If three successive terms of a G.P. with common ratio r(r\gt 1) are the lengths of the sides of a triangle and [r] denotes the greatest integer less than or equal to r , then 3[r]+[-r] is equal to \underline\hspace2cm .

Sequences and Series question

2024 · 1 Feb · Shift 2 · Q59

2024 · 1 Feb · Shift 2 · Q59

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Correct answer: 1

Let the three successive terms of the G.P. be $a,\ ar,\ ar^2 \qquad (r>1).$
Since these are the sides of a triangle, they must satisfy the triangle inequality. Because $r>1$ , the largest side is $ar^2$ . So the only nontrivial condition is $a+ar>ar^2.$
Divide by $a>0$ : $1+r>r^2.$ Rearranging, $r^2-r-1<0.$
The roots of $r^2-r-1=0$ are $r=\frac{1\pm\sqrt{5}}{2}.$ Since $r>1$ , we get $1<r<\frac{1+\sqrt{5}}{2}.$
Now $\frac{1+\sqrt{5}}{2}\approx 1.618,$ so $[r]=1.$
Also, since $1<r<2$ , we have $-2<-r<-1,$ hence $[-r]=-2.$
Therefore, $3[r]+[-r]=3(1)+(-2)=1.$

So the required integer is $\boxed{1}.$

2024 · 1 Feb · Shift 2 · Q59

2024 · 1 Feb · Shift 2 · Q59

View written solutionFree

Correct answer: 1

Let the three successive terms of the G.P. be $a,\ ar,\ ar^2 \qquad (r>1).$
Since these are the sides of a triangle, they must satisfy the triangle inequality. Because $r>1$ , the largest side is $ar^2$ . So the only nontrivial condition is $a+ar>ar^2.$
Divide by $a>0$ : $1+r>r^2.$ Rearranging, $r^2-r-1<0.$
The roots of $r^2-r-1=0$ are $r=\frac{1\pm\sqrt{5}}{2}.$ Since $r>1$ , we get $1<r<\frac{1+\sqrt{5}}{2}.$
Now $\frac{1+\sqrt{5}}{2}\approx 1.618,$ so $[r]=1.$
Also, since $1<r<2$ , we have $-2<-r<-1,$ hence $[-r]=-2.$
Therefore, $3[r]+[-r]=3(1)+(-2)=1.$

So the required integer is $\boxed{1}.$