Question

Need help with my math please. Determine the most probable next term in each list of numbers.

Answer 1

#23

• 1,8,27,64,125

Here we observe

• 1³,2³,3³,4³,5³

Next term

• 6³=216

#24

• 2,6,12,20,30,42

Here we observe

• 2,2+2(2),6+2(3),12+2(4),20+2(5),30+2(6)

Next term is

• 42+2(7)=56

#25

• 4,7,12,19,28,39
• 4+3,7+5,12+7,19+9,28+11

Next term

• 39+13=52

#24

• -1,2,-3,4,-5,6

Next term

• -7

#25

Next term is 5

As previously it had 4 consecutive 5's so this time it has 5 in a row

Answer 2

Answer:

23.  216

24.  56

25.  52

26.  -7

27.  5

Step-by-step explanation:

Question 23

Work out the differences between the terms until the differences are the same:

$1 \underset{+7}{\longrightarrow} 8 \underset{+19}{\longrightarrow} 27 \underset{+37}{\longrightarrow} 64 \underset{+61}{\longrightarrow} 125$

    $7 \underset{+12}{\longrightarrow} 19 \underset{+18}{\longrightarrow} 37 \underset{+24}{\longrightarrow} 61$

        $12 \underset{+6}{\longrightarrow} 18 \underset{+6}{\longrightarrow} 24$

As the third differences are the same, the sequence is cubic and will contain an n³ term.  The coefficient of n³ is always a sixth of the third difference.  Therefore, the coefficient of n³ = 1.

To work out the nth term of the sequence, write out the numbers in the sequence n³ and compare this sequence with the sequence in the question.

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} n&1 & 2 & 3&4&5 \\\cline{1-6} n^3 & 1 & 8 & 27 & 64 & 125 \\\cline{1-6} \sf sequence & 1 & 8 & 27 & 64 & 125 \\\cline{1-6}\end{array}$

Therefore, the nth term is:

$a_n=n^3$

So the next term in the sequence is:

$\implies a_6=6^3=216$

Question 24

Work out the differences between the terms until the differences are the same:

$2 \underset{+4}{\longrightarrow} 6 \underset{+6}{\longrightarrow} 12 \underset{+8}{\longrightarrow} 20 \underset{+10}{\longrightarrow} 30 \underset{+12}{\longrightarrow} 42$

     $4 \underset{+2}{\longrightarrow} 6 \underset{+2}{\longrightarrow} 8 \underset{+2}{\longrightarrow} 10 \underset{+2}{\longrightarrow} 12$

As the second differences are the same, the sequence is quadratic and will contain an n² term.  The coefficient of n² is always half of the second difference.  Therefore, the coefficient of n² = 1.

To work out the nth term of the sequence, write out the numbers in the sequence n² and compare this sequence with the sequence in the question.

$\begin{array}{| c | c | c | c | c | c | c |}\cline{1-7} n&1 & 2 & 3&4&5&6 \\\cline{1-7} n^2 & 1 & 4 & 9 & 16 & 25 &36\\\cline{1-7} \sf operation & +1 & +2 & +3 & +4 & +5 & +6 \\\cline{1-7} \sf sequence & 2 & 6 & 12 & 20 & 30 & 42 \\\cline{1-7}\end{array}$

Therefore, the nth term is:

$a_n=n^2+n$

So the next term in the sequence is:

$\implies a_7=7^2+7=56$

Question 25

Work out the differences between the terms until the differences are the same:

$4 \underset{+3}{\longrightarrow} 7 \underset{+5}{\longrightarrow} 12 \underset{+7}{\longrightarrow} 19 \underset{+9}{\longrightarrow} 28 \underset{+11}{\longrightarrow} 39$

     $3 \underset{+2}{\longrightarrow} 5 \underset{+2}{\longrightarrow} 7 \underset{+2}{\longrightarrow} 9 \underset{+2}{\longrightarrow} 11$

As the second differences are the same, the sequence is quadratic and will contain an n² term.  The coefficient of n² is always half of the second difference.  Therefore, the coefficient of n² = 1.

To work out the nth term of the sequence, write out the numbers in the sequence n² and compare this sequence with the sequence in the question.

$\begin{array}{| c | c | c | c | c | c | c |}\cline{1-7} n&1 & 2 & 3&4&5&6 \\\cline{1-7} n^2 & 1 & 4 & 9 & 16 & 25 &36\\\cline{1-7} \sf operation & +3 & +3 & +3 & +3 & +3 & +3 \\\cline{1-7} \sf sequence & 4&7&12&19&28&39 \\\cline{1-7}\end{array}$

Therefore, the nth term is:

$a_n=n^2+3$

So the next term in the sequence is:

$\implies a_7=7^2+3=52$

Question 26

Given numbers:

-1, 2, -3, 4, -5, 6

As the list of numbers does not increase or decrease, we cannot apply the same method as the previous questions.

$\begin{array}{| c | c | c | c | c | c | c |}\cline{1-7} n&1 & 2 & 3&4&5&6 \\\cline{1-7} \sf list & -1 & 2 & -3 & 4 & -5 & 6 \\\cline{1-7}\end{array}$

From comparing the position of the term to its number in the list, we can see that the nth term is the same as n, but the odd numbers are negative and the even numbers are positive.

Therefore, the rule is:

$\implies a_n=n \quad \textsf{where }n \textsf{ is an even natural number}$

$\implies a_n=-n \quad \textsf{where }n \textsf{ is an odd natural number}$

So the next term in the sequence is:

$\implies a_7=-7$

(since 7 is an odd natural number)

Question 27

Given numbers:

5, 3, 5, 5, 3, 5, 5, 5, 3, 5, 5, 5, 5, 3, 5, 5, 5, 5

The pattern of these numbers appears to be an ascending number of 5s with a 3 in between:

• One 5, followed by a 3
• Two 5s, followed by a 3
• Three 5s, followed by a 3
• Four 5s, followed by a 3

Therefore, the next number of 5s will be five.  This means the next number in the list will be 5, since there are only four 5s after the last 3.

Learn more about sequences here:

brainly.com/question/27953040