NCERT Solutions for Class 10 Maths Ex 5.2

# NCERT Solutions for Class 10 Maths Ex 5.2

In this post, you will find the NCERT solutions for class 10 maths ex 5.2. These solutions are based on the latest curriculum of NCERT Maths class 10.

## NCERT Solutions for Class 10 Maths Ex 5.2

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

Solution: (i) a = 7, d = 3, n = 8

Here, we have to find an.

Using the formula, an = a + (n –1)d

Putting the values of a, d and n, we get

an = 7 + (8 −1)3

= 7 + (7)3 = 7 + 21 = 28

(ii) a = –18, n = 10, an = 0

Here, we have to find d.

Using the formula, an = a + (n −1)d

Putting the values of a, n and an, we get

0 = –18 + (10 – 1)d

0 = −18 + 9d

18 = 9

= 2

(iii) d = –3, n = 18, an = −5

Here, we have to find a.

Using the formula, an = a + (n – 1)d

Putting the values of dan and n, we get

–5 = a + (18 –1) (–3)

−5 = + (17) (−3)

−5 = – 51

= 46

(iv) a = –18.9, d = 2.5, an = 3.6

Here, we have to find n.

Using the formula, an = a + (n –1)d

Putting the values of dan and n, we get

3.6 = –18.9 + (n –1) (2.5)

3.6 = −18.9 + 2.5− 2.5

2.5= 25

= 10

(v) a = 3.5, d = 0, n = 105

Here, we have to find an.

Using the formula, an = a + (n −1)d

Putting the values of a, d and n, we get

an = 3.5 + (105 −1) (0)

an = 3.5 + 104 × 0

an = 3.5 + 0

an = 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

(ii) 11th term of the AP: −3, −1/2, 2 …, is

(A) 28             (B) 22          (C) –38             (D) 48½

Solution: (i) 10, 7, 4 …

Here, first term (a) = 10, common difference (d) = 7 – 10 = 4 –7 = –3

And n = 30      [Since we have to find 30th term]

Using the formula, an = a + (n – 1)d

a30 = 10 + (30 −1) (−3) = 10 – 87 = −77

Therefore, the option (C) is correct.

(ii) −3, −1/2, 2 …

Here, first term (a) = –3, common difference (d) = −1/2 − (−3) = 2 − (−1/2) = 5/2

And n = 11       [Since we have to find 11th term]

Using the formula, an = a + (n – 1)d

a11 = −3 + (11 –1) 5/2 = −3 + 25 = 22

Therefore, the option (B) is correct.

3. In the following Aps, find the missing terms:

(i) 2, __, 26

(ii) __, 13, __, 3

(iii) 5, __, __, 9½

(iv) –4, __, __, __, __, 6

(v) __, 38, __, __, __, –22

Solution: (i) 2, __, 26

We know that the difference between consecutive terms in an A.P. is always equal.

Let the missing term be x.

Then, – 2 = 26 – x

2= 28

= 14

Therefore, the missing term is 14.

(ii) __, 13, __, 3

Let the missing terms be x and y.

Then, the sequence becomes x, 13, y, 3

We know that the difference between consecutive terms in an A.P. is always equal.

Then, – 13 = 3 – y

2= 16

= 8

Again, 13 – – 13

= 26

Putting the value of y, we get

+ 8 = 26

= 18

Therefore, the missing terms are 18 and 8.

(iii) 5, __, __, 9½

Here, first term (a) = 5, n = 4 and fourth term (a4) =

Using the formula, an = a + (n – 1)d, we get

a4 = 5 + (4 − 1)d

19/2 = 5 + 3d

19 = 2(5 + 3d)

19 = 10 + 6d

6d = 19 – 10

6d = 9

3/2

Therefore, we get the common difference (d) = 3/2

Second term = a + d = 5 + 3/2 = 13/2

Third term = second term + d = 13/2 + 3/2 = 16/2 = 8

Therefore, the missing terms are 13/2 and 8.

(iv) –4, __, __, __, __, 6

Here, first term (a) = –4, n = 6 and sixth term (a6) = 6

Using the formula, an = a + (n – 1)d, we get

a6 = −4 + (6 − 1)d

6 = −4 + 5d

5= 10

= 2

Therefore, the common difference (d) = 2

Second term = first term + d = a + d = –4 + 2 = –2

Third term = second term + d = –2 + 2 = 0

Fourth term = third term + d = 0 + 2 = 2

Fifth term = fourth term + d = 2 + 2 = 4

Therefore, the missing terms are –2, 0, 2 and 4.

(v) __, 38, __, __, __, –22

Here, the 2nd and the 6th terms are given.

Using the formula, an = a + (n – 1)d

Second term (a2) = + (2 − 1)

38 = d …. (1)

And sixth term (a6) = + (6 − 1)d

−22 = + 5d   …. (2)

Equations (1) and (2) are the linear equations in two variables.

Let us solve it using substitution method.

Using equation (1), we get = 38 – d  …. (3)

Putting the value of a in equation (2), we get

−22 = 38 – + 5d

4= −60

= −15

Putting this value of d in equation (1), we get

38 = – 15

= 53

Therefore, we get = 53 and = −15

First term (a) = 53

Third term = second term + d = 38 – 15 = 23

Fourth term = third term + d = 23 – 15 = 8

Fifth term = fourth term + d = 8 – 15 = –7

Therefore, the missing terms are 53, 23, 8 and –7.

4. Which term of the AP: 3, 8, 13, 18, …, is 78?

Solution: Here, first term (a) = 3, common difference (d) = 8 – 3 = 13 – 8 = 5

Let the nth term of the given AP is 78, then an = 78

Using the formula, an = a + (n – 1)d, we get

an = 3 + (− 1)5

78 = 3 + (− 1) 5

75 = 5− 5

80 = 5n

= 16

It means the 16th term of the given AP is 78.

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

(i) 7, 13, 19, …, 205

(ii) 18, 15½, 13, …, −47

Solution: (i) 7, 13, 19, …, 205

Here, first term (a) = 7, common difference (d) = 13 – 7 = 19 – 13 = 6

Let the number of terms in the given AP is n, then an = 205

Using the formula, an = a + (n – 1)d, we get

205 = 7 + (− 1)6

205 = 7 + 6– 6

205 = 6+ 1

204 = 6n

= 34

Therefore, there are 34 terms in the given AP.

(ii) 18, 15½, 13, …, −47

Here, first term (a) =18, common difference (d) =

Let the number of terms in the given AP is n, then an = −47

Using the formula, an = a + (n – 1)d, we get

−47 = 18 + (− 1)(−5/2)

−47 = 18 – (5/2)5/2

−94 = 36 − 5+ 5

5= 135

= 27

Therefore, there are 27 terms in the given AP.

6. Check whether –150 is a term of the AP: 11, 8, 5, 2…

Solution:  Let −150 be the nth term of AP: 11, 8, 5, 2 …, which means that an = −150

Here, first term (a) = 11, common difference (d) = 8 – 11 = –3

Using the formula, an = a + (n – 1)d, we get

−150 = 11 + (− 1) (−3)

−150 = 11 − 3+ 3

3= 164

= 164/3

But, n cannot be in fraction.

Therefore, our supposition is wrong.

Hence, −150 cannot be a term of the given AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:  Here, we are given that a11 = 38 and a16 = 73

Using the formula, an = a + (n – 1)d, we have

38 = + (11 − 1)(d)

38 = + 10….. (1)

And 73 = + (16 − 1)(d)

73 = + 15d ….. (2)

These are linear equations in two variables.

Let us solve these equations using substitution method.

From equation (1), we get

= 38 − 10d ….. (3)

Putting the value of a in equation (2), we get

73 = 38 − 10+ 15d

35 = 5d

d = 7

Putting the value of d in equation (1), we get

38 = + 10 × 7

= −32

Therefore, common difference (d) = 7 and first term (a) = –32

Using the formula, an = a + (n − 1)d, we get

a31 = −32 + (31 − 1)(7)

= −32 + 210 = 178

Therefore, the 31st term of the given AP 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: The AP consists of 50 terms and a50 = 106

It is also given that a3 = 12

Using the formula, an = a + (n – 1)d,  we get

a50 + (50 − 1)

106 = + 49….. (1)

And a3 + (3 − 1)d

12 = + 2d ….. (2)

These are linear equations in two variables.

Let us solve these equations using substitution method.

Using equation (1), we get

= 106 − 49d ….. (3)

Putting the value of a in the equation (2), we get

12 = 106 − 49+ 2d

47= 94

= 2

Putting the value of d in equation (3), we get

= 106 – 49(2)

= 106 – 98

= 8

Therefore, first term (a) = 8 and common difference (d) = 2

Let us use the formula, an = a + (n – 1)d to find the 29th term.

a29 = 8 + (29 − 1)2

= 8 + 56

= 64

Therefore, the 29th term of the given AP 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: It is given that the 3rd and the 9th terms of an AP are 4 and –8 respectively.

It means that, a3 = 4 and a9 = −8

Using the formula, an = a + (n – 1)d, we get

a3 = a + (3 – 1)d

4 = a + 2d ….. (1)

And, a9 = a + (9 – 1)d

–8 = a + 8d ….. (2)

These are linear equations in two variables.

Let us solve these equations using substitution method.

From equation (1), we get

= 4 − 2d ….. (3)

Putting the value of a in equation (2), we get

−8 = 4 − 2+ 8d

−12 = 6

= −2

Putting the value of d in equation (2), we get

−8 = + 8(−2)

−8 = – 16

= 8

Therefore, first term (a) = 8 and common difference (d) = −2

We want to know which term of the given AP is equal to zero.

Let the nth term of the given AP is 0.

Using the formula, an = a + (n − 1)d, we get

0 = 8 + (− 1) (−2)

0 = 8 − 2+ 2

0 = 10 − 2n

2= 10

= 5

Therefore, the 5th term of the given AP is 0.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution: It is given that a17 = a10 + 7 ….. (1)

Using the formula, an = a + (n – 1)d, we have

a17 = + 16d ….. (2)

a10 = + 9d ….. (3)

Putting the values of a17 and a10 from equations (2) and (3) in equation (1), we get

+ 16+ 9+ 7

7= 7

= 1

Therefore, the common difference is 1.

11. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?

Solution: Let us first calculate the 54th term of the given AP.

Here, first term (a) = 3 and common difference (d) = 15 – 3 = 12

Using the formula, an = a + (n – 1)d, we get

a54 + (54 − 1)

= 3 + 53(12)

= 3 + 636 = 639

We have to find that which term is 132 more than its 54th term.

Let us suppose it is nth term which is 132 more than the 54th term.

an = a54 + 132

3 + (− 1)12 = 639 + 132

3 + 12– 12 = 771

12– 9 = 771

12= 780

= 65

Therefore, the 65th term is 132 more than its 54th term.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms.

Solution: Let the first term of 1st AP = a and the first term of 2nd AP = a

It is given that their common difference is the same.

Let their common difference be d.

It is also given that the difference between their 100th terms is 100.

Using the formula, an = a + (n – 1)d, we get

+ (100 − 1)– [a′ + (100 − 1)d] = 100

+ 99− a′ − 99= 100

− a′ = 100 ….. (1)

We have to find the difference between their 1000th terms, which means we have to calculate:

+ (1000 − 1)– [a′ + (1000 − 1)d] = + 999− a′ − 999

– a

Putting the value of − a′ from equation (1) in the above equation, we get

+ (1000 − 1)– [a′ + (1000 − 1)d] = 100

Therefore, the difference between their 1000th terms is equal to 100.

13. How many three digit numbers are divisible by 7?

Solution: The first three-digit number divisible by 7 is 105 and the last three-digit number divisible by 7 is 994.

Therefore, we have AP of the form 105, 112, 119, …, 994

Let there are total n terms in the above AP.

Then, an = 994

We have to find n here.

First term (a) = 105 and common difference (d) = 112 – 105 = 7

Using the formula, an = a + (n – 1)d,  we get

994 = 105 + (− 1) (7)

994 = 105 + 7− 7

896 = 7

= 128

It means 994 is the 128th term of the above AP.

Therefore, there are 128 terms in the above AP.

Hence, there are 128 three-digit numbers which are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

Solution: The first multiple of 4 which lies between 10 and 250 is 12 and the last multiple of 4 which lies between 10 and 250 is 248.

Therefore, the AP is of the form: 12, 16, 20, …, 248

Let there are total n terms in the above AP.

Then, an = 248

We have to find n here.

Here, first term (a) = 12 and common difference (d) = 4

Using the formula, an = + (– 1)d, we get

248 = 12 + (− 1) (4)

248 = 12 + 4− 4

240 = 4

= 60

It means that 248 is the 60th term of the above AP.

Hence, we can say that there are 60 multiples of 4 which lie 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: Let us consider the first AP: 63, 65, 67, …

Here, first term (a) = 63 and common difference (d) = 65 – 63 = 2

Using the formula, an = a + (n – 1)d, we get

an = 63 + (− 1)(2) ….. (1)

Now, let us consider the second AP: 3, 10, 17, …

Here, first term (a) = 3 and common difference (d) = 10 – 3 = 7

Using the formula, an = a + (n 1)d, we get

an = 3 + (− 1)(7) ….. (2)

According to the question, we have

Equation (1) = Equation (2)

63 + (− 1)(2) = 3 + (− 1)(7)

63 + 2– 2 = 3 + 7− 7

65 = 5n

= 13

Therefore, the 13th terms of both the APs are equal.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution: Let the first term of the given AP be a and the common difference be d.

It is given that its 3rd term is equal to 16.

Using the formula, an = a + (n 1)d, we get

16 = + (3 − 1)d

16 = + 2d ….. (1)

It is also given that the 7th term exceeds the 5th term by 12.

According to this condition, we have

a7 = a5 + 12

+ (7 – 1)+ (5 – 1)+ 12                 [Using the formula, an = a + (n 1)d]

6= 4d + 12

2= 12

= 6

Putting the value of d in equation (1), we get

16 = + 2(6)

= 4

Therefore, the first term (a) = 4 and common difference (d) = 6

Thus, AP is 4, 4 + 6, 4 + 12, 4 + 18, …

Hence, AP is 4, 10, 16, 22, …

17. Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.

Solution: We have to find the 20th term of the given AP from the last.

Let us write the given AP from the last: 253, …, 13, 8, 3

Here, first term (a) = 253 and common difference (d) = 8 – 13 = –5

Using the formula, an = a + (n 1)d, we get

a20 = 253 + (20 − 1)(−5)

a20 = 253 + 19(−5) = 253 – 95 = 158

Therefore, the 20th term of the given AP from the last is 158.

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of 6th and 10th terms is 44. Find the first three terms of the AP.

Solution: It is given that the sum of the 4th and 8th terms of an AP is 24

Then, a4 + a8 = 24 ….. (1)

It is also given that the sum of the 6th and 10th terms is 44.

Then, a6 + a10 = 44 ….. (2)

Using the formula, an = a + (n  − 1)d in equation (1), we get

+ (4 − 1)+ [+ (8 − 1)d] = 24

+ 3+ 7d = 24

2+ 10d = 24

+ 5d = 12 ….. (3)

Using the formula, an = a + (n  − 1)d in equation (2), we get

+ (6 − 1)+ [+ (10 − 1)d] = 44

+ 5+ 9= 44

2+ 14= 44

+ 7=22 ….. (4)

Equations (3) and (4) are linear equations in two variables.

Let us solve these equations using substitution method.

From equation (3), we have = 12 − 5d ….. (5)

Putting the value of a in equation (4), we get

12 − 5+ 7= 22

12 + 2= 22

2= 10

= 5

Putting the value of d in equation (5), we get

= 12 – 5(5) = 12 – 25 = −13

Therefore, first term (a) = −13 and common difference (d) = 5

Therefore, the AP is –13, –8, –3, 2 …

Its first three terms are –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: Subba Rao’s starting salary =  5000

It means that the first term (a) = 5000

He gets an increment of  200 after every year.

Therefore, the common difference (d) = 200

His salary after one year = 5000 + 200 =  5200

His salary after two years = 5200 + 200 =  5400

Therefore, it is an AP of the form: 5000, 5200, 5400, 5600, …, 7000

We have to find in which year his income reaches  7000.

Let after n years, his salary reaches  7000.

Using the formula, an = a + (n  − 1)d, we get

7000 = 5000 + (− 1) (200)

7000 = 5000 + 200− 200

7000 – 5000 + 200 = 200n

2200 = 200n

= 11

It means after 11 years, Subba Rao’s income would be  7000.

Hence, his income reached  7000 in the year 2006.

20. Ramkali saved  5 in the first week of a year and then increased her weekly savings by  1.75. If in the nth week, her weekly savings become  20.75, find n.

Solution:  Ramkali saved  5 in the first week of a year. It means first term (a) = 5

Ramkali increased her weekly savings by  1.75.

Therefore, the common difference (d) = 1.75

Money saved by Ramkali in the second week = a + d = 5 + 1.75 =  6.75

Money saved by Ramkali in the third week = 6.75 + 1.75 =  8.5

Therefore, it is an AP of the form: 5, 6.75, 8.5, …, 20.75

We have to find in which week her weekly savings become  20.75.

Let in nth week, her weekly savings become  20.75.

Using the formula, an = a + (n  − 1)d, we get

20.75 = 5 + (− 1) (1.75)

20.75 = 5 + 1.75− 1.75

17.5 = 1.75n

= 10

It means that in the 10th week, her weekly savings become  20.75.