A, B and C are terms of an AP and 2^A , 2^B and 2^C are terms of a GP. Are there any integer examples for A, B and C?  

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If A,B,C are the terms of an AP then the middle terms is the arithmetical mean of the neighbor terms:

B = (A+C)/2

We'll cross multiply and we'll get:

2B = (A+B) (1)

If 2^A,2^B,2^C are the terms of a GP then the middle terms is the geometric mean of the neighbor terms:

2^B = sqrt(2^A*2^C)

We'll square raise both sides:

2^2B = 2^A*2^C

Since the bases of the exponentials from the right side are matching, we'll add the exponents:

2^2B = 2^(A+C)

Since the bases are matching, we'll apply one to one property:

2B = (A+B)

According to (1), the relation is true for any consecutive terms of an AP.

We'll choose the terms:

A = 2

We'll choose the common difference as d = 10

B = A + d

B = 2 + 10

B = 12

C = B + d

C = 12 + 10

C = 22

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

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Now, as A, B and C are terms of an AP we know that A+C = 2B.

It is also given that 2^A, 2^B and 2^C are in GP. So we have 2^A*2^C = (2^B) ^2

=> 2^ (A+C) = 2^2B

Now for 2^ (A+C) to be equal to 2^2B, A+C = 2B. This is true for all values of A, B and C as they are in AP

As an example if we take A = 2, B = 5 and C = 8 which are in an AP with first term 2 and common difference 3, 2^A, 2^B and 2^C are equal to 4, 32, 256 respectively which are in a GP with the first term 4 and common ratio 8.

Any three values in AP will make 2^A, 2^B and 2^C form a GP.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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Since A,  B and C are in AP,  the succesuve terms have the same common difference. So, B-A = C-B.

Therefore 2B = A+C.

Since 2^A, 2^B and 2^C are in GP,  the successive tems have the same common ratio. Therefore  2^B/2^A = 2^C / 2^A .

(2^B)^2 = 2^A * 2^c

2^2B = 2^(A+C).

Bases being same, we equate the powers.

2^2B = A+C.

Similarly for x^A , x^B, and x^C ,

(x^B)^2 = x^A*x^C

x^(2B) = x^(A+C) holds as 2B = A+C.

Therefore for any integers in A,B and C  in A P ,  we can choose a base x. Then  x^A, x^B and x^C  are  in GP.

Example:

We take x = 7.  A= 3,   B = 5 and C = 7.

7^3 , 7^5 and 7^7 are in GP with 7^2 as common ratio.

 

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