Let us define a relation R in R as aRb if a ≥ b. Then R is

The correct option is "reflexive, transitive but not symmetric".

Let us analyze the given relation $R$ defined on the set of real numbers $\mathbb{R}$ as:
$aRb \text{if and only if} a\ge b$

To determine the nature of the relation, we test it for three properties: reflexivity, symmetry, and transitivity.

1. Reflexivity:
A relation $R$ on a set $\mathbb{R}$ is reflexive if for every element $a\in \mathbb{R}$, we have $aRa$.
According to the definition of the relation, $aRa$ means:
$a\ge a$
Since any real number is always equal to itself, the inequality $a\ge a$ is mathematically true for all $a\in \mathbb{R}$.
Therefore, the relation $R$ is reflexive.

2. Symmetry:
A relation $R$ on a set $\mathbb{R}$ is symmetric if $aRb$ implies $bRa$ for all $a,b\in \mathbb{R}$.
Let us test this with an example. Suppose $a=5$ and $b=3$:
Since $5\ge 3$, we have $5R3$.
However, $3\ge 5$ is false, which means $3R5$ does not hold true.
Since $aRb$ does not guarantee $bRa$, the relation $R$ is not symmetric.

3. Transitivity:
A relation $R$ on a set $\mathbb{R}$ is transitive if whenever $aRb$ and $bRc$, then we must also have $aRc$ for all $a,b,c\in \mathbb{R}$.
Assume $aRb$ and $bRc$. This gives us two inequalities:
$a\ge b$ and $b\ge c$
Combining these two inequalities, we get:
$a\ge b\ge c\Rightarrow a\ge c$
This satisfies the definition $aRc$.
Therefore, the relation $R$ is transitive.

Combining our findings, the relation $R$ is reflexive, transitive but not symmetric.