a,b,c are the consecutive terms of a g.p. a+b+c=124a is the third term of an a.p. b is the thirteenth of an a.p. c is the fifteenth of an a.p....

Since a,b,c, are the consecutive terms of a geometric
progression, we'll apply the average theorem:

b^2  =a*c
(1)

From enunciation, we know
that:

c = a + 12d (2)

b = a +
10d (3)

d is the common difference of the arithmetical
progression, whose terms are a,b,c.

We'll substitute (2)
and (3) in (1):

(a + 10d)^2 = a(a +
12d)

We'll expand the
square:

a^2 + 20ad + 100d^2 = a^2 +
12ad

We'll eliminate a^2 both
sides:

20ad + 100d^2 - 12ad =
0

8ad + 100d^2 = 0

We'll
divide by 4:

2ad + 25d^2 =
0

We'll factorize by d:

d(2a +
25d) = 0

We'll set 2a + 25d =
0

a = -25d/2 (4)

We'll write
the condition from enunctiation, that:

a + b + c =
124

a + a + 12d + a + 10d =
124

3a + 22d = 124 (5)

We'll
substitute (4) in (5):

-75d/2 + 22d =
124

-75d + 44d = 248

-31d =
248

d = -8

3a = 124 +
176

a = 300/3

a =
100

Furthermore, substituting, b = 20 and c =
4.

The terms a,b,c are: a = 100, b = 20 and c
= 4.