When the consecutive terms in a sequence increase by a small constant value, we can use the addition rule to find more terms in that pattern.
Ex. \(1, \;3, \;5, \;7, \;9,......\)
When the consecutive terms in a sequence decrease by a small constant value, we can use the subtraction rule to find more terms in that pattern.
Ex. \(16, \;14, \;12, \;10, ...\)
When the increase in consecutive terms of a sequence is in multiples, we can use the multiplication rule to find more terms in that pattern.
Ex. \(3, \;6, \;12, \;24, ...\)
When the decrease in consecutive terms of a sequence is in multiples, we can use the division rule to find more terms in that pattern.
Ex. \(120, \;60, \;30, \; ...\)
A \(7\)
B \(8\)
C \(9\)
D \(10\)
For example:
Consider the following sequence:
\(5, \;11, \;23, \;47,....\)
Let's check it.
Let \(x=\) any term of the pattern
\(5×2=10\\ 11×2=22\\ 23×2=46\)
\((5×2)+1=11\\ (11×2)+1=23\\ (23×2)+1=47\\ \vdots\)
and so on.
Now, it is same as the given sequence.
So, the rule is \(2x+1\)
Multiply each term by \(2\) and add \(1\).
A \(138\)
B \(142\)
C \(130\)
D \(140\)
For example: Consider the following sequence:
\(382,\;190,\;94,\;46,...\)
Let's check it.
Let \(x =\) any term in the pattern
\(\dfrac {382}{2}=191\)
\(\dfrac {190}{2}=95\)
Rule: \(\dfrac {x}{2}-1\)
\(\text {First Term}\rightarrow382\)
\(\dfrac {382}{2}-1=191-1=190\leftarrow\;\text {Second Term}\)
\(\dfrac {190}{2}-1=95-1=94\leftarrow\;\text {Third Term}\)
\(\dfrac {94}{2}-1=47-1=46\leftarrow\;\text {Fourth Term}\)
Now, it is same as the given sequence.
Divide each term by \(2\) and then subtract \(1\) from it.
A \(6\)
B \(8\)
C \(10\)
D \(12\)
The number patterns in which the increase is in powers are called power patterns.
For example: Consider the following sequence:
\(2,\,4,\,8,\,16............\)
Now we have to find the next term in the given sequence.
We know,
\(2=2^1 \\ 4=2^2\\8= 2^3\\ 16=2^4\)
\(2^5=2×2×2×2×2=32\)
A \(81\)
B \(12\)
C \(51\)
D \(72\)
Example:
Consider the following sequence:
\(1, \;4, \;19, \;94,.......\)
Let's check it.
First three terms of the given sequence are \(1\), \(4\) and \(19\).
Let's try by multiplying each term by 5, starting with 1.
\(1×5=5\\ 4×5=20\\ 19×5=95\\ \vdots\)
But it is not same as the given sequence.
We can observe that the products obtained are 1 more than the corresponding terms of the given sequence.
So, multiply each term by 5 and then subtract 1 from it to obtain the next term.
\((1×5)-1=4\\ (4×5)-1=19\\ (19×5)-1=94\\ \vdots\)
and so on.
Now, it is same as the given sequence.
Multiply each term by \(5\) and subtract \(1\) from it.
A \(344\)
B \(270\)
C \(184\)
D \(427\)
For example: Consider the following sequence:
\(164,\;84,\;44,\;24,...\)
Let's check it.
Let \(x =\) any term in the pattern
\(\dfrac {164}{2}=82\)
\(\dfrac {84}{2}=42\)
\(\dfrac {44}{2}=22\)
Rule: \(\dfrac {x}{2}+2\)
\(\dfrac {164}{2}+2=84\)
\(\dfrac {84}{2}+2=44\)
\(\dfrac {44}{2}+2=24\)
and so on.
Now, it is same as the given sequence.
Divide each term by \(2\) and then add \(2\) to it.
A \(51\)
B \(34\)
C \(32\)
D \(22\)
Triangular numbers are the sum of successive counting numbers starting with \(1\).
Following is the triangular number sequence:
\(1,\;3,\;6,\;10,\;15,\;21,\;28,...\)
This can be represented as shown in the figure.
We can write a rule for the triangular numbers.
\(x_n=\dfrac {n(n+1)}{2}\)
Here, \(x_n\) is the value of the \(n^{th}\) term.
\(n\) is the number of terms.
Put \(n=1\) then \(x_1=\dfrac {1(1+1)}{2}=\dfrac {2}{2}=1\)
Put \(n=2\) then \(x_2=\dfrac {2(2+1)}{2}=\dfrac {6}{2}=3\)
Put \(n=3\) then \(x_3=\dfrac {3(3+1)}{2}=\dfrac {12}{2}=6\)
and so on.
Following is the square number sequence:
1, 4, 9, 16, 25,...............
This can be represented as shown.