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Arithmetic Progressions Exercise 5.1

HI Friends and Champs!

Before proceeding to next exercise. let us first understand, what is an arithmetic progression.

An arithmetic progression is a list of numbers, where the difference between any two neighbor numbers is same, for example see the following lists:

1 , 2 , 3 , 4 , 5 ,....................     difference is equal to 1 for any two neighboring numbers.

2 , 4 , 6 , 8 , 10 ...................    difference is 2

5 , 10 , 15 , 20 , 25 ................  difference is 5

5 , 5 , 5 , 5 , 5 ................  difference is 0

these types of list or series are called arithmetic progression.

Note: In arithmetic progression series you can find next number by adding the common difference in the preceding number.

Generally, we represent first number of the arithmetic progression series by "a" and the common difference by "d". Therefore, the general form of arithmetic progression or A.P. is:

a, a+d, a+2d, a+3d,.......

Finite A.P. : A.P. series which have last term .

Example: 1 , 2 , 3 , 4 , 5

Infinite A.P.: A. P. series which do not have last term.

Example: 1 , 2 , 3 , 4 , 5 , 6 ,7 , -------------------------------------  ∞ 

 

                                                                EXERCISE 5.1 

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? 

 (i) The taxi fare after each km when the fare is ` 15 for the first km and ` 8 for each additional km. 

Solution: Yes , 15,23,31,39,47,53,...........

 (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. 

Solution: Let  initial volume of cylinder = v

After 1st removal volume is v-(1/4)v = (3/4)v

After 2nd removal volume is (3/4)v-(1/4)(3/4)v 

                                                =(3/4)v - (3/16)v

                                                =(9/16)v = ( 3/4 )2 v

After 3rd removal volume is (9/16)v - (1/4)(9/16)v

                                                = (9/16)v - (9/64)v

                                                =(27/64)v = = ( 3/4 )3 v

so the series is :

v , ( 3/4 ), ( 3/4 )2v, ( 3/4 )3v ......  

which do not have same common difference, hence not an A.P.

 

 (iii) The cost of digging a well after every metre of digging, when it costs ` 150 for the first metre and rises by ` 50 for each subsequent metre. 

Solution: Yes 

150 , 200, 250, 300, 350, .......................................

 (iv) The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum. 

Solution:

Amount is 10000(1 +  8/100 ) , 10000(1+ 8/100 )2 , 10000(1+ 8/100 )3

which do not have same common difference, hence not an A. P. 

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:

Solution:

(i) a = 10 d=10     (ii) a = -2 d =0     (iii) a = 4 d = -3    (iv) a = 1 d = -1/2     (v) a = -1.25 d = -0.25

(i) a = 10 d=10

10, 20, 30, 40

(ii) a = -2 d =0

-2,  -2, -2, -2

(iii) a = 4 d = -3

4, 1, -2, -5

(iv) a = 1 d = -1/2

1, 1/2, 0, -1/2

(v) a = -1.25 d = -0.25

-1.25, -1.50, -1.75, -2.00

3. For the following APs, write the first term and the common difference: 

 (i) 3, 1, – 1, – 3, . . . (ii)– 5, – 1, 3, 7, . . . (iii) 1/3, 5/3, 9/3, 13/3. . . (iv) 0.6, 1.7, 2.8, 3.9, . . 

Solution :

(i) 3, 1, – 1, – 3, . . 

First Term a = 3

Common Diff. (d) = 1-3 = -2

(ii)– 5, – 1, 3, 7, . . .

a= -5 , d = -1 - (-5) =4

(iii) 1/3, 5/3, 9/3, 13/3. . .

a= 1/3 d = 5/3 - 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, . .

a= 0.6, d= 1.7 - 0.6 = 1.1

 4. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. 

 (i) 2, 4, 8, 16, . . .      (ii) 2, 5/2, 3, 7/2.....    (iii)– 1.2, – 3.2, – 5.2, – 7.2, . . . . . (iv)– 10, – 6, – 2, 2, . . . 

 (v) 3, 3+√2, 3+2√2, 3+3√2,....  (vi) 0.2, 0.22, 0.222, 0.2222, . . . (vii) 0, – 4, – 8, –12, . . . .

 (viii) -½,-½,-½,-½,-½....    (ix) 1, 3, 9, 27, . . .     (x) a, 2a, 3a, 4a, . . .    (xi) a,a2,a3,a4,.... (xii) √2,√8,√18,√32,...    (xiii)√3,√6,√9,√12,...    (xiv) 12,32,52,72,....    (xv) 12,52,72,73....

Solution:

 (i) 2, 4, 8, 16, . . .   

Common diff   is not same between neighbor terms, so not an AP/

(ii) 2, 5/2, 3, 7/2.....

AP,  Common difference (d)=1/2 , First Term (a) = 2    more three terms =4, 9/2, 5 

(iii)– 1.2, – 3.2, – 5.2, – 7.2, . . . . . 

AP, Common difference (d) = -2.0 and First Term (a) = -1.2 more three terms = -9.2, -11.2, -13.2

(iv)– 10, – 6, – 2, 2, . . . 

AP, Common difference (d) = 4 , First Term = -10 more three terms = 6, 10, 14

 (v) 3, 3+√2, 3+2√2, 3+3√2,....  

AP, Common difference (d) = √2, First Term (a) = 3  more three terms = 3+4√2, 3+5√2, 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222, . . . 

Common difference is not same , so not an AP. 

(vii) 0, – 4, – 8, –12, . . . .

AP, Common difference (d) = -4 , First Term = 0   more three terms = -16, -20, -24

 (viii) -½,-½,-½,-½,-½....    

AP, Common difference (d) = 0 , First Term = -½  more three terms = -½,-½,-½

(ix) 1, 3, 9, 27, . . .     

Common difference is not same, so not an AP.

(x) a, 2a, 3a, 4a, . . .    

AP, Common difference (d) = a, First Term (a) = a , more three terms =5a, 6a, 7a

(xi) a,a2,a3,a4,.... 

Common difference is not same, so not an AP.

(xii) √2,√8,√18,√32,...    

Common difference (d) = √2 , First Term (a) = √2 

more three terms = √32 + √2 = 4√2 + √2 = 5√2

            next term=5√2 + √2 =6√2 =√72

            next term=6√2 + √2 =7√2 =√98

 (xiii)√3,√6,√9,√12,...   

Common difference is not same, so not an AP.

 (xiv) 12,32,52,72,....    

Common difference is not same, so not an AP.

(xv) 12,52,72,73....

Common difference is not same, so not an AP.

 

Please share your valuable feedback in comment section.

 

 

 


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