CBSE Solutions For Class 10 Mathematics Chapter 5 Arithmetic Progression

Arithmetic Progression

Question 1. The nth term of a Sequence In defined as follows. Find the first four terms:

1. an=3n+1
Solution:

3n=3nt

Put n= 1,2,3,4, we get

a1 = 3×1+1=4

a2 = 3×2+1=7

a3 = 3×3 +1 = 10

a4=3×4+1= 13

The first four terms of the Sequence are 4,7, 10,13.

2. an= n2+3
Solution:

an=n2+3

Put n=1,2,3,4 we get

a1= (1)2+3 = 4

a2=(2)2 +3= 7

a3 = (3)2+3 = 12

a4=(4)2 +3= 19

The first four terms of the Sequence are 4, 7, 12, and 19.

3. an = n(n+1)
Solution:

an=n (n+1)

Put n= 1,2,3,4 we get

a1 = 1(1+1)=2

a2 = 2(2+1)=6

a3=3(3+1)= (2

a4 = 4(4+1)=20

The first four terms of the Sequence are 2,6,12,20

4.\(a_n=n+\frac{1}{n}\)
Solution:

⇒ \(a_n=n+\frac{1}{n}\)

Put n=1,2,3,4 we get

⇒ \(a_1=1+\frac{1}{1}=2 \)

⇒ \(a_2=2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2} \)

⇒ \(a_3=3+\frac{1}{3}=\frac{9+1}{3}=\frac{10}{3} \)

⇒ \(a_4=4+\frac{1}{4}=\frac{16+1}{3}=\frac{17}{3}\)

First four terms of the Sequence are 2,\( \frac{5}{2}, \frac{10}{3}, \frac{17}{3}\)

5. an= 3n
Solution:

an=3n

Put n=1,2,3,4 we get

a1=31=3

a2=32 = 9

a3=33=27

a4=34=81

The first four terms of the Sequence are 3, 9, 27, 81

Question 2. The nth term of a Sequence is (3n-7). Find its 20th term.
solution:

3n-7

n=20

3(20)-7

60-7

57

The 20th term of a Sequence is 57

Question 3. Which of the following are A.p.’s? If they form an A.P., find the Common difference ‘d’ and write three more terms:

1. -10, -6, -2, 2,…….
Solution:

Here a=-10,

d=-6-10=4

-10, -6, -2, 2,4,6,10,14

Yes d= 4, next three items = 6, 10, 14

2. 3,3+ √2, 3+252, 3+3√2,…..
Solution:

Here a=3

d=3+√2-3 =) √2

3, 3+ √2, 3+2√2, 3+3√2, 3+4√2,3+5√2

Yes d=52, next three terms = 3+3√2, 3+4√2,3+5√2.

3. 0, -4, -8,-12,
Solution:

Here a=0

d=0-4=-4

0,-4,-8,-12-16,-20,-24

Yes do, next three terms = -16,-20,-24

Question 4. For the following A.-P. ‘S write the first terms and Common differences:

1. 2,5, 8, 11,..
Solution:

2, 5, 8, 11,

Her first term a=2

Common difference = 5-2 = 3

2.  -5,-1, 3, 7,
Solution: -5,-1, 3, 7

Her first term a=-5

Common difference = -541 = 4

CBSE Solutions For Class 10 Mathematics Chapter 5 Arithmetic Progression

Question 5. write the first four terms of the Ap., when the first term ‘a’ and the Common difference ‘d’ are given as follows:

1.  a=5, d=3
solution:

a=5, d= 3

a1 =5

a2=5+3=8

a3=8+3=11

a4=11+3= 14

The first four terms are 5, 8, 11, 14

2.  a=-2,d=4
Solution:

a=-2, d=4

a1 =-2

a2=-2+4=2

a3=2+4=6

a4=6+4= 10

The first four terms are -2,2,6,10

Question 6. Find the 10th term of the AP: 1,3,5,7,..
Solution:

Here, a=1

d=3-1=2,

n = 10

an = a+ (n-1)

90= 1+ (10-1)2

⇒96=1+18

⇒ 210=19

10th term of the given Ap=19

Question 7. Find the 7th term of the AP 80, 77,74,71,
Solution:

Here a=80

d=77-80=-3,

n=7

a7 = a + (n-1)d

a7=80+(7-1)-3

a7=80-18

⇒ a7 = 62

7th term of the given A.p. =62

Question 8. Find the nth term of the A⋅p: -5, -3, -1, 1, —-
solution:

Here a = -5

d=-3-5=2

n = n

an = a+ (n-1) d

an=-5+ (n-1)2

⇒ an=-5+2n-2

⇒ an= 2n-7

nth term of the given A.p. = (2n-7)

Question 9. Which term of the A-P. 4, 8, 12, is 76?
Solution:

Here, a = 4,

d=8-4=4

Let an=76

=) 4+ (n-1)4=76

= 4+4n-4=76

4n=76

⇒ \(n=\frac{76}{4} \Rightarrow n=19\)

19thterm of the given A.P. is 76

Question 10. which term of the Ap. 36, 33, 30, is Zero?
Solution:

Here a=36

d=33-36=-3

let an = 0

36+ (n-1)-3=0

36-3n+3=0

-3n=-39

\(n=\frac{39}{3} \Rightarrow n=13\)

The 13th term of the given A.P. is zero

Question 11. which term of the\(\frac{3}{4}, 1, \frac{5}{4}, \ldots \text {. is } 12 ?\)
Solution:

⇒ \(\text { Here } a=\frac{3}{4} \)

⇒ \(d=1-\frac{3}{4}=\frac{1}{4} \)

⇒ \(\text { let } a_n=12 \)

⇒ \(\frac{3}{4}+(n-1) \frac{1}{4}=12 \)

⇒ \(\frac{3+(n-1)}{4}=12\)

3+n-1=48

⇒ n+2=48

⇒n=46

46th term of the given AP is 12.

Question 12. Find the number of terms in the Ap. 8, 12, 16,
solution:

Here a=8

d=12-8=) 4

let an=124

=) 8+ (n-1)4=124

=) 8+40-4=124

=) 4n=124-4

4n = 120

⇒ n = \(n=\frac{120}{4} \Rightarrow 30\)

30th term of the given A.P. is 124.

Question 13. Find the number of terms in the A.p. 75, 70, 65, 15
Solution:

Here a=75

d=70-75=-5

let an=15

75+ (n-1)-5=15

75-5n+5=15

70-5n=15

-517=15-70

-5n=-55 ⇒ n=11

11th term of the given A. p. is 15.

Question 14. Find the 10th term from the end of the A.P. 82, 79, 76, —-,4.
Solution:

Here aa 1=4

d=79-82=-3

n=10

(-(10-1)defined

– 4-(10-1)-3

=4727

=31

10th term from the end = 31

Question 15. Find the 16th term from the end of the A.P. 3,6,9,99
solution:

Here, 1=99

d= 6-3 = 3,

n=16

⇒ 99-(16-1) 3

⇒ 99-45

⇒ 54

16th  term from the end = 54

Question 16. Find the Sum of the following A.p.: 3,8, 13, to 20 terms
Solution:

S1 =3+8+13,

a1=3

d=8-3=5

n=20

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d]] \)

⇒ \(S_{20}=\frac{20}{2}[2(3)+(20-1) 5]\)

⇒ \(S_{20}=10[6+95]\)

⇒ \(S_{20}=10[101]\)

⇒ \(S_{20}=1010 \)

Question 17. Find the sum of the following A.p. 5: 1,4,7,– to 50 terms
Solution:

S = 1+4+7+—–50

a=1

d=4-1= 3

n=50

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d] \)

⇒ \(S_{50}=\frac{50}{2}[2(1)+(50-1) 3] \)

⇒ \(S_{50}=25[2+147] \)

⇒ \(S_{50}=25[(49]\)

⇒ \(S_{50}=3,725\)

Question 18. Find the Sum given below: 3+6+9+ …..+96
Solution:

S=3+6+9+…… +96

a1 =3, d= 6-3 =) 3

an=96

a1+ (n-1) d=96

3+(n-1)3=96

3+3n-3=96

3n=96

⇒ \(n=\frac{96}{3} \Rightarrow n=32\)

S1 = Sum of 32 terms with first 3 terms and last term 96

⇒ \(S_1=\frac{32}{2}[3+96]\)

⇒ \(S_1=\frac{32}{2}[99]\)

⇒ \(S_1=16[99]\)

⇒ \(S_1=1584\)

Question 19. Find the Sum given below! 2 + 4+ 6+.
solution:

S=2+4+6+—–+50

a1 =2,

d=4-2 =) 2 Q1=2,

an=50

a1+ (n-1)d=50

2+(n-1)2=50

2+2n=2=50

⇒ \(n=\frac{50}{2} \Rightarrow n=25\)

S1 = Sum of 25 terms with first 2 terms and last terms so

⇒ \(S_1=\frac{50}{2}[2+50] \)

⇒ \(S_1=\frac{50}{2}[526] \)

⇒ \(S_1=50[26] \)

⇒ \(S_1=650\)

Question 20. In an A.p.: given a=2, d= 3, Qn = 50, find n and Sn.
Solution:

Given a=2, d=3, an =50

a+(n-1)d=an

2+(n-1)3=50

2+3n-3=50

3n=5041

⇒ \(n=\frac{51}{3} \Rightarrow n=17 \)

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d] \)

⇒ \(5_n=\frac{17}{2}[2(2)+(17-1) 3] \)

⇒ \(S_n=\frac{17}{2}[2(2)+(17-1) 3] \)

⇒ \(S_n=\frac{17}{2}[4+48] \)

⇒ \(S_n=\frac{17}{2}\left[S_2\right] \)

⇒ \(S_n=17[26] \)

⇒ \(S_n=442\)

Question 21. Find the Value of x for which (x+2), 2x, (2x+3) are three consecutive terms of A.P.
solution:

(x+2), 2x, (2x+3) are three consecutive terms of A.p.

\(2 x=\frac{(x+2)+(2 x+3)}{2}\)

4x= x+2+2x+3

4x=3x+5

4x-3x=5

x=5

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