Question

Complete the following patterns:
\[\begin{align}
  & \left( i \right)\text{ 40,35,30, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
 & \left( ii \right)\text{ 0,2,4, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
 & \left( iii \right)\text{ 84,77,70, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
 & \left( iv \right)\text{ 4}\text{.4,5}\text{.5,6}\text{.6, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
 & \left( v \right)\text{ 1,3,6,10, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
 & \left( vi \right)\text{ 1,1,2,3,4,8,13,21, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
 & \left( vii \right)\text{ 1,8,27,64, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } \\
\end{align}\]

Answer 1

Hint: First of all, we should understand the nature of the pattern. The nature of the series can be found by observing the common relation between the terms. First, we should use all the given terms. Then, we should understand the relationship between the terms. Now we should find the remaining terms of the pattern. For example, the first set of terms are 40, 35, 30, …. Now observing them, we get that the terms are decreasing by 5 and it forms an AP. So, the common difference is -5. Now we will use \[{{t}_{n}}=a+\left( n-1 \right)d\] to find the terms, where a is the first term of the AP series, n is the number of terms, and d is a common difference. So, we can fill it as 40, 35, 30, 25, 20, 15.

Complete step-by-step answer:
Before solving the question, we should know that if a is the first of an A.P, d is the common difference of A.P, then \[{{t}_{n}}=a+\left( n-1 \right)d\].
Now let us consider
\[\left( i \right)\text{ 40,35,30, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\]
In this pattern, the first term is equal to 40, the second term is equal to 35 and the third term is equal to 30.
Let us assume the first term is equal to a.
\[\begin{align}
  & {{t}_{1}}=a=40......(1) \\
 & {{t}_{2}}-{{t}_{1}}=35-40=-5 \\
 & {{t}_{3}}-{{t}_{2}}=30-35=-5 \\
\end{align}\]
So, it is clear that the difference between second term and first term is equal to difference between third term and second term.
So, we can say that the common difference is equal to -5.
Let us assume the common difference is equal to d.
\[d=-5.....(2)\]
We know that if a is the first of an A.P, d is the common difference of A.P, then \[{{t}_{n}}=a+\left( n-1 \right)d\].
Now from the pattern, it is clear that we have to find the fourth term, fifth term and sixth term.
\[\begin{align}
  & {{t}_{4}}=40+\left( 4-1 \right)\left( -5 \right) \\
 & \Rightarrow {{t}_{4}}=40+3\left( -5 \right) \\
 & \Rightarrow {{t}_{4}}=40-15 \\
 & \Rightarrow {{t}_{4}}=25 \\
\end{align}\]
\[\begin{align}
  & {{t}_{5}}=40+\left( 5-1 \right)\left( -5 \right) \\
 & \Rightarrow {{t}_{5}}=40+4\left( -5 \right) \\
 & \Rightarrow {{t}_{5}}=40-20 \\
 & \Rightarrow {{t}_{5}}=20 \\
\end{align}\]
\[\begin{align}
  & {{t}_{6}}=40+\left( 6-1 \right)\left( -5 \right) \\
 & \Rightarrow {{t}_{6}}=40+5\left( -5 \right) \\
 & \Rightarrow {{t}_{6}}=40-25 \\
 & \Rightarrow {{t}_{6}}=15 \\
\end{align}\]
So, the filled pattern is
\[\left( i \right)\text{ 40,35,30,25,20,15}\]

Now let us consider
\[\left( ii \right)\text{ 0,2,4, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\]
In this pattern, the first term is equal to 0, the second term is equal to 2 and the third term is equal to 4.
Let us assume the first term is equal to a.
\[\begin{align}
  & {{t}_{1}}=a=0.....(1) \\
 & {{t}_{2}}-{{t}_{1}}=2-0=2 \\
 & {{t}_{3}}-{{t}_{2}}=4-2=2 \\
\end{align}\]
So, it is clear that the difference between second term and first term is equal to difference between third term and second term.
So, we can say that the common difference is equal to 2.
Let us assume the common difference is equal to d.
\[d=2....(2)\]
We know that if a is the first of an A.P, d is the common difference of A.P, then \[{{t}_{n}}=a+\left( n-1 \right)d\].
Now from the pattern, it is clear that we have to find the fourth term, fifth term and sixth term.
\[\begin{align}
  & {{t}_{4}}=0+\left( 4-1 \right)\left( 2 \right) \\
 & \Rightarrow {{t}_{4}}=0+3\left( 2 \right) \\
 & \Rightarrow {{t}_{4}}=0+6 \\
 & \Rightarrow {{t}_{4}}=6 \\
\end{align}\]
\[\begin{align}
  & {{t}_{5}}=0+\left( 5-1 \right)\left( 2 \right) \\
 & \Rightarrow {{t}_{5}}=0+4\left( 2 \right) \\
 & \Rightarrow {{t}_{5}}=0+8 \\
 & \Rightarrow {{t}_{5}}=8 \\
\end{align}\]
\[\begin{align}
  & {{t}_{6}}=0+\left( 6-1 \right)\left( 2 \right) \\
 & \Rightarrow {{t}_{6}}=0+5\left( 2 \right) \\
 & \Rightarrow {{t}_{6}}=0+10 \\
 & \Rightarrow {{t}_{6}}=10 \\
\end{align}\]
So, the filled pattern is
\[\left( ii \right)\text{ 0,2,4,6,8,10}\]
Before solving the question, we should know that if a is the first of an A.P, d is the common difference of A.P, then \[{{t}_{n}}=a+\left( n-1 \right)d\].

Now let us consider
\[\left( iii \right)\text{ 84,77,70, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\]
In this pattern, the first term is equal to 84, the second term is equal to 77 and the third term is equal to 70.
Let us assume the first term is equal to a.
\[\begin{align}
  & {{t}_{1}}=a=84......(1) \\
 & {{t}_{2}}-{{t}_{1}}=84-77=7 \\
 & {{t}_{3}}-{{t}_{2}}=77-70=7 \\
\end{align}\]
So, it is clear that the difference between second term and first term is equal to difference between third term and second term.
So, we can say that the common difference is equal to 7.
Let us assume the common difference is equal to d.
\[d=7.....(2)\]
We know that if a is the first of an A.P, d is the common difference of A.P, then \[{{t}_{n}}=a+\left( n-1 \right)d\].
Now from the pattern, it is clear that we have to find the fourth term, fifth term and sixth term.
\[\begin{align}
  & {{t}_{4}}=84+\left( 4-1 \right)\left( -7 \right) \\
 & \Rightarrow {{t}_{4}}=84+3\left( -7 \right) \\
 & \Rightarrow {{t}_{4}}=84-21 \\
 & \Rightarrow {{t}_{4}}=63 \\
\end{align}\]
\[\begin{align}
  & {{t}_{5}}=84+\left( 5-1 \right)\left( -7 \right) \\
 & \Rightarrow {{t}_{5}}=84+4\left( -7 \right) \\
 & \Rightarrow {{t}_{5}}=84-28 \\
 & \Rightarrow {{t}_{5}}=56 \\
\end{align}\]
\[\begin{align}
  & {{t}_{6}}=84+\left( 6-1 \right)\left( -7 \right) \\
 & \Rightarrow {{t}_{6}}=84+5\left( -7 \right) \\
 & \Rightarrow {{t}_{6}}=84-35 \\
 & \Rightarrow {{t}_{6}}=49 \\
\end{align}\]
So, the filled pattern is
\[\left( iii \right)\text{ 84,77,70,63,56,49}\]

Now let us consider
\[\left( iv \right)\text{ 4}\text{.4,5}\text{.5,6}\text{.6, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\]
In this pattern, the first term is equal to 4.4, the second term is equal to 5.5 and the third term is equal to 6.6.
Let us assume the first term is equal to a.
\[\begin{align}
  & {{t}_{1}}=a=4.4......(1) \\
 & {{t}_{2}}-{{t}_{1}}=5.5-4.4=1.1 \\
 & {{t}_{3}}-{{t}_{2}}=5.5-4.4=1.1 \\
\end{align}\]
So, it is clear that the difference between second term and first term is equal to difference between third term and second term.
So, we can say that the common difference is equal to 1.1.
Let us assume the common difference is equal to d.
\[d=1.1.....(2)\]
We know that if a is the first of an A.P, d is the common difference of A.P, then \[{{t}_{n}}=a+\left( n-1 \right)d\].
Now from the pattern, it is clear that we have to find the fourth term, fifth term and sixth term.
\[\begin{align}
  & {{t}_{4}}=4.4+\left( 4-1 \right)\left( 1.1 \right) \\
 & \Rightarrow {{t}_{4}}=4.4+3\left( 1.1 \right) \\
 & \Rightarrow {{t}_{4}}=4.4+3.3 \\
 & \Rightarrow {{t}_{4}}=7.7 \\
\end{align}\]
\[\begin{align}
  & {{t}_{5}}=4.4+\left( 5-1 \right)\left( 1.1 \right) \\
 & \Rightarrow {{t}_{5}}=4.4+4\left( 1.1 \right) \\
 & \Rightarrow {{t}_{5}}=4.4+4.4 \\
 & \Rightarrow {{t}_{5}}=8.8 \\
\end{align}\]
\[\begin{align}
  & {{t}_{6}}=4.4+\left( 5-1 \right)\left( 1.1 \right) \\
 & \Rightarrow {{t}_{6}}=4.4+4\left( 1.1 \right) \\
 & \Rightarrow {{t}_{6}}=4.4+5.5 \\
 & \Rightarrow {{t}_{6}}=9.9 \\
\end{align}\]
So, the filled pattern is
\[\left( iv \right)\text{ 4}\text{.4,5}\text{.5,6}\text{.6,7}\text{.7,8}\text{.8,9}\text{.9}\]


Let us consider
\[\left( v \right)\text{ 1,3,6,10, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\]
In this pattern, the first term is equal to 1, the second term is equal to 3, the third term is equal to 6 and the fourth term is equal to 10.
Let us assume the first term is equal to a.
\[\begin{align}
  & {{t}_{1}}=a=1......(1) \\
 & {{t}_{2}}-{{t}_{1}}=3-1=2 \\
 & {{t}_{3}}-{{t}_{2}}=6-3=3 \\
 & {{t}_{4}}-{{t}_{3}}=10-6=4 \\
\end{align}\]
Here, it is clear that the difference between two consecutive terms is not the same. So, we can not say that the given pattern is in A.P.
The difference between two consecutive terms is increasing.
So, let us consider the difference between consecutive terms as follows:
\[\begin{align}
  & {{d}_{1}}=2 \\
 & {{d}_{2}}=3 \\
 & {{d}_{3}}=4 \\
\end{align}\]
Here, it is clear that the difference between two consecutive terms are in A.P.
Because, the common difference between these terms is equal to 1.
So, we can say that
\[\begin{align}
  & {{d}_{4}}=5 \\
 & \Rightarrow {{t}_{5}}-{{t}_{4}}=5 \\
 & \Rightarrow {{t}_{5}}-10=5 \\
 & \Rightarrow {{t}_{5}}=15 \\
 & {{d}_{5}}=6 \\
 & \Rightarrow {{t}_{6}}-{{t}_{5}}=6 \\
 & \Rightarrow {{t}_{6}}-15=6 \\
 & \Rightarrow {{t}_{6}}=21 \\
 & {{d}_{6}}=7 \\
 & \Rightarrow {{t}_{7}}-{{t}_{6}}=7 \\
 & \Rightarrow {{t}_{7}}-21=7 \\
 & \Rightarrow {{t}_{7}}=28 \\
\end{align}\]
So, the filled pattern is
\[\left( v \right)\text{ 1,3,6,10,15,21,28}\].

Let us consider
\[\left( vi \right)\text{ 1,1,2,3,4,8,13,21, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\]
Here, the first term is 1, second term is 1, third term is 2, fourth term is 3, fifth term is 4, sixth term is 8, seventh term is 13 and eight term is 21.


If we observe, the above pattern is a Fibonacci series.
In this series, term \[{{t}_{n}}\]is equal to sum of \[{{t}_{n-1}}\] and \[{{t}_{n-2}}\].
\[\begin{align}
  & {{t}_{9}}={{t}_{8}}+{{t}_{7}}=21+13=34 \\
 & {{t}_{10}}={{t}_{9}}+{{t}_{8}}=34+8=42 \\
 & {{t}_{11}}={{t}_{10}}+{{t}_{9}}=42+34=76 \\
\end{align}\]
So, the filled pattern is

Now let us consider
\[\left( vii \right)\text{ 1,8,27,64, }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ , }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\]
Here the first term is equal to 1, second term is equal to 8, third term is equal to 27 and fourth term is equal to 64.
We know that
\[\begin{align}
  & {{1}^{3}}=1 \\
 & {{2}^{3}}=8 \\
 & {{3}^{3}}=27 \\
 & {{4}^{3}}=64 \\
\end{align}\]
So, we can say that \[{{t}_{n}}={{n}^{3}}\].
\[\begin{align}
  & {{t}_{5}}={{5}^{3}}=125 \\
 & {{t}_{6}}={{6}^{3}}=216 \\
 & {{t}_{7}}={{7}^{3}}=343 \\
\end{align}\]
So, the filled sequence is
\[\left( vii \right)\text{ 1,8,27,64,125,216,343}\]

Note: We should be careful while observing the pattern. We cannot apply A.P formulas for Fibonacci series and we cannot apply Fibonacci series concept for A.P. So, we should apply the respective formulae for each series. If a small mistake is done, we cannot find the correct terms of the pattern to get filled. So, the concept should be clear for a student to have a correct answer.