Answer :
Alright, let's analyze the sequence [tex]\( u_n = n(n-1) \)[/tex] and identify the positions for which the numbers 30 and 110 appear in this sequence.
1. Finding the position of 30 in the sequence [tex]\( u_n = n(n-1) \)[/tex]:
[tex]\[ u_n = n(n-1) \][/tex]
Let [tex]\( u_n = 30 \)[/tex]:
[tex]\[ n(n-1) = 30 \][/tex]
2. Solving the quadratic equation for 30:
[tex]\[ n^2 - n - 30 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]. The solutions for [tex]\( n \)[/tex] are found using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]:
[tex]\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{1 + 120}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm 11}{2} \][/tex]
This results in two solutions:
[tex]\[ n = \frac{1 + 11}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ n = \frac{1 - 11}{2} = \frac{-10}{2} = -5 \][/tex]
So, the positions where the number 30 appears in the sequence are [tex]\( n = 6 \)[/tex] and [tex]\( n = -5 \)[/tex].
3. Finding the position of 110 in the sequence [tex]\( u_n = n(n-1) \)[/tex]:
[tex]\[ u_n = n(n-1) \][/tex]
Let [tex]\( u_n = 110 \)[/tex]:
[tex]\[ n(n-1) = 110 \][/tex]
4. Solving the quadratic equation for 110:
[tex]\[ n^2 - n - 110 = 0 \][/tex]
Again, this is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -110 \)[/tex]. Using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -110 \)[/tex]:
[tex]\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-110)}}{2 \cdot 1} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{1 + 440}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{441}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm 21}{2} \][/tex]
This results in two solutions:
[tex]\[ n = \frac{1 + 21}{2} = \frac{22}{2} = 11 \][/tex]
[tex]\[ n = \frac{1 - 21}{2} = \frac{-20}{2} = -10 \][/tex]
So, the positions where the number 110 appears in the sequence are [tex]\( n = 11 \)[/tex] and [tex]\( n = -10 \)[/tex].
Summary:
- The number 30 is found in the sequence [tex]\( u_n = n(n-1) \)[/tex] at positions [tex]\( n = 6 \)[/tex] and [tex]\( n = -5 \)[/tex].
- The number 110 is found in the sequence [tex]\( u_n = n(n-1) \)[/tex] at positions [tex]\( n = 11 \)[/tex] and [tex]\( n = -10 \)[/tex].
1. Finding the position of 30 in the sequence [tex]\( u_n = n(n-1) \)[/tex]:
[tex]\[ u_n = n(n-1) \][/tex]
Let [tex]\( u_n = 30 \)[/tex]:
[tex]\[ n(n-1) = 30 \][/tex]
2. Solving the quadratic equation for 30:
[tex]\[ n^2 - n - 30 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]. The solutions for [tex]\( n \)[/tex] are found using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]:
[tex]\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{1 + 120}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm 11}{2} \][/tex]
This results in two solutions:
[tex]\[ n = \frac{1 + 11}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ n = \frac{1 - 11}{2} = \frac{-10}{2} = -5 \][/tex]
So, the positions where the number 30 appears in the sequence are [tex]\( n = 6 \)[/tex] and [tex]\( n = -5 \)[/tex].
3. Finding the position of 110 in the sequence [tex]\( u_n = n(n-1) \)[/tex]:
[tex]\[ u_n = n(n-1) \][/tex]
Let [tex]\( u_n = 110 \)[/tex]:
[tex]\[ n(n-1) = 110 \][/tex]
4. Solving the quadratic equation for 110:
[tex]\[ n^2 - n - 110 = 0 \][/tex]
Again, this is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -110 \)[/tex]. Using the quadratic formula:
[tex]\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -110 \)[/tex]:
[tex]\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-110)}}{2 \cdot 1} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{1 + 440}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm \sqrt{441}}{2} \][/tex]
[tex]\[ n = \frac{1 \pm 21}{2} \][/tex]
This results in two solutions:
[tex]\[ n = \frac{1 + 21}{2} = \frac{22}{2} = 11 \][/tex]
[tex]\[ n = \frac{1 - 21}{2} = \frac{-20}{2} = -10 \][/tex]
So, the positions where the number 110 appears in the sequence are [tex]\( n = 11 \)[/tex] and [tex]\( n = -10 \)[/tex].
Summary:
- The number 30 is found in the sequence [tex]\( u_n = n(n-1) \)[/tex] at positions [tex]\( n = 6 \)[/tex] and [tex]\( n = -5 \)[/tex].
- The number 110 is found in the sequence [tex]\( u_n = n(n-1) \)[/tex] at positions [tex]\( n = 11 \)[/tex] and [tex]\( n = -10 \)[/tex].
