All categories

3 years ago

Ind an expression for the nth term (t_n) in each of the following

Mathematics

1 answer:

Mazyrski [523]3 years ago

5 0

<h3>Answer:</h3>

$\boxed{\pink{\tt\longmapsto The \ nth \ term \ of \ the \ sequence\ is \ given \ by \ frac{1}{n}}}$

<h3>Step-by-step explanation:</h3>

A sequence is given to us is , and we need to find expression for nth term of the sequence. The given sequence is ,

$\bf 1 , \dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6}$

When we flip the numbers , we can clearly see that the number are in Harmonic Progression.

That is when we flip the numbers they are found to be in Arithmetic Progression .

And nth term of an Harmonic Progression is :-

$\boxed{\purple{\bf T_{(Harmonic \ Progression)} = \dfrac{1}{a+(n-1)d}}}$

Hence here

Common difference = 1
First term = 1 .

Substituting the respective values,

$\bf\implies T_{n}= \dfrac{1}{a+(n-1)d} \\\\\bf\implies T_{n} = \dfrac{1}{1+(n-1)1}\\\\\bf\implies T_n = \dfrac{1}{1+n-1}\\\\\bf\implies\boxed{\red{\bf T_{n}=\dfrac{1}{n}}}$

<h3>Hence the nth term of the given sequence is given by ¹/n . </h3>

You might be interested in

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago

How does knowing the double 6+6=12 help solve the near double 6+7=13?

Yakvenalex [24]

Because, if 6 + 6 = 12, then in 6 + 7, 7 can be splitted into 6 + 6 + 1, and so you find it easier...!!

7 0

3 years ago

Explain how you could use the distributive property and mental math to find 5 * 198.

bogdanovich [222]

5 * 198 = 990

198
* 5
-------
990

6 0

3 years ago

A regulation football field is 300 feet by 160

igomit [66]

Answer:

366.7424164 feet

Step-by-step explanation:

330x160

Use pythag theorem

a²+b²=c²

330²+160²=c²

c=366.7424164 feet

5 0

2 years ago

A nurse collected data about the average birth weight of babies in the hospital that month. Her data is shown using the dot plot

mixer [17]

Answer:

The values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

Step-by-step explanation:

The data provided is as follows:

X Frequency

8 0

8.1 0

8.2 2

8.3 2

8.4 2

8.5 3

8.6 0

8.7 1

8.8 0

8.9 1

9 0

So, the actual data is:

S = {8.2 , 8.2 , 8.3 , 8.3 , 8.4 , 8.4 , 8.5 , 8.5 , 8.5 , 8.7 , 8.9 }

A boxplot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

Minimum (shown at the bottom of the chart)
First Quartile (shown by the bottom line of the box)
Median (or the second quartile) (shown as a line in the center of the box)
Third Quartile (shown by the top line of the box)
Maximum (shown at the top of the chart).

The data set is arranged in ascending order.

The minimum value is, Min. = 8.2

The lower quartile is,

$[\frac{n+1}{4}]^{th}\ obs.=[\frac{11+1}{4}]^{th}\ obs.=3^{rd }\ obs. =8.3$ ,

Q₁ = 8.3.

The median value is,

$[\frac{n+1}{2}]^{th}\ obs.=[\frac{11+1}{2}]^{th}\ obs.=6^{th}\ obs.=8.4$

Median = 8.4

The upper quartile is,

$[\frac{3(n+1)}{4}]^{th}\ obs.=[\frac{3(11+1)}{4}]^{th}\ obs.=9^{th }\ obs. =8.5$ ,

Q₃ = 8.5.

The maximum value is, Max. = 8.9.

So, the values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.

5 0

3 years ago

Ind an expression for the nth term (t_n) in each of the following​

Ind an expression for the nth term (t_n) in each of the following