Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 5 - Arithmetic Progression

The given AP is

First term = 6, common difference =

a = 6, d =

The n^th term is given by

Hence, 37^th term is 69.

The given AP is

The first term = 5,

common difference =

a = 5,

The n^th term is given by

Hence, the 25^th term is -7.

Since the numbers  are in AP,

Therefore,

Hence, P = 2 and the numbers are 3, 7, 11.

Given AP is 16, 9, 2, -5.

First term, a = 16

Common difference, d = 9 - 16 = -7

The n^th term of the AP is given by

 (given)

Then,

Now,

Since , the given progression is in AP.

Thus, we have

(i) first term = -6

(ii) common difference = 4

(iii) 16^th term =  

In the given AP, we have a = 6 and d = (10 - 6) = 4

Suppose there are n terms in the given AP,.

Then,

Hence, there are 43 terms in the given AP.

In the given AP we have a = 41 and d = 38 - 41 = - 3

Suppose there are n terms in AP.

Then,

Hence, there are 12 terms in the given AP.

In the given AP, we have a = 3 and d = 8 - 3 = 5

Suppose there are n terms in given AP.

Then,

Hence, the 18^th term of given AP is 88.

In the given AP, we have a = 72 and d = 68 - 72 = - 4

Suppose there are n terms in given AP.

Then, 

Hence, the 19^th term in the given AP is 0.

In the given AP, we have

Suppose there are n terms in given AP, we have

Then,

Thus, the 14^th term in the given AP is 3.

Here, a = 8 and d = 14 - 8 = 6

Therefore, 41^st term is given by

Required term = 248 + 72 = 320

Let it be nth term.

Then,

Hence, 53^rd is the required term.

The given AP is 5, 15, 25....

a = 5, d = 15 - 5 = 10

Thus, the required term is 44^th term.

In the given AP let the first term = a,

And common difference = d

Then, T_n = a + (n - 1)d

T₁₀= a + (10 - 1)d, T₁₇ = a + (17 - 1)d and T₁₃ = a + (13 - 1)d

T₁₀ = a + 9d, T₁₇ = a + 16d and T₁₃ = a + 12d

Now, T₁₀ = 52 ⇒ a + 9d = 52.....(1)

and T₁₇ = T₁₃  + 20 ⇒ a + 16d = a + 12d + 20

⇒ 4d = 20 ⇒ d = 5

Putting d = 5 in (1), we get

a + 9 x 5 = 52 ⇒ a = 52 - 45 ⇒ a = 7

Thus, a = 7 and d = 5

So the required AP is 7, 12, 17, 22....

Here a = 7, d = (10 - 7) = 3, l = 184

And n = 8

Hence, the 8^th term from the end is 163.

Here a = 17, d = (14 - 17) = -3, l = -40

And n = 6

Now, n^th term from the end = [l - (n - 1)d]

Hence, the 6^th term from the end is -25.

In the given AP, let the first term = a common difference = d

So the required AP is 8, 6, 4, 2, 0......

Let the first term of given AP = a and common difference = d

Hence 25^th term is triple its 11^th term.

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

Then, n^th term,  

Now,

Now,

And,

Hence, proved.

Let a be the first term and d be the common difference

First AP is 63, 65, 67....

First term = 63, common difference = 65 - 63 = 2

n^thterm = 63 + (n - 1) 2 = 63 + 2n - 2 = 2n + 61

Second AP is 3, 10, 17 ....

First term = 3, common difference = 10 - 3 = 7

n^th term = 3 + (n - 1) 7 = 3 + 7n - 7 = 7n - 4

The two n^th terms are equal

2n + 61 = 7n - 4 or 5n = 61 + 4 = 65

Let a be the first term and d be the common difference.

p^th term = a +(p - 1)d = q    (given)-----(1)

q^th term = a +(q - 1) d = p   (given)-----(2)

Subtracting (2) from (1),

(p - q)d = q - p

(p - q)d = -(p - q)

d = -1

Putting d = -1 in (1)

Let a be the first term and d be the common difference

n^th term from the beginning = a + (n - 1)d-----(1)

n^th term from end= l - (n - 1)d ----(2)

adding (1) and (2),

sum of the n^th term from the beginning and n^th term from the end = [a + (n - 1)d] + [l - (n - 1)d] = a + l

The three-digit numbers divisible by 9 are 108, 117, 126, 135,......, 999.

Here,

a = 108 and d = 9

Now,

Hence, 100 three-digit numbers are divisible by 9.

Number of rose plants in first, second, third rows.... are 43, 41, 39... respectively.

There are 11 rose plants in the last row

So, it is an AP . viz. 43, 41, 39 .... 11

a = 43, d = 41 - 43 = -2, l = 11

Let n^th term be the last term

Hence, there are 17 rows in the flower bed.

Integers between 200 and 500 divisible by 8 are 208, 216, 224, 232, ……, 496.

Clearly, these numbers form an AP with

a = 208, d = 8 and l = 496.

Let it contain n terms.

Then,

Hence, there are 37 integers between 200 and 500 which are divisible by 8.

If are consecutive terms of an AP, then

Let the three numbers in AP be .

Then, sum of these numbers  

And, product of these numbers 

Given, sum = 15 and product = 105

Then,

And,

Hence, the required numbers are (3, 5, 7) or (7, 5, 3).

But sum = 24 and product = 440

Let the required numbers be (a - d), a, (a + d)

Sum of these numbers = (a - d) + a + (a + d) = 3a

sum of squares of these numbers =

Sum of numbers = 21, sum of squares of numbers = 165

3a = 21

a = 7

Thus, a = 7 and d =

Hence, the required numbers are (4, 7, 10) or (10, 7, 4).

Let the required angles be (a - 3d)°, (a - d) °, (a + d) ° and (a + 3d) °

Common difference = (a - d) - (a- 3d) = a - d - a + 3d = 2d

Common difference = 10°

2d = 10° = d = 5°

Sum of four angles of quadrilateral = 360°

First angle = (a - 3d)° = (90 - 3 × 5) ° = 75°

Second angle = (a - d)° = (90 - 5) ° = 85°

Third angle = (a + d)° = (90 + 5°) = 95°

Fourth angle = (a + 3d)° = (90 + 3 × 5)° = 105°

Let the required number be (a - 3d), (a - d), (a + d) and (a + 3d).

Sum of these numbers = (a - 3d) + (a - d)+ (a + d) + (a + 3d)

4a = 28 a = 7

Sum of the squares of these numbers

Hence, the required numbers are (4, 6, 8, 10) or (10, 8, 6, 4).

Let the three numbers in AP be .

Then, sum of these numbers  

Given, sum = 18

And,

When d = -9, the numbers are 15, 6, -3.

When d = 4, the numbers are 2, 6, 10.

Since the numbers a, 7, b, 23, c are in AP,

7 - a = b - 7 = 23 - b = c - 23

Consider b - 7 = 23 - b

⇒ 2b = 30

⇒ b = 15

Consider 7 - a = b - 7

⇒ a + b = 14

⇒ a + 15 = 14

⇒ a = -1

Consider 23 - b = c - 23

⇒ b + c = 46

⇒ 15 + c = 46

⇒ c = 31

Hence, a = -1, b = 15 and c = 31.

Given arithmetic series is as follows:

5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + … + (-5) + 81 + (-3)

= (5 + 9 + 13 + 17 + … + 81) + [(-41) + (-39) + (-37) + … + (-5) + (-3)] …..(i)

Let A₁ = 5 + 9 + 13 + + 17 + … + 81

Here,

a = 5, l = 81 and d = 9 - 5 = 4

Let A₁ series contain n₁ terms. Then,

Let A₂ = (-41) + (-39) + (-37) + … + (-5) + (-3)

Here,

a = -41, l = -3 and d = -39 - (-41) = 2

Let A₂ series contain n₂ terms. Then,

Thus, we have

5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + … + (-5) + 81 + (-3)

= (5 + 9 + 13 + + 17 + … + 81) + [(-41) + (-39) + (-37) + … + (-5) + (-3)]

(i) The n^th term is given by

(ii) Putting n = 1 in , we have

(iii) Putting n = 1 in , we have

Now, common difference, d =

The n^th term is given by

Putting n = 20 in , we have

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

Then,

Subtracting (ii) from (i),

Substituting the value of 'd' in (i),

Sum of first mn terms is given by

Here a = 21, d = (18 - 21) = -3

Let the required number of terms be n, then

sum of first 15 terms = 0

Odd numbers between 0 and 50 are 1, 3, 5, 7, ……, 49.

Clearly, these numbers form an AP with

a = 1, d = 2 and l = 49.

Let it contain n terms.

Then,

Required sum is given by

First 15 multiples of 8 are 8, 16, 24, ……, 120.

Clearly, these numbers form an AP with

a = 8, d = 8, l = 120 and n = 15

Required sum is given by

All numbers between 300 and 700 that are multiples of 9 are 306, 315, 324, 333, ..., 693

This is an AP in which a = 306, d = (315 - 306) = 9, l = 693

Let the number of these terms be n, then

Three-digit numbers divisible by 13 are 104, 117, 130, ……, 988.

Clearly, these numbers form an AP with

a = 104, d = 13 and l = 988.

Let it contain n terms.

Then,

Required sum is given by

First 100 even natural numbers divisible by 5 are 10, 20, 30, ……, 100 terms.

Clearly, these numbers form an AP with

a = 10, d = 10 and n = 100

Required sum is given by

Required sum of n terms is given by

First term 'a' of an AP = 2

The last term l = 29

common difference = 3

First term of an AP, a = 22

Last term = n^th term = - 11

Thus, n = 12, d = -3

Let the first term and common difference of 1^st AP be 'a' and 'd' respectively.

And, let the first term and common difference of 2^nd AP be 'A' and 'D' respectively.

Then

Ratio of 12^th terms

Putting n = 23 in (i),

The values of four prizes form an AP with common difference d = -20.

Let the value of first prize = a

And, n = 7

Now, sum of values of prizes = Rs. 700

Hence, the values of 7 prizes are Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60 and Rs. 40.

Let the first instalment be Rs. a and let the common difference be d.

Now, for n = 40, S₄₀ = 36000

When 30 instalments are paid, one-third is unpaid.

That means, two-rhird is paid.

So, for n = 30, S₃₀ =  

Now,

Hence, the value of first instalment is Rs. 510.

Saving done by a child on 1^st day = Rs. 5

Saving done by a child on 2^nd day = Rs. 5 + Rs. 5 = Rs. 10

Saving done by a child on 3^rd day = Rs. 10 + Rs. 5 = Rs. 15 and so on

We observe that the savings done by a child form an AP a = 5 and d = 5.

Let the child contributes five-rupee coins in a piggy bank for n days.

Total number of five-rupee coins in the piggy bank = 190

Therefore, total money saved by a child in n days = Rs. 5 × 190 = Rs. 950

Then,

Hence, a child put the five-rupee coins into piggy bank for 19 days and she saved Rs. 950.

For an AP,

First term = a

Second term = b

Then, common difference = b - a

Last term, l = c

Let 'c' be the n^th term.

Then, n^th term = 

Now, required sum is given by

Hence, proved.

Given AP is

First term

Common difference

Therefore, next term of AP

Given AP is

First term

Common difference

Therefore, next term of AP

The given AP is 21, 18, 15, ....

First term = 21, common difference = 18 - 21= - 3

Let n^th term be zero

a + (n - 1)d = 0or 21 + (n - 1)(-3) = 0

21 - 3n + 3 = 0

3n = 24

or 

Hence, 8^th term of given AP is 0.

Sum of n natural numbers = 1 + 2 + 3 + ... + n

Here a = 1, d = 2 - 1 = 1

Sum of even natural numbers = 2 + 4 + 6 + ... to n terms

a = 2, d = 4 - 2 = 2

First term of AP = a = p

Common difference = d = q

n^th term = a + (n 1)d

10^th term = p + (10 1)q

               = p + 9q

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

Given,

Given,

1 + 4 + 7 + 10 + ….. + x = 287

This forms an AP with

first term, a = 1

and common difference = 4 - 1 = 3

Let x be the n^th term.

Now,

Solving the quadratic equation,

Correct option: (d)

Here,

First term, a = 5

Common difference, d = 9 - 5 = 4

Let 185 be the n^th term.

Then,

Correct option: (d)

Given,

(2p + 1), 10 and (5p + 5) are three consecutive terms of an AP.

Therefore,

10 - (2p + 1) = (5p + 5) - 10

⇒ 9 - 2p = 5p - 5

⇒ 7p = 14

⇒ p = 2