JEE Main 2024 — Sequences And Series Question with Solution

To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $a{r}^2$, where $a$ is the first term and $r>1$. These three terms represent the lengths of the sides of a triangle.

According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:

$$1)a+ar>a{r}^2$$

$$2)a+a{r}^2>ar$$

$$3)ar+a{r}^2>a$$

Given that $r>1$, inequalities 2 and 3 will always hold because:

$$ar<a{r}^2\text{and}a<ar,$$

indicating that both $a+a{r}^2$ and $ar+a{r}^2$ will be greater than $ar$ and $a$ respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:

$$a+ar>a{r}^2$$

Simplifying this, we get:

$$a(1+r)>a{r}^2$$

Since $a$ is positive (as it represents the length of a side of a triangle), we can divide both sides of the inequality by $a$ without changing the direction of the inequality:

$$1+r>r^2$$

We can then subtract $r$ from both sides:

$$1>r^2-r$$

Simplifying the right side by factoring $r$:

$$1>r(r-1)$$

Given that $r>1$, the quantity $(r-1)$ is positive; hence, $r(r-1)$ is also positive. This means the actual value for $r$ to satisfy the inequality is within the interval $(1,\sqrt{2})$ because $r(r-1)$ increases with increasing $r$, and it would be 1 when $r=\sqrt{2}$. It should be greater than 1, and less than $\sqrt{2}$ such that $r^2-r$ stays below 1.

Now let’s consider the expressions $[r]$ and $[-r]$. The symbol $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function).

Since $1<r<\sqrt{2}$, $[r]=1$, because 1 is the greatest integer less than $r$ within that interval.

For $[-r]$, we need the greatest integer less than or equal to $-r$. Since $-r$ is negative and less than $-1$ (because $r>1$), $[-r]=-2$, as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\sqrt{2}$ and $-1$).

Now we can substitute these values into the expression:

$$3[r]+[-r]=3\cdot 1+(-2)=3-2=1$$

Therefore $3[r]+[-r]$ is equal to 1.