ARITHMETIC PROGRESSION

Definition of arithmetic progression

A sequence a₁, a₂, a₃,.......an is called an arithmetic progression or arithmetic sequence, if a (n +1) = an + d ,n  N where d is a constant. Here a is called the first term and the constant d is called the common difference.

An arithmetic sequence is also called an Arithmetic Progression (A.P.)

Consider the following pattern:

(i)   2, 6, 10, 14, 18, 22, ............

In the sequence each term after the first term differs from the previous term by a constant. The constant is same in the whole sequence.

here, a = 2 and common difference = 6 - 2 = 4

General form of arithmetic progression

 a, (a+d), (a+2d), (a+3d) ,.........................

here a = first term and d = common difference

How to find common difference in a.p

d = t - t

t₂ = second term, t₁ = first term

General term arithmetic sequence

an = a + (n - 1) d

How to find the total number of terms in an arithmetic sequence?

using the formula n = [(L- a)/d] + 1, we can find the total number of terms of an arithmetic sequence. Here L stands for last term.

Example problems of arithmetic progression

Question 1 :

The first term of an A.P is 6 and the common difference is 5. Find the A.P and its general term.

Solution :

The first term (a) = 6

common difference (d) = 5

General form of A.P is a, (a+d), (a+2d), ...................

a = 6,

a + d = 6 + 5 = 11

a + 2d = 6 + 2(5) = 6 + 10 = 16

The sequence A.P is 6,11,16,...............

General term :

an = a + (n - 1)d

an = 6 + (n - 1) 5 ==> 6 + 5n - 5 ==> 5n + 1

Hence, 6, 11,16,........... is the required sequence and 1 + 5n is the general term.

Question 2 :

Find the common difference and 15th term of an A.P 125 , 120 ,115 , 110 , ……….…. 

Solution :

First term (a) = 125

Common difference (d) = t2 – t1 ==> 120 – 125 ==>  -25

General term of an A.P (an) =  a + (n - 1) d

= 125 + (15 - 1) (-25) ==> 125 + 14 (-25) ==> 125 – 350

t₁₅   = -225

Therefore 15th term of A.P is -225

Question 3 :

Which term of the arithmetic sequence is 24  , 23 ¼ ,22 ½ , 21 ¾ , ………. Is 3?

Solution :

First term (a) = 24

Common difference = t2 – t1 ==> 23 ¼ – 24 ==> (93/4) – 24

d = -3/4

an =  a + (n - 1) d

Let us consider 3 as nth term

an = 3

3 = 24 + (n-1) (-3/4)

3 – 24 = (n-1)  (-3/4)

 (-21 x 4)/(-3) = n -1 ==> 84/3 = n -1 ==> 28 = n – 1 ==> n=29

Hence,3 is the 29th term of A.P.  

Question 4 :

Find the 12th term of the A.P √2 , 3 √2 , 5 √2 , …………

Solution :

First term (a) = √2, 

Common difference = 3 √2 - √2  ==> 2 √2

n = 12

General term of an A.P

an =  a + (n - 1) d

a₁₂ =  √2 + (12 - 1) (2√2)

=  √2 + 11 (2√2)

=  √2 + 22 √2

=  23 √2

Hence, 12th of A.P is 23 √2

Question 5 :

Find the 17th term of the A.P 4 , 9 , 14 ,…………

Solution :

First term (a) = 4

Common difference (d) = 9 - 4 ==> 5

n = 17

General term of an A.P

an =  a + (n - 1) d

an =  4 + (17 - 1) (5) ==> 4 + 16 (5) ==> 84

Hence, 17th of A.P is 84

Question 6 : 

How many terms are there in the following Arithmetic progressions?

-1,-5/6,-2/3,……………10/3

Solution :

First term (a) = -1

Common difference (d) = t2 – t1 =>(-5/6)–(-1) => d  = 1/6

n = [(L-a)/d] + 1

L = 10/3

n = 27

Hence, 27 terms are in the given A.P

Question 7 : 

How many terms are there in the following Arithmetic progressions?

7, 13, 19,……………205

Solution :

First term (a) = 7

Common difference  (d) = t2 – t1 =>13 – 7 = 6

L = 205

n = [(L - a)/d] + 1

n = [(205 - 7)/6] + 1 ==> (198/6) + 1 ==> 34

Hence, 34 terms are in the given A.P

Question 8 :

The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th term

Solution :

10th term = 41 ==>  a + 9 d = 41   ------- (1)

18th term = 73 ==> a + 17 d = 73   ------- (2)

Subtracting the second equation from first equation

 a + 17 d - (a + 9 d) = 73 - 41  

 a + 17d - a - 9d = 32 ==> 8d = 32 ==> d = 4

Substitute d = 4 in the first equation

a + 9 (4) = 41 ==> a + 36 = 41 ==> a = 5

Now, we have to find 27th term          

an =  a + (n - 1) d

here n = 27

=  5 + (27-1) 4 ==> 5 + 26 (4) ==> 5 + 104 ==> 109

Hence, 27th term of the sequence is 109.

Question 9 :

Find n so that the nth terms of the following two A.P’s are the same

1 , 7 ,13 ,19, ………………. and 100 , 95 , 90 ,………..

Solution :

an = a + (n - 1) d

nth term of the first sequence

a = 1   d = t₂-t₁ ==> 7-1 ==> d = 6

an = 1 + (n-1) 6 ==> 1 + 6 n – 6 ==> 6 n – 5   -----(1)

nth term of the second sequence

a = 100   d = t₂-t₁ ==> 95 - 100 ==> -5

an = 100 + (n-1) (-5) ==> 100 - 5 n + 5 ==> 105 - 5 n -----(2)

(1) = (2)

6 n – 5 = 105 – 5 n

6 n + 5 n = 105 + 5

11 n = 110 ==> 110/11 ==> 11

Hence, 11th terms of the given sequence are equal.

Question 10 :

How many two digit numbers are divisible by 13?

Solution :

10, 11, 12,………… 99

Now we need to find how many terms from this sequence are divisible by 13

The first two digit number divisible by 13 is 13; the next two digits number divisible by 13 is 26 and 39 so on. The last two digit numbers which are divisible by 13 is 91.

13 , 26, 39, …………….. 91

Now, we need to find how many terms are there in this sequence for that let us use formula for n.

n = [(L-a)/d] + 1

a = 13, d = 26 – 13 ==> 13,  L = 91

n = [(91 - 13)/13] + 1 ==> (78/13) + 1 ==> 6 + 1 ==>  7

7 two digit numbers are divisible by 13.

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