Not only Arithmetic, But also Geometric

Algebra Level 2

$a_1, a_2, a_3, \ldots \ldots \ldots$ The above progression of real numbers is both Arithmetic and Geometric.

One can easily see─ so is possible whenever all terms are equal, that is, $0 \neq a_1=a_2=a_3= \ldots \ldots \ldots$ , leaving $0$ as common difference for Arithmetic Nature, and $1$ as common ratio for Geometric Nature.

But is so possible without all terms being equal?

Yes No

1 solution

Muhammad Rasel Parvej
Jan 22, 2018

Let $a, b$ and $c$ be any three consecutive terms of the progression.
As the progression is arithmetic, $c+a=2b$ .
As the progression is geometric, $ca=b^2$ .
Now, $(c-a)^2 = (c+a)^2-4ca=(2b)^2-4b^2=4b^2-4b^2=0 \implies c-a=0 \implies c=a \implies \boxed{a=b=c}$ .

So, the answer is $\boxed{\text{No}}$ .

Can't there be only two different terms in GP/AP. Like for eg. can $2,3$ be a AP and a GP? As there is no third term to check for whether the difference is 'constant' or ratio is 'common'! This is same as saying $p\implies q$ is true whenever $p$ is false ie you have not enough terms to check for whether condition of AP/GP is satisfied or not?

Rishabh Jain - 3 years, 4 months ago

That's why I meant an infinite progression.

Muhammad Rasel Parvej - 3 years, 4 months ago

Not only Arithmetic, But also Geometric

1 solution

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