# Telangana SCERT Class 10 Maths Solution Chapter 6 Progressions Exercise 6.2

## Telangana SCERT Solution Class X (10) Maths Chapter 6 Progressions Exercise 6.2

### Exercise – 6.2

(Q1) Fill in the blanks in the following table, given that a is the first term, d the common difference and a the ∩th term of the A.P.

 SL No. a d n an (i) 7 3 8 28 (ii) -18 2 10 0 (iii) 46 -3 18 -5 (iv) -18.9 2.5 10 3.6 (v) 3.5 0 105 3.5

(i) a = 7, d = 3, n = 8, an =?

=> Solution:

In an A.P,

Given: a = 7, d = 3, n = 8,

an = a + (n – 1) d

a8 = 7 + (8-1) 3

= 7 + 7 X (3)

a8 = 7 + 21

= 28

(ii) Given : a = – 18, n = 10, an = 0, d =?

In an A.P.

An = a + (n – 1) d

0 = – 18 + (10 – 1) X d

0 = – 18 + 9d

0 = – 18 + 9d

18 = 9d

18/9 = d

d = 2

(iii) Given: a =?, d = -3, n = 18, an = – 5

In an A.P.

an = a + (n – 1) d

– 5 = a + (18 – 1) (-3)

– 5 = a + 17 (-3)

– 5 + 51 = a

46 = a

∴ a = 46

(iv) Given: a = 18.9, d = 2.5, n =? an = 36

In an A.P

an = a + (n – 1) d

3.6 = -18.9 + (n – 1) 2.5

3.6 + 18.9 = (n – 1) 2.5

22.5 = (n – 1) 2.5

22.5/2.5 = n – 1

9 = n – 1

9 + 1 = n

n = 10

(v) a = 3.5, d = 0, n = 105, an =?

=> Given: a = 3.5, d = 0, n = 105

an = a + (n – 1) d

= 3.5 + (105 – 1) 0

= 3.5 + (104) 0

= 3.5 + 0

an = 3.5

(Q2) Find the

(i) 30th term of the A.P, 10, (1’1)

=> Solution:

The given A.P is

10, 7, 4, —–

A = 10, d = 7 – 10 = – 3

We have, an = a + (n – 1) d

a30 = 10 + (30 – 1) (-3)

= 10 + 29 (-3)

a30 = – 77

(ii) 11th term of the A.P – 3, -1/2, 2

=> Solution:

The given A.P is,

– 3, -1/2, 2, —-

A = -3, d = -1/2 – (-3) = -1/2 + 3 = -1+6/2 = 5/2

We have, an = a + (n – 1) d

a11 = – 3 + (11 – 1) (5/2)

a11 = -3 + (10) (5/2)

= -3 + 5 (5)

= – 3 + 25

a11 = 22

(Q3) Find the respective term for the following A.P.S

(i) a1 = 2, a3 = 26 find a2

=> Solution:

Given: a1 = a = 2 and a3 = 26

a3 = 26

a + 2d = 26 [∵ a3 = a + 2d)

2 + 2d = 26

2d = 26 – 2

2d = 24

d = 24/2

d = 12

a2 = a + d = 2+12 = 14 a3 = a + 2d

= 18 + 2 (-5)

= 18 – 10

a3 = 8

∴ a1 = 18 and a3 = 8

(iii) a1 = 5, a4 = 9 1/2 find a2, a3

=> Solution:

Given: a1 = 5, a4 = 9 1/2

a1 = a = 5

a4 = 9 1/2 = 19/2

a + 3d = 19

5 + 3d = 19/2

3d = 19/2 – 5

= 19-10/2

3d = 9/2

d = 9/3X2 = 9/6

d = 3/2

a2 = a + d

= 5+3/2

= 10+3/2

a2 = 13/2

a3 = a + 2d

= 5 + 2 (3/2)

= 5 + 3

a3 = 8

∴ a2 = 13/2 and a3 = 8

(iv) a1 = – 4, a6 = 6 find a2, a3, a4, a5

=> Solution:

Given: a1 = – 4, a6 = 6

a1 = a = -4

a + 5d = 6   (a6 = a + 5d)

– 4 + 5d = 6

5d = 6+4

5d = 10

d = 10/5

d = 2

a2 = a + b

= – 4 + 2

a2 = – 2

a3 = a + 2b

= – 4 + 2 (2)

= – 4 + 4

a3 = 0

a4 = a + 3d

= – 4 + 3 (2)

= – 4 + 6

a4 = 2

a5 = a + 4d

= – 4 + 4 (2)

= – 4 + 8

a5 = 4

∴ a2 = -2, a3 = 0, a4 = 2 and a5 = 4 a3 = a + 2d

= 53 + 2 (-15)

= 53 – 30

a3 = 2.3

a4 = a + 3d

= 53 + 3 (-15)

a4 = 53 – 45

a4 = 8

a5 = a + 4d

= 53 + 4(-15)

a5 = 53 – 60

a5 = – 7

∴ a1 = 53, a3 = 23, a4 = 8 and a5 = – 7

(Q1) Which term of the A.P. 3, 8, 13, 18

=> Solution:

Given A.P is 3, 8, 13, 18,

Here, a = 3, d = a2 – a1

= 8 – 3

= 5

Let an = 78

An = a + (n – 1) d

78 = 3 + (n – 1) 5

78 – 3 = (n – 1) 5

75 = (n – 1) 5

75/5 = n – 1

15 = n – 1

15 + 1 = n

16 = n

∴ n = 16

∴ 78 is the 16th term.

(Q5) Find the number of terms in each of the following A.P.s

(i) 7, 13, 19 ——, 205

=> Solution: a = 7, d = a2 – a1, = 13 – 7 = 6

Let, an = 205

an = a + (n – 1) d

205 = 7 + (n – 1) 6

205 – 7 = (n – 1) 6

198 = (n – 1) 6

198/6 = n – 1

33 = n – 1

33+1 = n

34 = n

∴ n = 34

∴ There are 34 terms

(ii) 18, 15, 15 1/2, 13 —– -47

=> Solution:

a1 = 18, d = a2 – a1 = 15-18/2 = -5/2

Let an = -47

an = a + (n – 1) d

– 47 = 18 + (n – 1) (-5/2)

-47 – 18 = (n – 1) (-5/2)

-65 = (n – 1) (-5/2)

-65 X (-2/5) = n – 1

13 X 2 = n – 1

26 = n – 1

26 + 1 = n

27 = n

∴ n = 27

∴ There are 27 terms

(Q6) Check whether, – 150 is a term of the A.P 11, 8, 5, 2

=> Solution: Given A.P 11, 8, 5, 2,

A = 11, d = 8 – 11 = – 3

Let an = – 150

an = a + (n – 1) d

– 150 = 11 + (n – 1) (-3)

– 150 – 11 = (n – 1) (-3)

– 161 = (n – 1) (-3)

– 161/-3 = n – 1

161/3 = n – 1

161/3 + 1 = n

161/3 + 3 = n

164/3 = n

∴ n = 164/3

∴ – 150 is not a term of given A.P.  Substituting d = – 2 in equn (1)

a + 2d = 4

a + 2 (-2) = 4

a + (-4) = 4

a + 4 = 4

a = 4 + 4

a = 8

Let an = 0

an = a + (n – 1) d

0 = 8 + (n – 1) d

0 – 8 = 8 + (n – 1) (-2)

– 8 = (n – 1) (-2)

– 8 = (n – 1) (-2)

-8/-2 = n – 1

4 = n – 1

4 + 1 = n

∴ n = 5

∴ 5th term of the A.P. is ‘0’

(Q9) The 17th term of an A.P exceeds its 10th term by 7. Find the common difference.

=> Solution:

Given: 17th term of an A.P exceeds it is 10th term by 7.

a17 = a10 + 7

=> a + 16d = a + 9d + 7

=> 16d – 9d = 7

=> 7d = 7

=> d = 7/7

D = 1

∴ Common difference = 1

(Q10) Two A.P.s have the same common difference. The difference between their 100th terms is 100. What is the difference between their 100th terms?

=> Solution:

Let the 100th terms of the two A.P.s are

a1 + 99d1 and a2 + 99d2

Difference between them is 100

(a2 + 99d2) – (a1 + 99d1) = 100

a2 + 99d2 – a1 – 99d1 = 100

Since their common difference is same i.e. d1 = d2

a2 + 99d1 – a1 – 99d1 = 110

a2 – a1 = 100 —— (1)

Their 1000th terms are,

a1 + 99d1 and a2 + 99d2

Difference between from

= (a2 + 999d2) – (a1 + 999d1)

= a2 + 999d2 – a1 – 999d1

= a2 + 999d1 – a1 – 999d1 (d2 = d1)

= a2 – a1

= 100 (from (10))

(Q11) How many three digit numbers are divisible by 72

=> Solution:

The three digit numbers = 100, 101, 102, 103, 104, —– 999

Since the common difference is same it forms an A.P.

Here, a = 105, d = 112 – 105 = 7

Let an = 994

a + (n – 1) d = 994

105 + (n – 1)d = 994

(n – 1) d = 994 – 105

(n – 1) 7 = 889

(n – 1) = 889/7

(n – 1) = 127

n = 127 + 1

n = 128

∴ There are 128 three digit numbers which are divisible by 7.

(Q12) How many multiples of 4 lie between 10 and 250?

=> Solution:

Multiple of ‘4’ between and 250 à 12, 16, 20, 24, 248

Since the common difference is same. It forms an A.P.

A = 12, d = a2 – a1 = 16 – 12 = 4

Let an = 248

A + (n – 1) d = 248

12 + (n – 1) 4 = 248

(n – 1) 4 = 248 – 12

(n – 1) 4 = 236

n – 1 = 236/4

n – 1 = 59

n = 59+1

n = 60

There are 60 multiples of 4 lie in between 10 and 250.

(Q13) For what value of n, are the nth terms of two A.P.s 63, 65, 61

=> Solution:

The first A.P is 63, 65, 67 —–

a = 63, d = a2 – a1 = 65 – 63 = 2.

nth term an = a + (n – 1) d

an = 63 + (n – 1) 2 ——- (1)

The second A.P is 3, 10, 17, ——

A = 3, d = a2 – a1 = 10 – 3 = 7

nth term an = a + (n – 1) d

an = 3 + (n – 1) 7 ——– (2)

Since the nth terms are equal from equations (1) and (2)

63 + (n – 1) = 3 + (n – 1) 7

63 – 3 = (n – n7 – (n – 1) 2

60 = 7n – 7 – 2n + 2

60 = 5n – 5

60 + 5 = 5n

65 = 5n

65/5 = n

13 = n

∴ n = 13

For n = 13, the nth terms of the given two A.P.s are equal.

(Q14) Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

=> Solution:

The 3rd term is 16

a3 = 16

a + 2d = 16 ——- (1)

7th term exceeds the 5th term by 12.

i.e. a7 = a6 + 12

a + 6d = a + 4d + 12

6d – 4d = 12

2d = 12

D = 12/6

D = 6

Substitute d = 6 in equn (1), we get

a + 2 X 6 = 16

a + 11 = 16

a = 16 – 12

a = 4

a2 = a + d = 4+6 = 10

a3 = a+2d = 4 + 2(6) = 4 + 12 = 16

The A.P is 4, 10, 16, 22, ——

(Q16) The sum of the 4th and 8th terms of an A.P is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P

=> Solution:

Sum of 4th and 8th terms is 24.

a4 + a8 = 24

a + 3d + a + 7d = 24

2a + 10d = 24

2 (a + 5d) = 24

a + 5d = 24/2

a + 5d = 12 ——- (1) a1 = a = -13

a2 = a + d = -13 + 5 = -8

a3 = a + 2d = -13 + 2(5) = -13 +10 = -3

The first three terms of the A.P are,

a1 = -13, a2 = -8, a3 = -3

-13, -8 and -3

(Q15) Find the 20th term from the end of the A.P 3, 8, 13, —— 253

=> Solution:

The given A.P. is

3, 8, 13, —– 253

a = 3, d = a2 – a1 = 8 – 3 = 5

Let an = 253

a + (n – 1) d = 253

3 + (n – 1) d = 253

3 + (n – 1) 5 = 253

(n – 1) 5 = 253 – 3

(n – 1) 5 = 250

n – 1 = 250/5

n – 1 = 50

n = 50 + 1

n = 51

There are 51 terms

20th term from the end = (51 – 20) + 1

= 31 + 1

= 32nd term from the first

a32 = a + 31d

= 3 + 31(5)

= 3 + 155

a32 = 158

(Q17) Subba Rao started this job in 1995 at a monthly salary of RS 5000 and received an increment of RS 200 each year. In which year did his salary reach RS 7000?

=> Solution:

Subba Rao in 1995 in RS 5000

Rearly increment is RS 200

List of his salaries are RS 5000, 5200, 5400, —–

Since the common difference is done it forms an A.P

a = 5000, d = 5200 – 5000

d = 200

Let an = 7000

a + (n – 1) d = 7000

5000 + (n – 1) 200 = 7000

(n – 1) 200 = 7000 – 5000

(n – 1) 200 = 2000

n – 1 = 2000/200

n – 1 = 10

n = 10 + 1

n = 11

After 11 years, his salary becomes RS 7000 i.e. 1995 + 11 = 2006

In the year 2006 his salary becomes RS 7000.

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Updated: June 11, 2021 — 11:39 am