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# Arithmetic Progression Type III and IV questions

- March 23, 2018
- Posted by: allexamshelps
- Category: Mathematics

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974 total views, 5 views today

__ __**Arithmetic Progression or Arithmetic Sequence with practical examples:**

**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 12^{th} 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 12^{th} term from the last term i.e., -15, so we can use this logic:

##### Total terms =19

##### Nth term to be found= 12^{th}

##### 19-12=7

##### This means it is the 7^{th} 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 12^{th} 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.

##### (Please leave your comments, to encourage us, to let us know how good/bad we are doing it (Please support your comments with reasons)).

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