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### Practice Set 3.2 Arithmetic Progression Class 10th Mathematics Part 1 MHB Solution

Practice Set 3.2

1, 8, 15, 22, . . . Here a = square , t_1 = square , t_2 = square , t_3 = square…

3, 6, 9, 12, . . . Here t_1 = square , t_2 = square , t_3 = square , t_4 = square…

- 3, - 8, - 13, - 18, . . . Here t_3 = square , t_2 = square , t_4 = square , t_1 =…

70, 60, 50, 40, . . . Here t_1 = square , t_2 = square , t_3 = square therefore a =…

Decide whether following sequence is an A.P., if so find the 20th term of the…

Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this…

Find the 19th term of the following A.P. 7, 13, 19, 25, . . .

Find the 27th term of the following A.P. 9, 4, - 1, - 6, - 11, . . .…

Find how many three digit natural numbers are divisible by 5.

The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the…

11, 8, 5, 2, . . . In this A.P. which term is number - 151?

In the natural numbers from 10 to 250, how many are divisible by 4?…

In an A.P. 17th term is 7 more than its 10th term. Find the common difference.…

###### Practice Set 3.2

Question 1.

Write the correct number in the given boxes from the following A. P.

1, 8, 15, 22, . . .

Here

1, 8, 15, 22, . . .

First term a = 1

Second term t1 = 8

Third term t2 = 15

Fourth term t3 = 22

We know that d = tn + 1 – tn

Thus, t2 – t1 = 15 – 8 = 7

t3 – t2 = 22 – 15 = 7

Thus, d = 7

Question 2.

Write the correct number in the given boxes from the following A. P.

3, 6, 9, 12, . . .

Here

3,6,9,12, . . .

First term a = 3

Second term t1 = 6

Third term t2 = 9

Fourth term t3 = 12

We know that d = tn + 1 – tn

Thus, t2 – t1 = 9 – 6 = 3

t3 – t2 = 12 – 9 = 3

Thus, d = 3

Question 3.

Write the correct number in the given boxes from the following A. P.

– 3, – 8, – 13, – 18, . . .

Here

– 3, – 8, – 13, – 18, . . .

First term a = – 3

Second term t1 = – 8

Third term t2 = – 13

Fourth term t3 = – 18

We know that d = tn + 1 – tn

Thus, t2 – t1 = – 13 – ( – 8) = – 13 + 8 = – 5

t3 – t2 = – 18 – ( – 13) = – 18 + 13 = – 5

Thus, d = – 5

Question 4.

Write the correct number in the given boxes from the following A. P.

70, 60, 50, 40, . . .

Here

70, 60, 50, 40, . . .

First term a = 70

Second term t1 = 60

Third term t2 = 50

Fourth term t3 = 40

We know that d = tn + 1 – tn

Thus, t2 – t1 = 50 – 60 = – 10

t3 – t2 = 40 – 50 = – 10

Thus, d = – 10

Question 5.

Decide whether following sequence is an A.P., if so find the 20th term of the progression.

– 12, – 5, 2, 9, 16, 23, 30, . . .

Given A.P. is – 12, – 5, 2, 9, 16, 23, 30, . . .

Where first term a = – 12

Second term t1 = – 5

Third term t2 = 2

Common Difference d = t2 – t1 = 2 – ( – 5) = 2 + 5 = 7

We know that, nth term of an A.P. is

tn = a + (n – 1)d

We need to find the 20th term,

Here n = 20

Thus, t20 = – 12 + (20 – 1)× 7

t20 = – 12 + (19)× 7 = – 12 + 133 = 121

Thus, t20 = 121

Question 6.

Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this progression.

Given A.P. is 12, 16, 20, 24, . . .

Where first term a = 12

Second term t1 = 16

Third term t2 = 20

Common Difference d = t2 – t1 = 20 – 16 = 4

We know that, nth term of an A.P. is

tn = a + (n – 1)d

We need to find the 24th term,

Here n = 24

Thus, t24 = 12 + (24 – 1)× 4

t24 = 12 + (23)× 4 = 12 + 92 = 104

Thus, t24 = 104

Question 7.

Find the 19th term of the following A.P.

7, 13, 19, 25, . . .

Given A.P. is 7, 13, 19, 25, . . .

Where first term a = 7

Second term t1 = 13

Third term t2 = 19

Common Difference d = t2 – t1 = 19 – 13 = 6

We know that, nth term of an A.P. is

tn = a + (n – 1)d

We need to find the 19th term,

Here n = 19

Thus, t19 = 7 + (19 – 1)× 6

t19 = 7 + (18)× 6 = 7 + 108 = 115

Thus, t19 = 115

Question 8.

Find the 27th term of the following A.P.

9, 4, – 1, – 6, – 11, . . .

Given A.P. is 9, 4, – 1, – 6, – 11, . . .

Where first term a = 9

Second term t1 = 4

Third term t2 = – 1

Common Difference d = t2 – t1 = – 1 – 4 = – 5

We know that, nth term of an A.P. is

tn = a + (n – 1)d

We need to find the 27th term,

Here n = 27

Thus, t27 = 9 + (27 – 1)× ( – 5)

t27 = 9 + (26)× ( – 5) = 9 – 130 = – 121

Thus, t27 = – 121

Question 9.

Find how many three digit natural numbers are divisible by 5.

List of three digit number divisible by 5 are

100, 105,110,115,……….. 995

Let us find how many such number are there?

From the above sequence, we know that

tn = 995, a = 100

t1 = 105, t2 = 110

Thus, d = t2 – t1 = 110 – 105 = 5

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

we can find value of “n”

Thus, on substituting all the value in formula we get,

995 = 100 + (n – 1)× 5

⇒ 995 – 100 = (n – 1)× 5

⇒ 895 = (n – 1) × 5

⇒ n = 179 + 1 = 180

Question 10.

The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th term of that A.P.

Given: t11 = 16 and t21 = 29

To find: t41

Using nth term of an A.P. formula

tn = a + (n – 1)d

we will find value of “a” and “d”

Let, t11 = a + (11 – 1) d

⇒ 16 = a + 10 d …..(1)

t21 = a + (21 – 1) d

⇒ 29 = a + 20 d …..(2)

Subtracting eq. (1) from eq. (2), we get,

⇒ 29 – 16 = (a – a) + (20 d – 10 d)

⇒ 13 = 10 d

Substitute value of “d” in eq. (1) to get value of “a”

⇒ 16 = a + 13

⇒ a = 16 – 13 = 3

Now, we will find the value of t41 using nth term of an A.P. formula

⇒ t41 = 3 + 4 × 13 = 3 + 52 = 55

Thus, t41 = 55

Question 11.

11, 8, 5, 2, . . . In this A.P. which term is number – 151?

By, given A.P. 11, 8, 5, 2, . . .

we know that

a = 11, t1 = 8, t2 = 5

Thus, d = t2 – t1 = 5 – 8 = – 3

Given: tn = – 151

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

we can find value of “n”

Thus, on substituting all the value in formula we get,

– 151 = 11 + (n – 1)× ( – 3)

⇒ – 151 – 11 = (n – 1)× ( – 3)

⇒ – 162 = (n – 1) × ( – 3)

⇒ n = 54 + 1 = 55

Question 12.

In the natural numbers from 10 to 250, how many are divisible by 4?

List of number divisible by 4 in between 10 to 250 are

12, 16,20,24,……….. 248

Let us find how many such number are there?

From the above sequence, we know that

tn = 248, a = 12

t1 = 16, t2 = 20

Thus, d = t2 – t1 = 20 – 16 = 4

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

we can find value of “n”

Thus, on substituting all the value in formula we get,

248 = 12 + (n – 1)× 4

⇒ 248 – 12 = (n – 1)× 4

⇒ 236 = (n – 1) × 4

⇒ n = 59 + 1 = 60

Question 13.

In an A.P. 17th term is 7 more than its 10th term. Find the common difference.

Given: t17 = 7 + t10 ……(1)

In t17, n = 17

In t10, n = 10

By using nth term of an A.P. formula,

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth term

Thus, on using formula in eq. (1) we get,

⇒ a + (17 – 1)d = 7 + (a + (10 – 1)d)

⇒ a + 16 d = 7 + (a + 9 d)

⇒ a + 16 d – a – 9 d = 7

⇒ 7 d = 7

Thus, common difference “d” = 1

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