Arithmetic progression

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 a₁ 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 = t₂ - t₁

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.

After having gone through the stuff given above, we hope that the students would have understood "Arithmetic progression".

Apart from the stuff given above, if you want to know more about "Arithmetic progression", please click here

Apart from the stuff "Arithmetic progression" given in this section, if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...

You can also visit our following web pages on different stuff in math.

ALGEBRA