# PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

## PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2

Question 1

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

AP:

Solution:

(i) Here a = 7, d = 3, n = 8

∵ a_{n} = a + (n – 1)d

∴ a_{8} = 7 + (8 – 1)3

= 7 + 21 = 28.

(ii) Here a = – 18, n = 10, a_{n} = 0

∵ a_{n} = a + (n – 1)d

∴ a_{10} = – 18 + (10 – 1)d

or 0 = – 18 + 9d .

or 9d = 18

d = 18/2 = 2.

(iii) Here d = – 3, n = 18, a_{n} = – 5

∵ a_{n} = a + (n – 1)d

∴ a_{18} = a + (18 – 1)(-3)

or -5 = a – 51

or a = – 5 + 51 = 46.

(iv) Here a = – 18.9, d = 2.5 a_{n} = 3.6

∵ a_{n} = a + (n – 1)d

∴ 3.6 = – 18.9 + (n – 1) 2.5

or 3.6 + 18.9 = (n – 1) 2.5

or (n – 1) 2.5 = 22.5

or n – 1 = 22.5/2.5

or n = 9 + 1 = 10.

(v) Here a = 3.5, d = 0, n = 105

∵ a_{n} = a + (n – 1) d

∴ a_{n} = 3.5 + (105 – 1) 0

a_{n} = 3.5 + 0 = 3.5.

Question 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

Solution:

(i) Given A.P. is 10, 7, 4 ……………

T_{1} = 10, T_{2} = 7, T_{3} = 4

T_{2} – T_{1} = 7 – 10 = – 3

T_{3} – T_{2} = 4 – 7 = – 3

∵ T_{2} – T_{1} = T_{3} – T_{2} = – 3 = d(say)

∵ T_{n} = a + (n – 1) d

Now, T_{30} = 10 + (30 – 1)(-3)

= 10 – 87 = – 77

∴ Correct choice is (C).

Question 3.

In the following APs, find the missing terms in the boxes:

(i) 2, , 26

(ii), 13, , 3

(iii) 5,, , 9

(iv) – 4, , , , , 6

(v) , 38, , , , – 22

Solution:

Let a be the first term and ‘d’ be the common difference of given A.P.

(i) Here T_{1} = a = 2

and T_{3} = a + 2d = 26

or 2 + 2d = 26

or 2d = 26 – 2 = 24

or d = 12

∴ Missing term = T_{2} = a + d = 2 + 12 = 14.

(ii) Here, T_{2} = a + d = 13 ……………(1)

and T_{4} = a + 3d = 3 …………….(3)

Now, (2) – (1) gives

Substitute this value of d in (1), we get

a – 5 = 13

a = 13 + 5 = 18.

∴ T_{1} = a = 18

T_{3} = a + 2d = 18 + 2(-5)

= 18 – 10 = 8.

(iv) Here T_{1} = a = —

T_{6} = a + 5d = 6

or -4 + 5d = 6

or 5d = 6 + 4

or 5d = 10

or d = 10/2 = 2

Now, T_{2} = a + d = -4 + 2 = -2

T_{3} = a + 2d = – 4 + 2(2)

= – 4 + 4 = 0

T_{4} = a + 3d = – 4 + 3(2)

= – 4 + 6 = 2

T_{5} = a + 4d = – 4 + 4(2)

= – 4 + 8 = 4

(v) Here T_{2} = a + d = 38 ………….(1)

and T_{6} = a + 5d = -22 ……………(2)

Now, (2) – (1) gives

Substitute this value of d in (1), we get

a + (-15) = 38

a = 38 + 15 = 53

∴ T_{1} = a = 53

T_{3} = a + 2d = 53 + 2(-15) = 53 – 30 = 23.

T_{4} = a + 3d = 53 + 3(-15) = 53 – 45 = 8

T_{5} = a + 4d = 53 + 4(-15) = 53 – 60 = – 7.

Question 4

Which term of the A.P. 3, 8, 13, 18, …………… is 78?

Solution:

Given A.P. is 3, 8, 13, 18, ………….

T_{1} = 3, T_{2} = 8, T_{3} = 13, T_{4} = 18

T_{2} – T_{1} =8 – 3=5

T_{3} – T_{2}= 13 – 8=5

T_{2} – T_{1} = T_{3} – T,= 5 = d (say)

Using, T_{n} = a + (n – I) d

or 78 = 3 + (n – 1) 5

or 5(n – 1) = 78 – 3 = 75

or n – 1 = 15

or n = 15 + 1 = 16

Hence, 16th term of given AP. is 78.

Question 5.

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

(i) 7, 13, 19,…, 205

(ii) 18, 15 1/2, 13, ………….., – 47

Solution:

(i) Given A.P. is 7, 13, 19, …………..

T_{1} = 7, T_{2} = 13, T_{3} = 19

T_{2} – T_{1} = 13 – 7 = 6

T_{3} – T_{2} = 19 – 13 = 6

T_{2} – T_{1} = T_{3} – T_{2} = 6 = d(say)

Using formula, T_{n} = a + (n – 1) d

205 = 7 + (n – 1) 6

or (n – 1) 6 = 205 – 7 = 198

or (n – 1) = 198/6

or n – 1 = 33

n = 33 + 1 = 34

Hence, 34th term of an AP. is 205.

Question 6.

Is – 150 a term of 11, 8, 5, 2….? why?

Solution:

Given sequence is 11, 8, 5, 2, ………..

T_{1} = 11, T_{2} = 8, T_{3} = 5, T_{4} = 2

T_{2} – T_{1} = 8 – 11 = – 3

T_{3} – T_{2} = 5 – 8 = – 3

T_{4} – T_{3} = 2 – 5 = – 3

T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = – 3 = d (say).

Let – 150 be any term of given A.P.

then T_{n} = – 150

a+(n – 1)d = – 150

or 11 +(n – 1)(- 3) = – 150

or (n – 1)( – 3) = – 150 – 11 = – 161

Question 7.

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

Solution:

Let ‘a’ and 4d’ be the first term and common difference of given A.P.

Given that T_{11} = 38

a +(11 – 1) d = 38

[∵ T_{n} = a + (n – 1) d]

a + 10 d = 38

and T_{16} = 73

a + (16 – 1) d = 73

[∵ T_{n} = a + (n – 1) d]

a + 15 d = 73

Now, (2) – (1) gives

Substitute this value of d in (1), we get

a + 10 (7) = 38

or a + 70 = 38

or a = 38 – 70 = – 32

Now, T_{31} = a + (31 – 1) d

= – 32 + 30 (7) = – 32 + 210 = 178.

Question 8.

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 291h term.

Solution:

Let ‘a’ and ‘d’ be the first term and common difference of given A.P.

Given that, T_{3} = 12

a + (3 – 1) d = 12

∵ T_{n} = a + (n – 1) d

or a + 2d = 12 ………………(1)

and Last term = T_{50} = 106

a + (50 – 1) d = 106

∵ T_{n} = a + (n – 1) d

a + 49 d = 106 ……………(2)

Now, (2) – (1) gives

Substitute this value of d in (1), we get

a + 2(2) = 12

or a + 4 = 12

or a + 12 – 4 = 8

Now, T_{29} = a + (29 – 1) d

= 8 + 28 (2) = 8 + 56 = 64.

Question 9.

If the 3rd and 9th tenus of an A.P. are 4 and – 8 respectively, which term of this A.P. is zero.

Solution:

Let ‘a’ and ‘d’ be the first term and common difference of given AP.

Given that: T_{3} = 4

a + (3 – 1) d = 4

∵ T_{n} = a + (n – 1) d

a + 2d = 4 …………..(1)

and T_{9} = – 8

a + (9 – 1) d = 8

and T_{9} = – 8

a + (9 – 1)d = 8

∵ T_{n} = a + (n – 1) d

or a + 8d = – 8

Now, (2) – (1) gives

Substitute this value of d in (1), we get

a + 2(- 2) = 4

or a – 4 = 4

or a = 4 + 4 = 8

Now, T_{n} = 0 (Given)

a + (n – 1) d = 0

or 8 + (n – 1)(- 2)=0

or -2 (n – 1) = – 8

or n – 1 = 4

or n = 4 + 1 = 5

Hence, 5th term of an AP. is zero.

Question 10.

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

Solution:

Let ‘a’ and ‘d’ be the first term and common difference of given A.P.

Now, T_{17} = a + (17 – 1) d = a + 16 d

and T_{10} = a + (10 – 1) d = a + 9 d

According to question

T_{17} – T_{10} = 7

(a + 16 d) – (a + 9 d) = 7

or a + 16 d – a – 9 d = 7

7 d = 7

or d = 7/7 = 1

Hence, common difference is 1.

Question 11.

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

Solution:

Let ‘a’ and ‘d’ be the first term and common difference of given A.P.

Given A.P. is 3, 15, 27, 39, …

T_{1} = 3, T_{2} = 15, T_{3} = 27, T_{4} = 39

T_{2} – T_{1} = 15 – 3 = 12

T_{3} – T_{2} = 27 – 15 = 12

:. d=T_{2} – T_{1} = T_{3} – T_{2} =12

Now, T_{54} = a + (54 – 1) d

= 3 + 53 (12) = 3 + 636 = 639

According to question

T = T_{54} + 132

a + (n – 1)d = 639 + 132

3 + (n – 1)(12) = 771

(n – 1) 12 = 771 – 3 = 768

or n – 1 = 768/12 = 64

or n = 64 + 1 = 65

Hence, 65th term of A.P. is 132 more than its 54th term.

Question 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 ‘a’ and ‘d’ be the first term and common difference of first AP.

Also, ‘A’ and ‘d’ be the first term and common difference of second A.P.

According to question

[T_{100} of second A.P.] – [T_{100} of first A.P.] = 100

or[A +(100 – 1)d] – [a +(100 – 1)d] = 100

or A + 99d – a – 99d = 100

or A – a = 100

Now, [T_{1000} of second A.P.] – [T_{1000} of first A.P.]

= [A + (1000 – 1) d) – (a + (1000 – 1) d]

= A + 999 d – a – 999 d

= A – a = 100 [Using (I)].

Question 13.

How many three-digits numbers are divisible by 7?

Solution:

Three digits numbers which divisible by 7 are 105, 112, 119 , 994

Here a = T_{1} = 105, T_{2} = 112, T_{3} = 119 and T_{n} = 994

T_{2} – T_{1} = 112 – 105=7

T_{3} – T_{2} = 119 – 112=7

∴ d = T_{2} – T_{1} = T_{3} – T_{2} = 7

Given that, T_{n} = 994

a + (n – 1) d = 994

or 105 + (n – 1) 7 = 994

or (n – 1) 7 = 994 – 105

or (n – 1) 7 = 889

or n – 1 = 889/7 = 127

or n = 127 + 1 = 128.

Hence, 128 terms of three digit number are divisible by 7.

Question 14.

How many multiples of 411e between 10 and 250?

Solution:

Multiples of 4 lie between 10 and 250 are 12, 16, 20, 24, … 248

Here a = T_{1} = 12, T_{2} = 16, T_{3} = 20 and T_{n} = 248

T_{2} – T_{1} = 16 – 12 = 4

T_{3} – T_{2} = 20 – 16 = 4

∴ d = T_{2} – T_{1} = T_{3} – T_{2} = 4

Given that, T_{n} = 248

a + (n – 1) d = 248

or 12 + (n – 1)4 = 248

or 4(n – 1) = 248 – 12 = 236

or n – 1 = 236/4 = 59

or n = 59 + 1 = 60

Hence, there are 60 terms which are multiples of 4 lies between 10 and 250.

Question 15.

For what value of n, are the n terms of two A.P.s 63, 65, 67, …………. and 3, 10, 17, …………….. equal?

Solution:

Given A.P. is 63, 65, 67, ……………..

Here a = T_{1} = 63, T_{2} = 65, T_{3} = 67

T_{2} – T1 = 65 – 63 = 2

T_{3} – T_{2} = 67 – 65 = 2

∴ d = T_{2} – T_{1} = T_{3} – T_{2} = 2

and second A.P. is 3, 10, 17, …

Here a = T_{1} = 3, T_{2} = 10, T_{3} = 17

T_{2} – T_{1} = 10 – 3 = 7

T_{3} – T_{2} = 17 – 10 = 7

According to question.

[nth term of first A.P.] = [nth term of second A.P.]

63 + (n – 1)2 = 3 + (n – 1) 7

or 63 + 2n – 2 = 3 + 7n – 7

or 61 + 2n = 7n – 4

or 2n – 7n = – 4 – 61

– 5n = – 65

n = 65/5 = 13.

Question 16.

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

Solution:

Let ‘a’ and ‘d’ be the first term and common difference of given A.P.

Given that T_{3} = 16

a + (3 – 1) d = 16

a + 2d = 16

According to question

T_{7} – T_{5} = 12

[a + (7 – 1) d] – [a + (5 -1) d] = 12

a + 6 d – a – 44 = 12

2d = 12

d = 12/2 = 6

Substitute this value of d in (1), we get

a + 2(6) = 16

a = 16 – 12 = 4 .

Hence, given A.P. are 4, 10, 16, 22, 28, ………….

Question 17.

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

Solution:

Given A.P. is 3, 8, 13, …………., 253

Here, a = T_{1} = 3, T_{2} = 8, T_{3} =13 and T_{n} = 253

T_{2} – T_{1} = 8 – 3 = 5

T_{3} – T2 = 13 – 8 = 5

∴ d = T_{2} – T_{1} = T_{3} – T_{1} = 5

Now, T_{n} = 253

3 + (n – 1)5 = 253

∵ T_{n} = a + (n – 1) d

(n – 1) 5 = 250

n-1 = 250/5 = 50

n = 50 + 1 = 51

20th term from the end of AP = (Total number of terms) – 20 + 1

= 51 – 20 + 1 = 32nd term

∴ 20th term from the end of AP

= 32nd term from the starting

= 3 + (32 – 1)5

∵ Tn = a + (n – 1)d

= 3 + 31 × 5

= 3 + 155 = 158.

Question 18.

The sum of the 4th and 8th term of an AP is 24 and the sum of the 6’ and 10th terms is 44. FInd the first three terms of the A.P.

Solution:

Let ‘a’ and ‘d’ be the first term and common difference of given A.P.

According to 1st condition

T_{4} + T_{8} = 24

a + (4 – 1) d + a + (8 – 1) d = 24

∵ T_{n} = a + (n – 1) d

or 2a + 3d + 7d = 24

2a + 10d = 24

a + 5d = 12 …………(1)

According to 2nd condition

T_{6} + T_{10} = 44

a + (6 – 1) d + a +(10 – 1) d = 44

∵ T_{n} = a + (n – 1) d

2a + 5d + 9d = 44

2a + 14d = 44

a + 7d = 22

Now (2) – (1) gives

Substitute this value of d in (I). we get

a + 5(5) = 12

a + 25 = 12

a = 12 – 25 = -13

T_{1} = a = -13

T_{2} = a + d = 13 + 5 = -8

T_{2} = a + 2d = – 13 + 2(5) = – 13 + 10 = -3

Hence, given A.P. is -13, -8, -3, ……………

Question 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

Annual increment = ₹ 200

Let ‘n’ denotes number of years.

∴ first term = a = ₹ 5000

Common diflerence = d = ₹ 200

and T_{n} = ₹ 7000

5000 + (n – 1) 200 = 7000

∵ T_{n} = a + (n – 1) d

(n – 1) 200 = 7000 – 5000

or (n – 1) 200 = 2000

or n – 1 = 2000/200 = 10

or n = 10 + 1 = 11

Now, in case of year the sequence are 1995. 1996, 1997, 1998, ……………

Here a = 1995, d = 1 and n = 11

Let T_{n} denotes the required year.

∴ T_{n} = 1995 + (11 – 1) 1

= 1995 + 10 = 2005

Hence, in 2005, Subba Rao’s salary becomes 7000.

Question 20.

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

Solution:

Amount saved in first week = ₹ 5

Increment in saving every week = ₹ 1.75

It is clear that, it form an A.P. whose terms are

T_{1} = 5, d = 1.75

∴ T_{2} = 5 + 1.75 = 6.75

T_{3} = 6.75 + 1.75 = 8.50

Also. T_{n} = 20. 75 (Given)

5 + (n – 1) 1.75 = 20.75

∵ T_{n} = a + (n – 1) d

or (n – 1) 1.75 = 20.75 – 5

or (n – 1) 1.75 = 15.75

or (n – 1) = 1575/100×100/175

or n – 1 = 9

or n = 9 + 1 = 10

Hence, in 10th week, Ramkali’s saving becomes ₹ 20.75.

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