Three numbers in AP and GP

Number Theory Level 2

Can three different unequal numbers a, b and c be together both in an Arithmetic Progression and Geometric Progression.

None of the above No It depends upon the situation in some cases they can be , in some other cases they cannot be. Yes

Srinivasa Gopal by Srinivasa Gopal , 53, India

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2 solutions

Marta Reece
Jul 16, 2018

Numbers in a geometric progression are located on an exponential curve.

Numbers in a arithmetic progression are located on a straight line.

The only way three numbers on an exponential curve can be in a straight line is if the curve is a straight line, that is y = a 1 x = a y=a1^x=a .

But then they are identical, which is not allowed. So there is no way.

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Srinivasa Gopal
Jul 16, 2018

Let the three numbers in an Arithmetic progression be a-d, a and a+d where d is the common difference.

Since these three numbers are also in a Geometric Progression they would be of the form a/r, a and ar where r is the common ratio of the GP

Adding the three numbers expressed as an AP yields a sum of 3a.

Adding the three numbers expressed as GP yields a/r + a + ar = a(1 + 1/r + r)

So 3a = a ( 1 + 1/r + r)

Hence 3 = 1 +1/r + r

1/r + r = 2 Which implies that r = 1 and d = 0 . The only 3 set of numbers for which this property holds true is when all the three numbers are equal.

Since it is asked if three different unequal numbers can have the property of being both in AP and GP , the proposition is false.

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