Question

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Complete the sequence

a) 36, 34, 30, 28, 24, __

b) 4, 9, 25, 49, 121, __

c) 5, 15, 45, 135, __
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Answer 1

Answer:

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Step-by-step explanation:

(a)

Between the first two numbers there is a decrease of 2 and then of 4 and this pattern continues.

36 - 34 = 2

34 - 30 = 4

30 - 28 = 2

28 - 24 = 4

Let the number be x

24 - x = 2

Thus x = 22

So the next number in the series is 22.

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(b)

The pattern is the squares of prime numbers.

2² = 4

3² = 9

5² = 25

7² = 49

11² = 121

Similarly, 13² = 169

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(c)

In the sequence, 5, 15, 45, 135 , x we can see the following pattern

15 = 5 x 3

45 = 15 x 3

135 = 45 x 3

X = 135 x 3

=> X = 405

Hence , next term is 405.

Answer 2

Answer

a) 36, 34, 30, 28, 24, 22

b) 4, 9, 25, 49, 121, 169

c) 5, 15, 45, 135, 405

Keys

A group of numbers decided by some order is called a sequence.

Solution

a) It's two separate sequences, so let us divide.

For an odd number term, the numbers starting from 36 decrease by 6.

$a_{2n-1}=42-6n$

For an even number term, the numbers starting from 34 decrease by 6.

$a_{2n}=40-6n$

b) *Square of prime numbers

Each term is a perfect square of a prime number. So, $a_{n}=(p_{n})^{2}$ where $p_{n}$ stands for the nth prime number.

c) Each term is getting multiplied by 5.

$a_{n}=\dfrac{5}{3} \times 3^{n}$

More information

Let's find out how to find all terms until the last one.

We use sum of series here, which is called sigma.

$\displaystyle{\sum^{n}_{k=1}a_{k}}$ means sum of $a_{k}$ from $k=1$ until $k=n$.

For a), we have two different cases.

Remember an odd term and even term is different? Then, you can see it is important whether the series ends in an odd or even term.

[I] $\displaystyle{\sum^{n}_{k=1}a_{k}}$ ends in an even term.

Series sum

$=\displaystyle{\sum^{n}_{k=1}a_{2k-1}}+\displaystyle{\sum^{n}_{k=1}a_{2k}}$

$=n\times \dfrac{(36+42-6n)}{2} +n\times \dfrac{(34+40-6n)}{2}$

$=-6n^2+76n$

[II] $\displaystyle{\sum^{n}_{k=1}a_{k}}$ ends in an odd term.

Subtract even number term from the previous result.

Series sum

$=-6n^2+82n-40$

For c), we have a geometric sum.

The sum of the series is $\dfrac{5(3^{n}-1)}{2}$.