# NCERT Solution Exercise 5.1 Arithmetic progressions Class-10

Arithmetic progressions: Basic concepts

Sequence: Some numbers arranged in a definite order, according to a definite rule, are said to form a sequence.

ex   (I) 1, 2, 3,………

(II) 100, 70, 40, 10 ………

(III) -1·0, -1·5, -2·0, -2·5, …….

Arithmatic progressions (AP):

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

→ this fixed number is called the common difference of A.P.

→ common differences can be positive, negative or zero

Exercise – 5·1

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

(I) the taxi fare after each km when the fare is rupees 15 for the first km and rupees 8 for each additional km.

Solution:

yes, it is in A.P.

15, 23, 31, ……… each succeeding term is obtained by adding 8 in its preceding.

(II) the amount of air present in a cylinder when a vacuum pump removes 1/4 of the additional km.

Solution:

No. volume: v, 3/4v, (3/4)² …….

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

Solution:

yes, 150, 200, 250 —– succeeding term is obtained by adding 50 in its preceding.

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

Solutions:

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

(I) a=10 , d=10

Solution:

a2 = a+d = 10+10 =20

a= a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+d = 40+10 = 50

(II) a=-2,d=0

Solution:

a2 = a+d = -2+0 =-2

a= a2+d = -2+0 = -2

a4 = a3+d = -2+0 =-2

a5 = a4+d = -2+0 =-2

(III) a=4 , d=-3

Solution:

a2 = a+d = 4-3 = 1

a= a2+d =1-3 = -2

a4 = a3+d =-2-3 =-5

a5 = a4+d = -5-3 =-8

(IV) a=-1 , d=1/2

Solution:

a2 = a+d = -1+1/2 = -2+1/2=-1/2

a= a2+d =-1/2+1/2=0

a4 = a3+d =0+1/2=1/2

a5 = a4+d =1/2+1/2 = 2/2=1

(V) a=-1·25 , d=-0·25

Solution:

a2 = a+d =-1·25+(-0.25)= -1·25-0·25=-1·50

a= a2+d =-1·50+(-0·25) = -1·50-0·25=-1·75

a4 = a3+d =-1·75+(-0·25) = -1·75-0·25=-2·00

a5 = a4+d =-2·00+(-0·25) = -2·00-0·25=-2·25

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

(I) 3,1,-1,-3, …..

Solution:

first term =a=3

common difference =1-3 = -2

(II) -5,-1,3,7, …..

Solution:

first term =-5

common difference = -1-5 = -6

(III) 1/3,5/3,9/3,13/3, …..

Solution:

first term = 1/3

common difference = 5/3-1/3=4/3

(IV) 0·6,1·7,2·8,3·9, …..

Solution:

first term =a = 0·6

common difference = 1·7-0·6 =1·1

Q.4 Which of the following are APS? if they form an A.P., find the common difference d write three more terms.

(I) 2,4,8,16,……..

Solution:

first term (a) = 2

d1=4-2=2

d2=8-4=4

d3=16-8=8

d1≠d2≠d3

No above sequence is not in A.P.

(II) 2,5/2,3,7/2,…..

Solution:

first term a=2

d1=5/2-2=5-4/2=1/2

d2=3-5/2=6-5/2=1/2

d3=7/2-3=7-6/2=1/2

d=d1=d2=d3

common difference d=-1/2

yes, it is in A.P.

a5=a4+d=7/2+1/2=8/2=4

a6=a5+d=4+1/2=8+1/2=9/2

a7=a6+d=9/2+1/2=10/2=5

a8=a7+d=5+1/2=10+1/2=11/2

Hence next three terms = 4,9/2,5,11/2,……

(III) -1·2,-3·2,-5·2,-7·2,…….

Solution:

d1=-3·2-(-1·2)=-3·2+1·2=-2·0

d2=-5·2-(-3·2)=-5·2+3·2=-5·2+3·2=-2·0

d3=-7·2-(-5·2)=-7·2+5·2=-2·0

here d=d1=d2=d3=-2·0

Hence yes above sequence form an A.P. next three terms

a5=a4+d=-7·2+(-2·0)=-7·2-2·0=-9·2

a6=a5+d=-9·2+(-2·0)=-9·2-2·0 =-11·2

a7=a6+d=-11·2+(-2·0)=-11·2-2·0=-13·2

(IV) -10,-6,-2,2…….

d1=-6-(-10)=-6+10=4

d2=-2-(-6)=-2+6=4

d3=2-(-2)=2+2=4

d=d1=d2=d3=4

yes, it is in A.P.

common difference =4

next three terms.

a5=a4+d=2+4=6

a6=a5+d=6+2=8

a7=a6+d=8+2=10

(V) 3,3+√2,3+2√2,3+3√2,……

d1=3+√2 – 3 = √2

d2=3+2√2 – (3+√2)= 3+2√2 – 3 -√2 = √2

d3=3+3√2 – (3+2√2) = 3+3√2 – 3 -2√2 = √2

d=d1=d2=d3=√2

Yes above sequence form an A.P. and the next three terms are:

a5=a4+d= 3+3√2+√2 = 3+4√2

a= a5+d= 3+4√2+√2 = 3+5√2

a7=a6+d= 3+5√2 + √2 = 3+6√2

(VI) 0.2,0.22,0.222,0.2222,…….

d1= 0.22 – 0.2= 0.20

d2= 0.222 – 0.22= 0.002

d1≠d2

No above sequence is not in A.P.

(VII) 0,-4,-8,-12,…..

d1=-4 – 0 = -4

d2= -8-(-4) = -8 + 4=-4

d3=-12-(-8)=-12+8=-4

d=d1=d2=d3=-4

Yes above sequence form an A.P. and the next three terms are:

a5=a4+d= -12+(-4)=-12-4 = -16

a= a5+d= -16+(-4)= -16-4 =-20

a7=a6+d= -20+(-4)=-20-4=-24

(VIII) -1/2,-1/2,-1/2,-1/2

d1=-1/2 – (-1/2)=-1/2+1/2=0

d2= -1/2 – (-1/2)=-1/2+1/2=0

d3=-1/2 – (-1/2)=-1/2+1/2=0

d=d1=d2=d3=0

Yes above sequence form an A.P. and the next three terms are:

a5=a4+d= -1/2+0=-1/2

a= a5+d= -1/2+0=-1/2

a7=a6+d= -1/2+0=-1/2

(IX) 1,3,9,27,…….

d1=3-1=2

d2=9-3=6

d1≠d2

No above sequence is not in A.P.

(X) a,2a,3a,4a,…….

d1=2a-a=a

d2=3a – 2a=a

d3=4a-3a=a

d=d1=d2=d3=a

Yes above sequence form an A.P. and the next three terms are:

a5=a4+d= 4a+a=5a

a= a5+d=5a+a=6a

a7=a6+d=6a+a=7a

(XI) a,a²,a³,a4,……

d1=a² – a =a(a-1)

d2=a³ – a² =a²(a-1)

d1≠d2

No above sequence is not in A.P.

(XII)√2,√8,√18,√32,…

d1=√8-√2=2√2-√2=√2

d2=√18-√8=√(3×3×2)-√(2×2×2)=3√2-2√2=√2

d3=√32-√18=√(4×4×2)-3√2=4√2-3√2=√2

d=d1=d2=d3=√2

Yes above sequence form an A.P. and the next three terms are:

a5=a4+d=√32+√2=√64

a= a5+d=√64+√2=√128

a7=a6+d=√128+√2=√256

(XIII)√3,√6,√9,√12,….

d1=√6-√3

d2=√9-√6

d1≠d2

No above sequence is not in A.P.

(XIV) 1²,3²,5²,7²,…..

d1=3²- 1²=9-1=8

d2=5²-3²=25-9=16

d3=7²- 5²=49-25=24

d1≠d2≠d3

No above sequence is not in A.P.

(XV) 1²,3²,5²,73,…..

d1=3²- 1²=9-1=8

d2=5²-3²=25-9=16

d3=5²- 73=25-73=-48

d1≠d2≠d3

No above sequence is not in A.P