# 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. find a,b,c?

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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.**