NCERT Solutions for Class 10 Math Exercise 5 Arithmetic Progressions | Arithmetic Series

NCERT Solutions for Class 10 Math Exercise 5 Arithmetic Progressions Arithmetic Series class 10th maths chapter 5 class 10 maths. Here We learn what is in class 10 maths solution chapter 5 arithmetic progression and how to solve questions with easiest method. In this chapter we solve the question of NCERT Solutions for Class 10 Maths chapter 5 exercise 5.1, class 10 maths chapter 5 exercise 5.2, class 10 maths chapter 5 exercise 5.3 and class 10 maths chapter 5 exercise 5.4. chapter 5 class 10 maths Arithmetic Series
10 maths ncert solutions Chapter 5 arithmetic progression are part of NCERT Solutions for Class 10 Maths PDF . Here we have given NCERT Solutions for ncert class 10, ncert solutions for class 10 maths. Ncert solutions for class 10 maths chapter 5 arithmetic progression with formula and solution. chapter 5 class 10 maths Arithmetic Series

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ncert solutions for class 10 maths Chapter 5 arithmetic progression
class 10th maths Arithmetic Series

Exercise – 5
arithmetic progression Arithmetic Series

ncert solutions for class 10 maths Chapter 5 arithmetic progression
class 10th maths solution ex 5.1

Arithmetic Progression (A.P.) – 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.

Common Difference (d) – This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero.
It is denoted by ‘d‘.

Denote the first term of an AP by a1, second term by a2, . . . , nth term by an and the common difference by d. Then the AP becomes a1, a2, a3, . . . , an.
So, a2a1 = a3a2 = . . . = anan – 1 = d.

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.
(ii) The amount of air present in a cylinder when a vacuum pump removes \(\displaystyle \frac{1}{4}\) of the air remaining in the cylinder at a time.
(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.
(iv) The amount of money in the account every year, when $ 10000 is deposited at compound interest at 8 % per annum.

Solutions :-
(i) The taxi fare after each km when the fare is $ 15 for the first km and $ 8 for each additional km.
This list will be in A. P.
first kilometer fare = $ 15
for each additional km = $ 8
so, second kilometer fare = 15 + 8 = $ 23
third kilometer fare = 23 + 8 = $ 31
So, 15, 23, 31, . . . let us form an A. P. because the common difference of all is 8.

(ii) The amount of air present in a cylinder when a vacuum pump removes \(\displaystyle \frac{1}{4}\) of the air remaining in the cylinder at a time.
Let the initial volume of air in the cylinder = V
the first time the pump air comes out = \(\displaystyle \frac{V}{4}\)
remaining air in the cylinder = \(\displaystyle \left( {V-\frac{V}{4}} \right)=\frac{{3V}}{4}\)
The second time the pump will blow out \(\displaystyle \frac{1}{4}\) of \(\displaystyle \frac{3V}{4}\) = \(\displaystyle \frac{3V}{16}\) the air.
so the remaining air = \(\displaystyle \frac{{3V}}{4}-\frac{{3V}}{{16}}=\frac{{12V-3V}}{{16}}=\frac{{9V}}{{16}}={{\left( {\frac{3}{4}} \right)}^{2}}V\)
Then a1 = V
a2 = \(\displaystyle \frac{3V}{4}\)
a3 = \(\displaystyle {{\left( {\frac{3}{4}} \right)}^{2}}V\)
Difference of two consecutive terms =
a2a1 = \(\displaystyle \frac{{3V}}{4}-V=-\frac{V}{4}\)  
a2a2 = \(\displaystyle {{\left( {\frac{3}{4}} \right)}^{2}}V-\frac{{3V}}{4}=\frac{3}{4}V\left( {\frac{3}{4}-1} \right)=\frac{3}{4}V\times \left( {-\frac{1}{4}} \right)=-\frac{3}{{16}}V\)
The difference of two consecutive terms is not fixed, so the volume of air is not in 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.
Cost of first meter = $ 150
Second meter excavation cost = 150 + 50 = $ 250
Third meter excavation cost = 250 + 50 = $ 300
Arithmetic Progression (A.P.) − 150, 200, 250, . . .
Common Difference = a2a1 = 200 – 150 = 50
a3a2 = 250 − 200 = 50
So all have the same common difference, so the series is A. P.

(iv) The amount of money in the account every year, when $ 10000 is deposited at compound interest at 8 % per annum.
First year amount = $ 10000
Rate of compound interest = 8%
Amount in account after one year a1 = \(\displaystyle 10000\left( {1+\frac{8}{{100}}} \right)\)
Amount in account after two year a2 = \(\displaystyle 10000{{\left( {1+\frac{8}{{100}}} \right)}^{2}}\)
Amount in the account after three years a3 = \(\displaystyle 10000{{\left( {1+\frac{8}{{100}}} \right)}^{3}}\)
common difference = a2a1
\(\displaystyle \begin{array}{l}=10000{{\left( {1+\frac{8}{{100}}} \right)}^{2}}-10000\left( {1+\frac{8}{{100}}} \right)\\=10000\left( {1+\frac{8}{{100}}} \right)\left( {1+\frac{8}{{100}}-1} \right)\\=10000\left( {1+\frac{8}{{100}}} \right)\left( {\frac{8}{{100}}} \right)\end{array}\)
a3a2
\(\displaystyle \begin{array}{l}=10000{{\left( {1+\frac{8}{{100}}} \right)}^{3}}-10000{{\left( {1+\frac{8}{{100}}} \right)}^{2}}\\=10000{{\left( {1+\frac{8}{{100}}} \right)}^{2}}\left( {1+\frac{8}{{100}}-1} \right)\\=10000{{\left( {1+\frac{8}{{100}}} \right)}^{2}}\left( {\frac{8}{{100}}} \right)\end{array}\)
a2a1a3a2 is not equal, Hence it is not in A.P.

2. Write 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 (ii) a = − 2, d = 0
(ii) a = 4, d = − 3 (iv) a = − 1, d = \(\displaystyle \frac{1}{2}\)
(v) a = − 1.25, d = − 0.25  

(i) a = 10, d = 10
First Term a1 = 10
Common Difference d = 10
Second Term a2 = a1 + d = 10 + 10 = 20
Third Term a3 = a2 + d = 20 + 10 = 30
Fourth Term a4 = a3 + d = 30 + 10 = 40
So, first four term of A. P. are = 10, 20, 30, 40

(ii) a = − 2, d = 0
First Term a1 = − 2
Common Difference d = 0
Second Term a2 = a1 + d = − 2 + 0 = − 2
Third Term a3 = a2 + d = − 2 + 0 = − 2
Fourth Term a4 = a3 + d = − 2 + 0 = − 2
So, first four term of A. P. are = − 2, − 2, − 2, − 2

(iii) a = 4, d = − 3
First Term a1 = 4
Common Difference d = 3
Second Term a2 = a1 + d = 4 − 3 = 1
Third Term a3 = a2 + d = 1 − 3 = − 2
Fourth Term a4 = a3 + d = − 2 − 3 = − 5
So, first four term of A. P. are = 4, 1, − 2, − 5

(iv) a = − 1, d = \(\displaystyle \frac{1}{2}\)
First Term a1 = − 1
Common Difference d = \(\displaystyle \frac{1}{2}\)
Second Term a2 = a1 + d = \(\displaystyle -1+\frac{1}{2}=-\frac{1}{2}\)
Third Term a3 = a2 + d = \(\displaystyle -\frac{1}{2}+\frac{1}{2}=0\)
Fourth Term a4 = a3 + d = \(\displaystyle 0+\frac{1}{2}=\frac{1}{2}\)
So, first four term of A. P. are = − 1, \(\displaystyle -\frac{1}{2}\), − 0, \(\displaystyle \frac{1}{2}\)

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

(i) 3, 1, − 1, − 3 . . . (ii) − 5, − 1, 3, 7 . . . 
(iii) \(\displaystyle \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3},…\) (iv) 0.6, 1.7, 2.8, 3.9, . . .

Solutions –
(i) 3, 1, − 1, − 3 . . .

First term for A. P. a1 = 3
Common Difference d = a2a2 = 1 − 3 = − 2

(ii) − 5, − 1, 3, 7 . . .
First term for A. P. a1 = − 5
Common Difference d = a2a2 = − 1 − (− 5) = − 1 + 5 = 4

(iii) \(\displaystyle \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3},…\)
First term for A. P. a1 = \(\displaystyle \frac{1}{3}\)
Common Difference d = a2a2 = \(\displaystyle \frac{5}{3}-\frac{1}{3}=\frac{{5-1}}{3}=\frac{4}{3}\)

(iv) 0.6, 1.7, 2.8, 3.9, . . .
First term for A. P. a1 = 0.6
Common Difference d = a2a1 = 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, \(\displaystyle \frac{5}{2}\), 3, \(\displaystyle \frac{7}{2}\), . . .
(iii) − 1.2, − 3.2, − 5.2, − 7.2 . . .  (iv) − 10, − 6, − 2, 2 . . .
(v) 3, 3 + \(\displaystyle \sqrt{2}\), 3 + 2\(\displaystyle \sqrt{2}\), 3 + 3\(\displaystyle \sqrt{2}\) . . . (vi) 0.2, 0.22, 0.222, 0.2222 . . .
(vii) 0, − 4, − 8, − 12 . . . (viii) \(\displaystyle -\frac{1}{2},\,-\frac{1}{2},\,-\frac{1}{2},\,-\frac{1}{2}…\)
(xi) 1, 3, 9. 27 . . .  (x) a, 2a, 3a, 4a . . .
(xi) a, a2, a3, a4 . . . (xii) \(\displaystyle \sqrt{2},\,\sqrt{8},\,\sqrt{{18}},\,\sqrt{{32}}…\)
(xiii) \(\displaystyle \sqrt{3},\,\sqrt{6},\,\sqrt{9},\,\sqrt{{12}}…\) (xiv) 12, 32, 52, 72 . . .
(xv) 12, 52, 72, 73 . . .  

Solutions –
(i) 2, 4, 8, 16 . . .
Common Difference d = a2a1 = 4 − 2 = 2
a3a2 = 8 − 4 = 4
Because a2a1a3a2, hence there is no A. P. from the given list of numbers.

(ii) 2, \(\displaystyle \frac{5}{2}\), 3, \(\displaystyle \frac{7}{2}\), . . .
Common Difference d = a2a1 = \(\displaystyle \frac{5}{2}-2=\frac{{5-4}}{2}=\frac{1}{2}\)
a3a2 = \(\displaystyle 3-\frac{5}{2}=\frac{{6-5}}{2}=\frac{1}{2}\)
The given list of numbers is a A. P. which d is = \(\displaystyle \frac{1}{2}\) 
because a2a1 = a3a2 , hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
\(\displaystyle \frac{7}{2}+\frac{1}{2}=\frac{{7+1}}{2}=\frac{8}{2}=4\) 
a6 = a5 + d
\(\displaystyle 4+\frac{1}{2}=\frac{{8+1}}{2}=\frac{9}{2}\)
a7 = a6 + d
\(\displaystyle \frac{9}{2}+\frac{1}{2}=\frac{{9+1}}{2}=\frac{{10}}{2}=5\)
So, next three terms are 4, \(\displaystyle \frac{9}{2}\) and 5.

(iii) − 1.2, − 3.2, − 5.2, − 7.2 . . . 
Common Difference d = a2a1 = − 3.2 − (− 1.2) = − 3.2 + 1.2 = − 2
a3a2 = − 5.2 − (− 3.2) = − 5.2 + 3.2 = − 2
The given list of numbers is a A. P. which d is = − 2
because a2a1 = a3a2 , hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
− 7.2 + (− 2) = − 7.2 − 2 = − 9.2 
a6 = a5 + d
− 9.2 + (− 2) = − 9.2 − 2 = − 11.2 
a7 = a6 + d
− 11.2 + (− 2) = − 11.2 − 2 = − 13.2 
So, next three terms are − 9.2, − 11.2 and − 13.2

(iv) − 10, − 6, − 2, 2 . . .
Common Difference d = a2a1 = − 6 − (− 10) = − 6 + 10 = 4
a3a2 = − 2 − (− 6) = − 2 + 6 = 4
The given list of numbers is a A. P. whose common difference dis = 4
because a2a1 = a3a2 , hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
2 + 4 = 6 
a6 = a5 + d
6 + 4 = 10 
a7 = a6 + d
10 + 4 = 14
So, next three terms are 6, 10 and 14

(v) 3, 3 + \(\displaystyle \sqrt{2}\), 3 + 2\(\displaystyle \sqrt{2}\), 3 + 3\(\displaystyle \sqrt{2}\) . . .
Common Difference d = a2a1 = \(\displaystyle 3+\sqrt{2}-3=\sqrt{2}\)
a3a2 = \(\displaystyle 3+2\sqrt{2}-3+\sqrt{2}=\sqrt{2}\)
The given list of numbers is a A. P. whose common difference d is = \(\displaystyle \sqrt{2}\) 
because a2a1 = a3a2, hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
\(\displaystyle 3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2}\)
a6 = a5 + d
\(\displaystyle 3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}\)
a7 = a6 + d
\(\displaystyle 3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}\)
So, next three terms are \(\displaystyle 3+4\sqrt{2}\), \(\displaystyle 3+5\sqrt{2}\) and \(\displaystyle 3+6\sqrt{2}\)

(vi) 0.2, 0.22, 0.222, 0.2222 . . .
Common Difference d = a2a1 = 0.22 − 0.2 = 0.02
a3a2 = 0.222 − 0.22 = 0.002
The given list of numbers is not a A. P. whose common difference d is different.
Because a2a1a3a2, hence there is not a A. P. from the given list of numbers.

(vii) 0, − 4, − 8, − 12 . . .
Common Difference d = a2a1 = − 4 − 0 = − 4
a3a2 = − 8 − (− 4) = − 8 + 4 = − 4
The given list of numbers is a A. P. whose common difference d is = − 4 
because a2a1 = a3a2 , hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
− 12 + (− 4) = − 12 − 4 = − 16 
a6 = a5 + d
− 16 + (− 4) = − 16 − 4 = − 20
a7 = a6 + d
− 20 + (− 4) = − 20 − 4 = − 24
So, next three terms are − 16, − 20 and − 24

(viii) \(\displaystyle -\frac{1}{2},\,-\frac{1}{2},\,-\frac{1}{2},\,-\frac{1}{2}…\)
Common Difference d = a2a1 = \(\displaystyle -\frac{1}{2}-\left( {\frac{1}{2}} \right)=-\frac{1}{2}+\frac{1}{2}=0\)
a3a2 = \(\displaystyle -\frac{1}{2}-\left( {\frac{1}{2}} \right)=-\frac{1}{2}+\frac{1}{2}=0\)
The given list of numbers is a A. P. whose common difference d is = 0 है।   
because a2a1 = a3a2, hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
\(\displaystyle -\frac{1}{2}+0=-\frac{1}{2}\)
a6 = a5 + d
\(\displaystyle -\frac{1}{2}+0=-\frac{1}{2}\)
a7 = a6 + d
\(\displaystyle -\frac{1}{2}+0=-\frac{1}{2}\)
So, next three terms are \(\displaystyle -\frac{1}{2}\), \(\displaystyle -\frac{1}{2}\) and \(\displaystyle -\frac{1}{2}\)

(xi) 1, 3, 9. 27 . . . 
Common Difference d = a2a1 = 3 − 1 = 2
a3a2 = 9 − 3 = 6
The given list of numbers is not a A. P. whose common difference d is different.
Because a2a1a3a2, hence there is not a A. P. from the given list of numbers.

(x) a, 2a, 3a, 4a . . .
Common Difference d = a2a1 = 2aa = a
a3a2 = 3a − 2a = a
The given list of numbers is a A. P. whose common difference d is = a है।   
because a2a1 = a3a2 , hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
4a + a = 5a
a6 = a5 + d
5a + a = 6a
a7 = a6 + d
6a + a = 7a
So, next three terms are 5a, 6a and 7a

(xi) a, a2, a3, a4 . . .
Common Difference d = a2a1 = a2a = a (a − 1)
a3a2 = a3a2 = a2 (a − 1)
The given list of numbers is not a A. P. whose common difference d is different.
Because a2a1a3a2, hence there is not a A. P. from the given list of numbers.

(xii) \(\displaystyle \sqrt{2},\,\sqrt{8},\,\sqrt{{18}},\,\sqrt{{32}}…\)
Common Differenc d = a2 − a1 = \(\displaystyle \sqrt{8}-\sqrt{2}=\sqrt{2}\)
a3a2 = \(\displaystyle \sqrt{18}-\sqrt{8}=\sqrt{2}\)
The given list of numbers is a A. P. whose common difference d is = \(\displaystyle \sqrt{2}\) है।   
because a2a1 = a3a2, hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
\(\displaystyle \sqrt{{32}}+\sqrt{2}=\sqrt{{50}}\)
a6 = a5 + d
\(\displaystyle \sqrt{{50}}+\sqrt{2}=\sqrt{{72}}\)
a7 = a6 + d
\(\displaystyle \sqrt{{72}}+\sqrt{2}=\sqrt{{98}}\)
So, next three terms are \(\displaystyle \sqrt{{50}}\), \(\displaystyle \sqrt{{72}}\) and \(\displaystyle \sqrt{{98}}\)

(xiii) \(\displaystyle \sqrt{3},\,\sqrt{6},\,\sqrt{9},\,\sqrt{{12}}…\) 
Common Difference d = a2a1 = 32 − 12 = 23
a3a2 = 52 − 32 = 24
The given list of numbers is not a A. P. whose common difference d is different.
Because a2a1a3a2, hence there is not a A. P. from the given list of numbers.

(xv) 12, 52, 72, 73 . . .
Common Difference d = a2a1 = 52 − 12 = 24 
a3a2 = 72 − 52 = 24
The given list of numbers is a A. P. whose common difference d is = 24 है।   
because a2a1 = a3a2, hence there is a A. P. from the given list of numbers. The next three terms of which will be as follows –
a5 = a4 + d
73 + 24 = 97
a6 = a5 + d
97 + 24 = 121
a7 = a6 + d
121 + 24 = 145
So, next three terms are 97, 121 and 145

ncert solutions for class 10 maths Chapter 5 arithmetic progression Arithmetic Series
class 10th maths solution Arithmetic Series ex 5.2

The nth term of an A. P. with the first term a and the common difference d can be found by an = a + (n − 1) d.
Note: – An is also called the general term of A.P.
If an A.P. has m terms, then am denotes its last term, which is sometimes also denoted by l.

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

  a d n an
(i) 7 3 8 . . .
(ii) − 18 . . . 10 0
(iii) . . . − 3 18 − 5
(iv) − 18.9 2.5 . . . 3.6
(v) 3.5 0 105 . . .

Solutions :-
(i) a = 7, d = 3, n = 8, an = ?
nth term of A. P. an = a + (n − 1) d
an = 7 + (8 − 1) 3
an = 7 + 7 × 3
an = 7 + 21
an = 28
8th term of A.P. is 28

(ii) a = − 18, d = ?, n = 10, an = 0
nth term of A. P. an = a + (n − 1) d
0 = − 18 + (10 − 1) d
0 = − 18 + (9) d
0 = − 18 + 9d
− 9d = − 18 (by transposition)
d = \(\displaystyle \frac{{18}}{9}\)
d = 2
Common Difference of A.P., d is = 2 

(iii) a = ?, d = − 3, n = 18, an = − 5
nth term of A. P. an = a + (n − 1) d
− 5 = a + (18 − 1) − 3
− 5 = a + (17) − 3
− 5 = a + 17 × (− 3) (by transposition)
− 5 = a + (− 51)
a = − 51 + 5
a = − 46
a = 46
First term of A.P., a is = 46

(iv) a = − 18.9, d = 2.5, n = ?, an = 3.6
nth term of A. P. an = a + (n − 1) d
3.6 = − 18.9 + (n − 1) 2.5
3.6 = − 18.9 + (2.5n − 2.5)
3.6 = − 18.9 + 2.5n − 2.5
− 2.5n = − 18.9 − 2.5 − 3.6 (by transposition)
− 2.5n = − 25
n = \(\displaystyle \frac{{-25}}{{-2.5}}\)
n = 10
nth term of A.P., n is = 10

(v) a = 3.5, d = 0, n = 105, an = ?
nth term of A. P. an = a + (n − 1) d
an = 3.5 + (105 − 1) 0
an = 3.5 + 104 × 0
an = 3.5 + 0
an = 3.5
105th term of A.P. is 3.5

2. Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7, 4, . . . , is

(A) 97 (B) 77 (C) − 77 (D) − 87

Solutions :-
a = 10, d = a2a1 = 7 − 10 = − 3, n = 30, an = ?
nth term of A. P. an = a + (n − 1) d
an = 10 + (30 − 1) − 3
an = 10 + (29) − 3
an = 10 + 29 × (− 3)
an = 10 + (− 87)
an = − 77
So, correct option is (C)

(ii) 11th term of the AP: – 3, \(\displaystyle -\frac{1}{2}\), 2, . . ., is

(A) 28 (B) 22 (C) − 38 (D) \(\displaystyle -48\frac{1}{2}\)

Solutions :-
a = − 3, n = 30, an = ?
d = a2a1 = \(\displaystyle -\frac{1}{2}-(-3)=-\frac{1}{2}+3=\frac{{-1+6}}{2}=\frac{5}{2}=2.5\)
nth term of A. P. an = a + (n − 1) d
an = − 3 + (11 − 1) 2.5
an = − 3 + (10) × 2.5
an = − 3 + 25
an = 22
So, correct option is (B)

3. In the following APs, find the missing terms in the boxes :
(i) 2, . . . , 26
(ii) . . . , 13, . . . , 3
(iii) 5, . . . , . . . , \(\displaystyle 9\frac{1}{2}\)
(iv) − 4, . . . , . . . , . . . , . . . , 6
(v) . . . , 38, . . . , . . . , . . . , − 22

Solutions :-

(i) 2, . . . , 26 
a = 2, a3 = 26, n = 3
nth term of A. P. an = a + (n − 1) d
26 = 2 + (3 − 1) d
26 = 2 + 2d
2d = 26 − 2
2d = 24
d = \(\displaystyle \frac{{24}}{2}\)
d = 12
because to find the second term of the A.P.
So, a2 = a1 + d
a2 = 2 + 12
a2 = 14 answer

(ii) . . . , 13, . . . , 3
because the term between 13 and 3 is = \(\displaystyle \frac{{13+3}}{2}=\frac{{16}}{2}\) = 8
So, d = a4a3 = 3 − 8 = − 5
a3a2 = 8 − 13 = − 5
a2a1 = − 5
13 − a1 = − 5
a1 = 13 + 5 
a1 = 18
Hence A.P. is 18, 13, 8, 3

(iii) 5, . . . , . . . , \(\displaystyle 9\frac{1}{2}\)
a = 5, a4 = \(\displaystyle 9\frac{1}{2}\)
Because a4 = a + 3d
9.5 = 5 + 3d
3d = 9.5 − 5
3d = 4.5
d = \(\displaystyle \frac{{4.5}}{3}\)
and a2 = a + d
a2 = 5 + 1.5
a2 = 6.5
Hence Arithmetic Progressions (A.P.) is 5, 6.5, 8, 9.5

(iv) − 4, . . . , . . . , . . . , . . . , 6
a = − 4, a6 = 6
Because a6 = a + 5d
6 = − 4 + 5d
5d = 6 + 4
5d = 10
d = \(\displaystyle \frac{{10}}{{5}}\)
d = 2
The next terms of Arithmetic Progressions (A.P.) are =
a2 = a + d = − 4 + 2 = − 2
a3 = a + 2d = − 4 + 4 = 0
a4 = a + 3d = − 4 + 6 = 2
a5 = a + 4d = − 4 + 8 = 4
Hence Arithmetic Progressions (A.P.) is − 4, − 2, 0, 2, 4, 6

(v) . . . , 38, . . . , . . . , . . . , − 22
Let’s first term is a1 = 38 and 5th term is a5 = − 22
Because a5 = a + 4d
− 22 = 38 + 4d
4d = − 22 − 38
4d = − 60
d = − 15
If 2nd term is 38, then first term is =
a = a2 − d
a = 38 + 15
a = 53
Now 3rd and 4th terms –
a3 = a + 2d
a3 = 53 + 2 (− 15) 
a3 = 53 − 30
a3 = 23
a4 = a + 3d
a4 = 53 − 45
a4 = 8
5th term –
a5 = a + 4d
a5 = 43 − 60
a5 = − 7
Hence Arithmetic Progressions (A.P.) is 53, 38, 23, 8, − 7, − 22

4. Which term of the AP : 3, 8, 13, 18, . . . , is 78?
Solution :-
a = 3, d = a2a1 = 8 − 3 = 5, an = 78, n = ?
an = a + (n − 1) d
78 = 3 + (n − 1) 5
78 = 3 + 5n − 5
78 = 3 − 5 + 5n
78 = − 2 + 5n (by transposition)
5n = 78 + 2
5n = 80
n = \(\displaystyle \frac{{80}}{{2}}\)
n = 16
16th term of Arithmetic Progressions (A.P.): 3, 8, 13, 18, . . . is 78

5. Find the number of terms in each of the following APs :

(i) 7, 13, 19, . . . , 205 (ii) 18, \(\displaystyle 15\frac{1}{2}\), 13, . . . , – 47

(i) 7, 13, 19, . . . , 205
a = 7
d = a2a1 = 13 − 7 = 6
an = 205
n = ?
an = a + (n − 1) d
205 = 7 + (n − 1) 6
205 = 7 + 6n − 6
205 = 7 − 6 + 6n
205 = 1 + 6n
6n = 205 − 1 (by transposition)
6n = 204 
n = \(\displaystyle \frac{{204}}{{6}}\)
n = 34
So, there are 34 term in Arithmetic Progressions (A.P.)  7, 13, 19, . . . , 205

(ii) 18, \(\displaystyle 15\frac{1}{2}\), 13, . . . , – 47
a = 18
d = a2a1 = \(\displaystyle 15\frac{1}{2}-18=\frac{{31}}{2}-18=\frac{{31-36}}{2}=\frac{{-5}}{2}=-2.5\)
an = − 47
n = ?
an = a + (n − 1) d
− 47 = 18 + (n − 1) − 2.5
− 47 = 18 + (− 2.5n + 2.5)
− 47 = 18 – 2.5n + 2.5
2.5n = 18 + 2.5 + 47 (by transposition)
2.5n = 67.5
n = \(\displaystyle \frac{{67.5}}{{2.5}}\)
n = 27
So, there are 27 term in Arithmetic Progressions (A.P.) 18, \(\displaystyle 15\frac{1}{2}\), 13, . . . , – 47

6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
a = 11
d = a2a1 = 8 − 11 = − 3
an = − 150
n = ?
an = a + (n − 1) d
− 150 = 11 + (n − 1) − 3
− 150 = 11 − 3n + 3
− 150 = 14 − 3n
3n = 14 + 150
3n = 164
n = \(\displaystyle \frac{{164}}{{3}}\)
No, – 150 is not a term of Arithmetic Progressions (A.P.) , 11, 8, 5, 2, . . . , Because no positive integer number is obtained.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
a11 = a + (n − 1) d
38 = a + (11 − 1) d
38 = a + 10d …………..(1)
Similarly –
a16 = a + (n − 1) d
73 = a + (17 − 1) d
73 = a + 16d …………..(2)
by solving the pair of equations (1) and (2) –
\(\displaystyle \begin{array}{l}\,\,\,\,38=a+10d\\\,\,\,\,73=a+16d\\\underline{{-\,\,\,\,\,\,\,\,\,-\,\,\,\,-\,\,\,\,\,\,\,\,\,}}\\-35=\,\,\,\,\,\,\,\,-5d\end{array}\)
d = \(\displaystyle \frac{{-35}}{{-5}}\)
d = 7
To find the first term –
a11 = a + (11 − 1) d
38 = a + (10) 7
38 = a + 70
a = 38 − 70
a = − 32
because we want to find 31th term so, 
a31 = a + (n − 1) d
a31 = − 32 + (31 − 1) 7 
a31 = − 32 + 30 × 7
a31 = − 32 + 210
a31 = 178
So, 31th term of Arithmetic Progressions (A.P.) is 178

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution :-
a3 = a + (n − 1) d
12 = a + (3 − 1) d
12 = a + 2d …………..(1)
Similarly –
a50 = a + (n − 1) d
106 = a + (50 − 1) d
106 = a + 49d …………..(2)
by solving the pair of equations (1) and (2) –
\(\displaystyle \begin{array}{l}\,\,\,\,12=a+2d\\\,\,106=a+49d\\\underline{{-\,\,\,\,\,\,\,\,\,\,-\,\,\,\,-\,\,\,\,\,\,\,\,}}\\-94=\,\,\,\,\,\,\,-47d\end{array}\)
− 47d = − 94
d = \(\displaystyle -\frac{{94}}{{47}}\)
d = 2
To find the first term –
a3 = a + (n − 1) d
12 = a + (3 – 1) 2
12 = a + 2 × 2
12 = a + 4
a = 12 – 4
a =  8
because we want to find 29th term so, 
a29 = a + (n − 1) d
a29 = 8 + (29 − 1) 2 
a29 = 8 + 28 × 2
a29 = 8 + 56
a29 = 64
So, 29th term of Arithmetic Progressions (A.P.) is 64

9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
Solution :-
3rd term of Arithmetic Progressions (A.P.) is a3 = 4
9th term of Arithmetic Progressions (A.P.) is a9 = – 8
a3 = a + (n − 1) d
4 = a + (3 − 1) d
4 = a + 2d …………..(1)
Similarly –
a9 = a + (n − 1) d
− 8 = a + (9 − 1) d
− 8 = a + 8d …………..(2)
by solving the pair of equations (1) and (2) –
\(\displaystyle \begin{array}{l}\,\,\,\,4=a+2d\\\,-8=a+8d\\\underline{{+\,\,\,\,\,\,\,-\,\,\,\,\,-\,\,\,\,\,\,}}\\\,\,12=\,\,\,\,\,-6d\end{array}\)
d = \(\displaystyle \frac{{-12}}{{6}}\)
d = – 2
To find the first term –
a3 = a + (n − 1) d
4 = a + (3 − 1) − 2
4 = a + 2 × (− 2)
4 = a − 4 (by transposition)
a = 4 + 4
a = 8
We want to find nth term so,
an = a + (n − 1) d
0 = 8 + (n − 1) − 2 
0 = 8 − 2n + 2 (by transposition)
2n = 10
n = 10 /2
n = 5
So, 0 will be the 5th term of the Arithmetic Progressions (A.P.)

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution :-
Let’s 10th term of Arithmetic Progressions (A.P.) = x
and 17th term of Arithmetic Progressions (A.P.) = x + 7
a10 = a + (n − 1) d
x = a + (10 − 1) d
x = a + 9 × d
– 9d = ax …………..(1)
Similarly –
a17 = a + (n − 1) d
x + 7 = a + (17 − 1) d
x + 7 = a + 16 × d
x + 7 = a + 16d (by transposition)
− 16d = ax − 7 …………..(2)
by solving the pair of equations (1) and (2) –
\(\displaystyle \begin{array}{l}\,\,\,\,\,\,-9d=a-x\\\,\,\,\,-16d=a-x-7\\\underline{{\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,+\,\,\,\,\,+\,\,\,\,\,\,}}\\\,\,\,\,\,\,\,\,\,\,7d=\,\,\,\,\,\,\,\,\,\,\,\,\,\,+7\end{array}\)
d = \(\displaystyle \frac{{7}}{{7}}\)
d = 1
So, the common difference of Arithmetic Progressions (A.P.) is 1

11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Solution :-
a = 3
d = a2a1 = 15 − 3 = 12
54th term of Arithmetic Progressions (A.P.) = ?
a54 = a + (n − 1) d
a54 = 3 + (54 − 1) 12
a54 = 3 + 53 × 12
a54 = 3 + 636
a54 = 639
132 more than 54th term
So, 639 + 132 = 771
xth term of Arithmetic Progressions (A.P.) ax= ?
ax = a + (n − 1) d
771 = 3 + (x − 1) 12
771 = 3 + 12x − 12
771 = 3 + 12x − 12 
12x = 771 + 12 − 3 (by transposition)
12x = 783 − 3
12x = 780
x = \(\displaystyle \frac{{780}}{{12}}\)
x = 65
So, 65th term Arithmetic Progressions (A.P.) 3, 15, 27, 39, . . . will be more than 132 of 54th term.

13. How many three-digit numbers are divisible by 7?
Solution :-
3 digit first number divided by seven = 105
d = 7
So, A.P.: 105, 112, 119, . . . , 994
3 digit last number divided by seven = 994
So, a = 105
d = 7
nth term = 994
an = a + (n − 1) d
994 = 105 + (n – 1) 7
994 = 105 + 7n – 7 
7n = 994 − 105 + 7 (by transposition)
7n = 1001 − 105
7n = 896
n = \(\displaystyle \frac{{896}}{7}\)
n = 128
Hence, 128 three digit numbers are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?
Solution :-
First multiple of 4 greater than 10 = 12
Multiple of 4 less than 250 = 248
Common Difference d = 4
a = 12
nth term = 248
an = a + (n − 1) d
248 = 12 + (n − 1) 4
248 = 12 + 4n − 4 
4n = 248 − 12 + 4 (by transposition)
4n = 252 − 12
4n = 240
n = \(\displaystyle \frac{{240}}{4}\)
n = 60
So there are 60 multiples of 4 between 10 and 250.

15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Solution :-
For first Arithmetic Progressions (A.P.) –
a = 63
d = a2a1 = 65 − 63 = 2
Suppose that xth term are equal, then 
an = a + (n − 1) 2
ax = 63 + (x − 1) 2
ax = 63 + 2x − 2
For second Arithmetic Progressions (A.P.) –
a = 3
d = a2a1 = 10 − 3 = 7
Suppose that xth term are equal, then 
an = a + (n − 1) d
ax = 3 + (x − 1) 7
ax = 3 + 7x − 7
\(\displaystyle \begin{array}{l}\,\,\,\,{{a}_{x}}=63+2x-2\\\,\,\,\,{{a}_{x}}=\,\,\,\,3+7x-7\\\underline{{-\,\,\,\,\,\,\,\,\,\,-\,\,\,\,-\,\,\,\,\,\,\,+\,\,\,\,\,\,}}\\\,\,\,\,0\,\,\,\,=60-5x+5\end{array}\)
5x = 60 + 5 (by transposition)
5x = 65
x = \(\displaystyle \frac{{65}}{5}\)
x = 13
Hence, both the APs 63, 65, 67, . , , and 3, 10, 17, . , , The 13th term of is equal.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution :-
a3 = 16
Common Difference = d
So, a7 = a1 + 6d
a5 = a1 + 4d
According to Question –
a7 = a5 + 12
So, a1 + 6d = a1 + 4d + 12
6d – 4d = 12
2d = 12
d = 6
Then, a3 = a1 + 2d
16 = a1 + 12
a1 = 16 – 12
a1 = 4
So that A.P. will be 4, 10, 16, . . .

17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
Solution :-
d = a2a1 = 8 − 3 = 5
So, second last term = 253 – 5 = 248
reversing the arithmetic progression
Suppose that A.P.: 253, 248, 243, . . .
So, now a = 253
d = a2a1 = 253 − 248 = – 5
20th term = x
an = a + (n − 1) d
ax = 253 + (20 − 1) − 5
ax = 253 + (− 100 + 5)
ax = 253 − 100 + 5
ax = 253 − 100 + 5
ax = 258 − 100
ax = 158
So The 20th term from the last term in A.P.: 3, 8, 13, . . ., 253 will be 158.

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP
Solution :-
Sum of 4th and 8th terms of Arithmetic Progressions (A.P.) = 24
a4 = a1 + 3d
a8 = a1 + 7d
According to Question –
a1 + 3d + a1 + 7d = 24
2a1 + 10d = 24 . . . . . . . . . . . . (1)

Sum of 6th and 10th terms = 44
a6 = a1 + 5d
a10 = a1 + 9d
According to Question –
a1 + 5d + a1 + 9d = 44
2a1 + 14d = 44 . . . . . . . . . . . (2)
From equations (1) and (2) –
\(\displaystyle \begin{array}{l}\,\,\,\,2{{a}_{1}}+10d=24\\\,\,\,\,2{{a}_{1}}+14d=44\\\underline{{-\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,}}\\\,\,\,\,0\,\,\,\,\,\,-4d\,\,\,\,=-20\end{array}\)
d = \(\displaystyle \frac{{-20}}{{-4}}\)
d = 5
Put the value of d in Eqn. (2) –
2a1 + 14 × 5 = 44
2a1 + 70 = 44
2a1 = 44 − 70
2a1 = − 26
a1 = \(\displaystyle \frac{{-26}}{{2}}\)
a1 = − 13
So, first term is = − 13
Second term is = a1 + d = − 13 + 5 = − 8
Third term is = a1 + 2d = − 13 + 2 × 5 = − 13 + 10 = − 3
So, the first three terms of Arithmetic Progressions (A.P.) will be – 13, − 8 and − 3.

19. Subba Rao started work in 1995 at an annual salary of $ 5000 and received an increment
of $ 200 each year. In which year did his income reach  7000?
Solution :-
Initial pay a1 = $ 5000
Increment d = $ 200
Salary after n years = $ 7000
So, an = a + (n − 1) d
7000 = 5000 + (n − 1) 200
7000 = 5000 + 200n − 200
200n = 7000 + 200 − 5000 (by transposition)
200n = 7200 − 5000
200n = 2200
n = \(\displaystyle \frac{{2200}}{{200}}\)
n = 11 year
So after 11th year (1995 + 11) in 2006 the salary will be $ 7000.

20. Ramkali saved $ 5 in the first week of a year and then increased her weekly savings by $ 17.50. If in the nth week, her weekly savings become  207.50, find n.
Solution :-
Saving of first week a1 = $ 50
Saving of every week d = $ 17.5
After nth week, her weekly savings is = $ 207.50
So, an = a + (n − 1) d
207.50 = 50 + (n − 1) × 17.5
207.50 = 50 + 17.5n − 17.5
17.5n = 207.50 + 17.5 − 50 (by transposition)
17.5n = 225 − 50
17.5n = 175
n = \(\displaystyle \frac{{175}}{{17.5}}\)
n = 10
So nth week = 10 weeks

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