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NNADVOKAT [17]
3 years ago
12

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference​

Mathematics
1 answer:
allsm [11]3 years ago
8 0

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}

so that

a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d

a_n=a_1+(n-1)d

33=8+(n-1)d

21=(n-1)d

n has to be an integer, which means there are 4 possible cases.

Case 1: n-1=1 and d=21. But

\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123

Case 2: n-1=21 and d=1. But

\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123

Case 3: n-1=3 and d=7. But

\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123

Case 4: n-1=7 and d=3. But

\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123

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Luke and Inga are riding the Prince Charming Carousel at Disney World. Luke is on a horse 17 feet from the center. Inga is on a
wlad13 [49]

Answer:

Luke's angular speed = 188.49 rad/sec

Inga's angular speed = 188.49 rad/sec

Step-by-step explanation:

Data provided in the question:

Distance between the center and Luke = 17 feet

Distance between the Inga and center = 25 feet

Frequency of rotation, f = 30 rpm

Now,

Angular speed, ω is given as = 2πf

thus,

ω = 2π × 30

or

ω = 188.49 rad/s

here the angular speed is independent of radius

Therefore,

Luke's angular speed = 188.49 rad/sec

Inga's angular speed = 188.49 rad/sec

7 0
3 years ago
Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (
IrinaVladis [17]

Answer:

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14  and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median  of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = \dfrac{9+9}{2}

median = \dfrac{18}{2}

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = \dfrac{12+13}{2}

Median = \dfrac{25}{2}

Median = 12.5

Low             Q1                Median              Q3                 High

6                  9                     11                      12.5                14

The interquartile range = Q3 - Q1

The interquartile range =  12.5 - 9

The interquartile range = 3.5

7 0
3 years ago
Meh needs some help once again
Ymorist [56]
1:1 because you simplify
7 0
2 years ago
13. Simplify –3√ 2 + 3√ 8 A. 0 B. 3√ 2 C. 9√ 2 D. –3√ 2
valina [46]

Answer:

B. 3√2.

Step-by-step explanation:

  1. -3√2 + 3√4x2
  2. -3√2 + 6√2
  3. 3√2

These are all of the steps to completely, find the correct answer to your question.

Hope this helps!!!

Kyle.

7 0
2 years ago
What is the range of the values shown on the table? What does the range represent?
hjlf

Answer:

Option D)R: {0 ≤ y ≤ 360}; The range represents the number of miles the car can travel

Step-by-step explanation:

The table in the attached figure

Let

x -----> the amount of gas used in gallons (independent variable)

y ----> the number of miles the car can travel (dependent variable)

In this problem

The domain is the interval -----> [0,12]

0\ gal \leq x \leq 12\ gal

The range is the interval ----> [0,360]

0\ miles \leq y \leq 360\ miles

3 0
3 years ago
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