Intro, Lecture 4

Arithmetic Progression or Arithmetic Sequence

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Arithmetic Progression or Arithmetic Sequence: Practical Examples (Type I and Type II)

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 Arithmetic Progression or Arithmetic Sequence with practical examples:

 

 

Practical Examples of A.P (Type 1):

 

If the 4th term of an A.P is 8 and the 7th term is 14, then find the three terms of this A.P.

 

Solution:

 

For the 4th term, we can write    (using the formula we developed in the previous post: learn here
{ a }_{ 1 }+3c.d=8 and
For the 7th term, we can write
{ a }_{ 1 }+6c.d=14 or
{ a }_{ 1 }=14-6c.d

 

Now { a }_{ 1 } is equal to 14-6c.d, this means we can use this value of { a }_{ 1 } in the first equation to find the common or constant difference (c.d).

 

The law of mathematics says whenever we have two unknowns in a single equation, we need two equations of the same type to find their values. In our example, we have two unknowns. First is { a }_{ 1 } and the second is the common or constant difference (c.d). This is why, we have made { a }_{ 1 } is equal to 14-6c.d. On solving, we will directly find the value of the constant difference (c.d).
{ a }_{ 1 }+3c.d=8
In the place of { a }_{ 1 }, we have to put -6c.d.
14-6c.d+3c.d=8
14-3c.d=8
-3c.d= 8-14
-3c.d= -6
c.d= \frac { 6 }{ 3 } =2

 

Now any of the two above equations can be used to find the value of { a }_{ 1 }.

 

For example, if we use the first equation by putting the value of c.d=2
{ a }_{ 1 }+3\times2 =8
{ a }_{ 1 }+6 =8
{ a }_{ 1 }=8-6
{ a }_{ 1 }= 2
So { a }_{ 1 }= 2

 

For example, if we use the second equation using the same method as above to do the same, the answer will not change. You can check it.

 

Now we have { a }_{ 1 }= 2, and c.d=2
So we can generate the first three terms of this A.P

 

2, 4, 6
Hence this is the answer.

 

 

Practical Examples of A.P (Type II)

 

 

Find out whether the term 217 is in the following A.P or not.
7, 14, 21, 28…

 

The examples have been intentionally kept simple to make understanding easier.

 

First find the common difference, c.d     
   
 ( It was made clear in the previous post that how to find the common difference, and other prerequisite knowledge required to understand and solve problems given in the Arithmetic Progression.) Learn here 
14-7=7
21-14=7
Hence  the common difference, c.d=7 and the first term is { a }_{ 1 }= 7
{ a }_{ n }={ a }_{ 1 }+(n-1)c.d
Where n is unknown and all other terms are known.

 

Note: If we had two unknowns here, we would have needed two equations of the same type to solve.

 

By putting values:
217=7+(n-1)7
217=7+7n-7
217=7+7n-7
217=7n
\frac { 217 }{ 7 } =31
Hence, 217 is in this A.P and it is the 31st term.

 

Practical Examples of A.P (Type II (i)):

 

Similarly, if we are to find whether 285 is  a term of the following A.P

7, 14, 21, 28…

Or not

 

Then, using the same RHS values as in the above method :  
7+(n-1)7
 (Because both A.Ps are the same, only the left-hand side variable is changed).

 

285 = 7+(n-1)7
285= 7+7n-7
285=7+7n-7
285=7n
\frac { 285 }{ 7 } =4.71
The term 285 is not on the list of this A.P because the common/constant difference, in our case, is a positive integer which is 7.

 

The terms in an A.P are obtained with a speed of the common difference/ constant difference. If the common difference is a negative integer, then the A.P will move in the left direction and the terms of that A.P sooner or later will become negative. If the common/constant difference is a real number(Fractions or otherwise), then it can also move with a difference of few decimal points or any other common difference depending on the value of that real number which is the common difference in that A.P.

 

But, in our case, the c.d is a positive integer, each subsequent(next) term will appear with this difference. That is, it has to move with a difference of a positive integer i.e., 7.

 

Hence the answer is that 285 is not on the list of this A.P.
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