Answered

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amc 12a 2014] let a < b < c be three integers such that a, b, c is an arithmetic progression and a, c, b is a geometric progression. what is the smallest possible value of c?

Sagot :

The smallest possible value of c is 1 in this case.

If the integer a, b and c are in arithmetic progression.

Then we can write,

b = (a+c)/2

If the integer a, b and c are in geometric progression,

Then, we can write,

b = √(ac)

Now, we can write,

a+b = 2√(ac)

Squaring both sides,

a²+c²+2ac = 4ac

(a-c)² = 0

So,

a = c.

If a = c.

Then,

b = (a+a)/2

b = a

So, a = b = c.

Smallest possible positive integer is 1.

So, the smallest possible value of c is 1.

To know more about Arithmetic progression, visit,

https://brainly.com/question/6561461

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