 12

when a=4 b= 6

Value of a and b are given above

oof cheaters cheat because they’re lazy that’s why i cheat xd

: )

step-by-step explanation:

is there any photos or whatever?