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zlopas [31]
2 years ago
6

Here is a rule for generating a sequence; 3n+10 What is the 150th term of the sequence generate by this rule?​

Mathematics
1 answer:
slava [35]2 years ago
3 0

Answer:

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

<h3><em>Topic </em><em>-</em><em> </em><em>Arithmetic</em><em> </em><em>Progression </em><em>(</em><em> </em><em>AP </em><em>)</em></h3>

  • Given - <u>a </u><u>general </u><u>term </u><u>for </u><u>a </u><u>sequence.</u>

  • To find - <u>the </u><u>1</u><u>5</u><u>0</u><u>t</u><u>h</u><u> </u><u>term </u><u>of </u><u>the </u><u>given </u><u>sequence</u>

so let's start ~

General term , A_{n} = 3n + 10

let's consider certain values of n to get some information with us which will help us solve the problem further !

let's first consider , n = 1 ! then ,

3n  + 10 = 3(1)  + 10 \\ \dashrightarrow \:  13

now , let's consider n = 2

3n + 10 = 3(2) + 10 = 6 + 10 \\ \dashrightarrow \: 16

then , let's consider n = 3

3n  + 10 = 3(3) + 10 = 9 + 10 \\ \dashrightarrow \: 19

hence ,

now we've with us an AP which is as follows -

13 \: , \: 16 \: , \: 19 \: ......

from this Arithmetic Progression ,

we can know that

a = first term = 13

d = common difference = 16 - 13 = 19 - 16 = <u>3</u>

now ,

A _{n} = a + (n - 1)d \\ \implies \: A _{150} = 13 + (149)(3) \\ \implies \: A _{150} = 13 + 447 \\  \pink{\implies \: A  _{150} = 460}

hope helpful :D

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The mean absolute deviation of the scuba diver’s depth over the entire 25 min is 2.4 meters

<h3>What is mean absolute deviation?</h3>

It is defined as the measure to show the variation in data set in other words between the mean and every data value, the distance known as the MAD.

We have depths of the scuba diver, in meters

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The frequency of the musical note C4 is about 261.63 Hz. What is the frequency of the note a perfect fifth below C4?
kolbaska11 [484]

Answer:

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B- 174.42 Hz

Step-by-step explanation:

Here we note that to get the "perfect fifth" of a musical note  we have to play a not that is either 1.5 above or 1.5 below the note to which we reference. Therefore to get the frequency of the note a perfect fifth below C4 which is about 261.63 Hz, we have

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