The sum of three different numbers which are consecutive terms of a geometric progression is 28. The numbers are also the first, third and seventh term of an arithmetic progression. Find the three numbers

Answer 1

Not sure

Answer 2

Please answer it quickly it is argent

Answer 3

Use your definitions.
a + ar + ar^2 = 28
a(1+r+r^2) = 28

for the AP information:
the first terms match ----> a=a
the 2nd of the GP = 3rd of AP
ar = a+2d
ar-a = 2d
a(r-1) = 2d **

the 3rd of the GP = 7th of the AP
ar^2 = a + 6d
ar^2 - a = 6d
a(r^2 - 1) = 6d ***

divide *** by **

a(r^1 - 1) / a(r - 1) = 6d / 2d
r+1 = 3
r = 2
then in a(1+r+r^2) = 28
a(1+2+4) = 28
7a = 28
a = 4

now in **
a(r-1) = 2d
4(1) = 2d
d = 2

so the GP is 4, 8, 16, ...
the AP is 4, 6, 8, 10, 12, 14, 16, ...,

notice the first terms match,
the 2nd of the GP is the 3rd of the AP
the 3rd of the GP is the 7th of the AP

I know this is a long solution, and I am sure there is a shorter version, but
I just followed by train of thought, and used your basic definitions.