QJEEM24Ma1481: JEE Main 2024 Mathematics - Class 11 JEEM24Ma1481 Solution: If three successive terms of a G.P. with common ratio are the…

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 $ar^2$ , 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:

$1) \ a + ar > ar^2 \\\\$

2) \ a + ar^2 > ar

3) \ ar + ar^2 > a

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

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

indicating that both $a + ar^2$ and $ar + ar^2$ 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 > ar^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.