Intro, Lecture 5

Arithmetic Progression Type III and IV questions

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Arithmetic Progression or Arithmetic Sequence: Practical Examples (Type III and Type IV)

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 Arithmetic Progression or Arithmetic Sequence with practical examples:

 

 

Practical Examples of A.P type III and type IV

Type III

 

How many two digits number are divisible by 4?

 

Solution:

 

Note: A division is just the opposite of subtraction. If you divide 10/2 and get 5 as an answer, quotient, you actually have subtracted 2 from 10 five times:
10-2=8, 8-2=6, 6-2=4, 4-2=2 and 2-2=0
If the numerator,10, is completely divisible by the denominator,2, you get a zero,0, in the last attempt or else you get any number in decimal.

 

 

Taken in this way, this question is actually asking how many numbers we have up to 99 with a  common/constant difference of 4.
This forms an A.P because now we have a first term and a common/constant difference.

 

Learn here:
what is an A.P     and
Other two types of questions.
The first number, { a }_{ 1 } =12 because it is the first number with two digits which is divisible by 4, and no other number before is divisible by 4.

 

The last term is 96 because 97, 98 and 99 are not divisible by 4. The number 100 has three digits.

 

12, 16, 20…96

 

It is a finite A.P because it has a last term.

 

Now using the same methodology and formula we learnt to solve, detailed in the previous post Learn here.  If not clicked on the previous link.

 

{ a }_{ L}={ a }_{ 1 }+(n-1)c.d

 

Where

 

{ a }_{ 1 } = the first term

 

c.d= { a }_{ 2 }-{ a }_{ 1 }= { a }_{ 3 }-{ a }_{ 2}

 

n= is either the nth term in the A.P or n is the number of terms in an A.P ( click on links given in this post to learn precisely)

 

Since we are putting the last term on the left-hand side, so N is the number of terms in this A.P.

 

And { a }_{ L } = the last term
96 = 12 + (n-1) 4
96 = 12 + 4n – 4
96 = 8 + 4n
96-8=4n
88/4=n
N=22

 

Hence there are 22 two digits number which are divisible by 4.

Type IV

 

Find the 12th term from the last term in the following A.P
21, 19, 17… -15

 

Firstly find the common difference:

 

c.d= { a }_{ 2 }-{ a }_{ 1 }= { a }_{ 3 }-{ a }_{ 2}

 

Common difference =19-21=17-19= -2

 

The first term, { a }_{ 1 } =21
The last term, { a }_{ L } = the last term= -15
We are putting the last term on the left-hand side because we have to find the number of terms in this A.P first.
Now use the formula, there will be no need to memorize the following formula if you learn it here.
{ a }_{ L}={ a }_{ 1 }+(n-1)c.d

 

-15 = 21 + (n-1) -2
-15  = 21 + (-2n) +2
-15 = 23 + (-2n)
-15 -23 = -2n
-38 = -2n
19=n
N=19

 

 

Hence there are 19 terms in this A.P which exist with a difference of -2 from 21 to -15.

 

Since we have to find the 12th term from the last term i.e., -15, so we can use this logic:

 

Total terms =19

 

Nth term to be found= 12th
19-12=7

 

This means it is the 7th term from the first term i.e., 21. And, we have to find the value of the 7th term.

 

But in the formula we use (n-1). This is why we will use 7+1=8.

 

So n=8

 

{ a }_{ n}={ a }_{ 1 }+(n-1)c.d
{ a }_{ 8}= 21+(8-1)-2
{ a }_{ 8}= 21+7/times-2
{ a }_{ 8}= 21-14
{ a }_{ 8}= 7

 

Hence the 12th term is 7 from the last term.

Short-Cut Method :

 

If we reverse the order of this A.P, the calculation will become easier.

 

-15, -13, -11…21

 

Now the first term is= -15 which was the last term previously.

 

21 is the last term which was the first term previously.

 

c.d= { a }_{ 2 }-{ a }_{ 1 }= { a }_{ 3 }-{ a }_{ 2}

 

Common difference = -13-(-15)=(-11)-(-13)= +2

 

Now the common difference which is 2 is positive, not negative.

 

Now execute the formula with changed terms

 

{ a }_{ L}={ a }_{ 1 }+(n-1)c.d
{ a }_{ 12}= -15 + (12-1) 2
{ a }_{ 12} = -15 + 22
{ a }_{ 12}  = 7

 

Hence 7 is the 12th term in this A.P which exist with a difference of +2 from -15 to 21.

 

 
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