RD Sharma Solutions for Class 11 Maths Chapter 19 Arithmetic Progressions

RD Sharma Solutions Class 11 Maths Chapter 19 – Free PDF Download Updated for (2023-24)

RD Sharma Solutions for Class 11 Maths Chapter 19 – Arithmetic Progressions are provided here for students to study and prepare for their board exams. A sequence is called an arithmetic progression if the difference between a term and the previous term is always the same. For students who are aspiring to get a good grip on the CBSE Class 11 Mathematics, the RD Sharma Class 11 Maths is the right tool for the purpose.

Chapter 19 – Arithmetic Progressions contains seven exercises, and RD Sharma Solutions provides comprehensive answers to the questions present in each exercise. Experts at BYJU’S have prepared the RD Sharma Solutions based on the students grasping abilities for a thorough understanding of concepts. To boost interest among students in this chapter, RD Sharma Class 11 Maths Solutions PDF links are given below for easy access. Now, let us have a look at the concepts discussed in this chapter.

• Sequence
• Arithmetic Progression (A.P.)
• The general term of an A.P
• Selection of terms in an A.P
• Sum to n terms of an A.P
• Properties of arithmetic progressions
• Insertion of arithmetic means
• Applications of A.P

RD Sharma Solutions for Class 11 Maths Chapter 19 – Arithmetic Progressions

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EXERCISE 19.1 PAGE NO: 19.4

1. If the n^th term of a sequence is given by a_n = n² – n+1, write down its first five terms.

Solution:

Given:

a_n = n² – n+1

By using the values n = 1, 2, 3, 4, 5, we can find the first five terms.

When n = 1:

a₁ = (1)² – 1 + 1

= 1 – 1 + 1

= 1

When n = 2:

a₂ = (2)² – 2 + 1

= 4 – 2 + 1

= 3

When n = 3:

a₃ = (3)² – 3 + 1

= 9 – 3 + 1

= 7

When n = 4:

a₄ = (4)² – 4 + 1

= 16 – 4 + 1

= 13

When n = 5:

a₅ = (5)² – 5 + 1

= 25 – 5 + 1

= 21

∴ The first five terms of the sequence are 1, 3, 7, 13, 21.

2. A sequence is defined by a_n = n³ – 6n² + 11n – 6, n ∈ N. Show that the first three terms of the sequence are zero, and all other terms are positive.

Solution:

Given:

a_n = n³ – 6n² + 11n – 6, n ∈ N

By using the values n = 1, 2, 3, we can find the first three terms.

When n = 1:

a₁ = (1)³ – 6(1)² + 11(1) – 6

= 1 – 6 + 11 – 6

= 12 – 12

= 0

When n = 2:

a₂ = (2)³ – 6(2)² + 11(2) – 6

= 8 – 6(4) + 22 – 6

= 8 – 24 + 22 – 6

= 30 – 30

= 0

When n = 3:

a₃ = (3)³ – 6(3)² + 11(3) – 6

= 27 – 6(9) + 33 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0

This shows that the first three terms of the sequence are zero.

Now, let’s check for when n = n.

a_n = n³ – 6n² + 11n – 6

= n³ – 6n² + 11n – 6 – n + n – 2 + 2

= n³ – 6n² + 12n – 8 – n + 2

= (n)³ – 3×2n(n – 2) – (2)³ – n + 2

By using the formula, {(a – b)³ = (a)³ – (b)³ – 3ab(a – b)}

a_n = (n – 2)³ – (n – 2)

Here, n – 2 will always be positive for n > 3

∴ a_n is always positive for n > 3

3. Find the first four terms of the sequence defined by a₁ = 3 and a_n = 3a_n–1 + 2 for all n > 1.

Solution:

Given:

a₁ = 3 and a_n = 3a_n–1 + 2, for all n > 1

By using the values n = 1, 2, 3, 4, we can find the first four terms.

When n = 1:

a₁ = 3

When n = 2:

a₂ = 3a_2–1 + 2

= 3a₁ + 2

= 3(3) + 2

= 9 + 2

= 11

When n = 3:

a₃ = 3a_3–1 + 2

= 3a₂ + 2

= 3(11) + 2

= 33 + 2

= 35

When n = 4:

a₄ = 3a_4–1 + 2

= 3a₃ + 2

= 3(35) + 2

= 105 + 2

= 107

∴ The first four terms of the sequence are 3, 11, 35, 107.

4. Write the first five terms in each of the following sequences:
(i) a₁ = 1, a_n = a_n–1 + 2, n > 1

(ii) a₁ = 1 = a₂, a_n = a_n–1 + a_n–2, n > 2

(iii) a₁ = a₂ =2, a_n = a_n–1 – 1, n > 2

Solution:

(i) a₁ = 1, a_n = a_n–1 + 2, n > 1

By using the values n = 1, 2, 3, 4, 5, we can find the first five terms.

Given:

a₁ = 1

When n = 2:

a₂ = a_2–1 + 2

= a₁ + 2

= 1 + 2

= 3

When n = 3:

a₃ = a_3–1 + 2

= a₂ + 2

= 3 + 2

= 5

When n = 4:

a₄ = a_4–1 + 2

= a₃ + 2

= 5 + 2

= 7

When n = 5:

a₅ = a_5–1 + 2

= a₄ + 2

= 7 + 2

= 9

∴ The first five terms of the sequence are 1, 3, 5, 7, 9.

(ii) a₁ = 1 = a₂, a_n = a_n–1 + a_n–2, n > 2

By using the values n = 1, 2, 3, 4, 5, we can find the first five terms.

Given:

a₁ = 1

a₂ = 1

When n = 3:

a₃ = a_3–1 + a_3–2

= a₂ + a₁

= 1 + 1

= 2

When n = 4:

a₄ = a_4–1 + a_4–2

= a₃ + a₂

= 2 + 1

= 3

When n = 5:

a₅ = a_5–1 + a_5–2

= a₄ + a₃

= 3 + 2

= 5

∴ The first five terms of the sequence are 1, 1, 2, 3, 5.

(iii) a₁ = a₂ =2, a_n = a_n–1 – 1, n > 2

By using the values n = 1, 2, 3, 4, 5, we can find the first five terms.

Given:

a₁ = 2

a₂ = 2

When n = 3:

a₃ = a_3–1 – 1

= a₂ – 1

= 2 – 1

= 1

When n = 4:

a₄ = a_4–1 – 1

= a₃ – 1

= 1 – 1

= 0

When n = 5:

a₅ = a_5–1 – 1

= a₄ – 1

= 0 – 1

= -1

∴ The first five terms of the sequence are 2, 2, 1, 0, -1.

5. The Fibonacci sequence is defined by a₁ = 1 = a₂, a_n = a_n–1 + a_n–2 for n > 2. Find (a_n+1)/a_n, for n = 1, 2, 3, 4, 5.

Solution:

Given:

a₁ = 1

a₂ = 1

a_n = a_n–1 + a_n–2

When n = 1:

(a_n+1)/a_n = (a₁₊₁)/a₁

= a₂/a₁

= 1/1

= 1

a₃ = a_3–1 + a_3–2

= a₂ + a₁

= 1 + 1

= 2

When n = 2:

(a_n+1)/a_n = (a₂₊₁)/a₂

= a₃/a₂

= 2/1

= 2

a₄ = a_4–1 + a_4–2

= a₃ + a₂

= 2 + 1

= 3

When n = 3:

(a_n+1)/a_n = (a₃₊₁)/a₃

= a₄/a₃

= 3/2

a₅ = a_5–1 + a_5–2

= a₄ + a₃

= 3 + 2

= 5

When n = 4:

(a_n+1)/a_n = (a₄₊₁)/a₄

= a₅/a₄

= 5/3

a₆ = a_6–1 + a_6–2

= a₅ + a₄

= 5 + 3

= 8

When n = 5:

(a_n+1)/a_n = (a₅₊₁)/a₅

= a₆/a₅ = 8/5

∴ Value of (a_n+1)/a_n when n = 1, 2, 3, 4, 5 are 1, 2, 3/2, 5/3, 8/5

EXERCISE 19.2 PAGE NO: 19.11

1. Find:
(i) 10^th term of the A.P. 1, 4, 7, 10, …..

(ii) 18^th term of the A.P. √2, 3√2, 5√2, …

(iii) nth term of the A.P 13, 8, 3, -2, ….

Solution:

(i) 10^th term of the A.P. 1, 4, 7, 10, …..

Arithmetic Progression (AP) whose common difference is = a_n – a_n-1 where n > 0

Let us consider, a = a₁ = 1, a₂ = 4 …

So, Common difference, d = a₂ – a₁ = 4 – 1 = 3

To find the 10^th term of A.P, firstly, find a_n

By using the formula,

a_n = a + (n-1) d

= 1 + (n-1) 3

= 1 + 3n – 3

= 3n – 2

When n = 10:

a₁₀ = 3(10) – 2

= 30 – 2

= 28

Hence, the 10^th term is 28.

(ii) 18^th term of the A.P. √2, 3√2, 5√2, …

Arithmetic Progression (AP) whose common difference is = a_n – a_n-1 where n > 0

Let us consider, a = a₁ = √2, a₂ = 3√2 …

So, Common difference, d = a₂ – a₁ = 3√2 – √2 = 2√2

To find the 18^th term of A.P, firstly, find a_n

By using the formula,

a_n = a + (n-1) d

= √2 + (n – 1) 2√2

= √2 + 2√2n – 2√2

= 2√2n – √2

When n = 18:

a₁₈ = 2√2(18) – √2

= 36√2 – √2

= 35√2

Hence, the 10^th term is 35√2

(iii) nth term of the A.P 13, 8, 3, -2, ….

Arithmetic Progression (AP) whose common difference is = a_n – a_n-1 where n > 0

Let us consider, a = a₁ = 13, a₂ = 8 …

So, Common difference, d = a₂ – a₁ = 8 – 13 = -5

To find the n^th term of A.P, firstly, find a_n

By using the formula,

a_n = a + (n-1) d

= 13 + (n-1) (-5)

= 13 – 5n + 5

= 18 – 5n

Hence, the n^th term is 18 – 5n

2. In an A.P., show that a_m+n + a_m–n = 2a_m.

Solution:

We know the first term is ‘a’, and the common difference of an A.P is d.

Given:

a_m+n + a_m–n = 2a_m

By using the formula,

a_n = a + (n – 1)d

Now, let us take LHS: a_m+n + a_m-n

a_m+n + a_m-n = a + (m + n – 1)d + a + (m – n – 1)d

= a + md + nd – d + a + md – nd – d

= 2a + 2md – 2d

= 2(a + md – d)

= 2[a + d(m – 1)] {∵ a_n = a + (n – 1)d}

a_m+n + a_m-n = 2a_m

Hence proved.

3. (i) Which term of the A.P. 3, 8, 13,… is 248?

(ii) Which term of the A.P. 84, 80, 76,… is 0?

(iii) Which term of the A.P. 4, 9, 14,… is 254?

Solution:

(i) Which term of the A.P. 3, 8, 13,… is 248?

Given A.P is 3, 8, 13,…

Here, a₁ = a = 3, a₂ = 8

Common difference, d = a₂ – a₁ = 8 – 3 = 5

We know, a_n = a + (n – 1)d

a_n = 3 + (n – 1)5

= 3 + 5n – 5

= 5n – 2

Now, to find which term of A.P is 248,

Put a_n = 248

∴ 5n – 2 = 248

= 248 + 2

= 250

= 250/5

= 50

Hence, the 50^th term of the given A.P. is 248.

(ii) Which term of the A.P. 84, 80, 76,… is 0?

Given A.P is 84, 80, 76,…

Here, a₁ = a = 84, a₂ = 88

Common difference, d = a₂ – a₁ = 80 – 84 = -4

We know, a_n = a + (n – 1)d

a_n = 84 + (n – 1)-4

= 84 – 4n + 4

= 88 – 4n

Now, to find which term of A.P is 0,

Put a_n = 0

88 – 4n = 0

-4n = -88

n = 88/4

= 22

Hence, the 22^nd term of the given A.P. is 0.

(iii) Which term of the A.P. 4, 9, 14,… is 254 ?

Given A.P. is 4, 9, 14,…

Here, a₁ = a = 4, a₂ = 9

Common difference, d = a₂ – a₁ = 9 – 4 = 5

We know, a_n = a + (n – 1)d

a_n = 4 + (n – 1)5

= 4 + 5n – 5

= 5n – 1

Now, to find which term of A.P is 254,

Put a_n = 254

5n – 1 = 254

5n = 254 + 1

5n = 255

n = 255/5

= 51

Hence, the 51^st term of the given A.P. is 254.

4. (i) Is 68 a term of the A.P. 7, 10, 13,…?

(ii) Is 302 a term of the A.P. 3, 8, 13,…?

Solution:

(i) Is 68 a term of the A.P. 7, 10, 13,…?

Given A.P. is 7, 10, 13,…

Here, a₁ = a = 7, a₂ = 10

Common difference, d = a₂ – a₁ = 10 – 7 = 3

We know, a_n = a + (n – 1)d [where a is the first term or a₁ and d is the common difference and n is any natural number]

a_n = 7 + (n – 1)3

= 7 + 3n – 3

= 3n + 4

Now, to find whether 68 is a term of this A.P. or not,

Put a_n = 68

3n + 4 = 68

3n = 68 – 4

3n = 64

n = 64/3

64/3 is not a natural number.

Hence, 68 is not a term of the given A.P.

(ii) Is 302 a term of the A.P. 3, 8, 13,…?

Given A.P is 3, 8, 13,…

Here, a₁ = a = 3, a₂ = 8

Common difference, d = a₂ – a₁ = 8 – 3 = 5

We know, a_n = a + (n – 1)d

a_n = 3 + (n – 1)5

= 3 + 5n – 5

= 5n – 2

To find whether 302 is a term of this A.P. or not,

Put a_n = 302

5n – 2 = 302

5n = 302 + 2

5n = 304

n = 304/5

304/5 is not a natural number

Hence, 304 is not a term of the given A.P.

5. (i) Which term of the sequence 24, 23 ¼, 22 ½, 21 ¾ is the first negative term?

Solution:

Given:

AP: 24, 23 ¼, 22 ½, 21 ¾, … = 24, 93/4, 45/2, 87/4, …

Here, a₁ = a = 24, a₂ = 93/4

Common difference, d = a₂ – a₁ = 93/4 – 24

= (93 – 96)/4

= – 3/4

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

We know, a_n = a + (n – 1) d

a_n = 24 + (n – 1) (-3/4)

= 24 – 3/4n + ¾

= (96+3)/4 – 3/4n

= 99/4 – 3/4n

Now we need to find the first negative term.

Put a_n < 0

a_n = 99/4 – 3/4n < 0

99/4 < 3/4n

3n > 99

n > 99/3

n > 33

Hence, the 34^th term is the first negative term of the given AP.

(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, … is (a) purely real (b) purely imaginary?

Solution:

Given:

AP: 12 + 8i, 11 + 6i, 10 + 4i, …

Here, a₁ = a = 12 + 8i, a₂ = 11 + 6i

Common difference, d = a₂ – a₁

= 11 + 6i – (12 + 8i)

= 11 – 12 + 6i – 8i

= -1 – 2i

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

a_n = 12 + 8i + (n – 1) -1 – 2i

= 12 + 8i – n – 2ni + 1 + 2i

= 13 + 10i – n – 2ni

= (13 – n) + (10 – 2n) i

To find the purely real term of this A.P., the imaginary part has to be zero.

10 – 2n = 0

2n = 10

n = 10/2

= 5

Hence, 5^th term is purely real.

To find the purely imaginary term of this A.P., the real part has to be zero.

∴ 13 – n = 0

n = 13

Hence, the 13^th term is purely imaginary.

6. (i) How many terms are in A.P. 7, 10, 13,…43?

Solution:

Given:

AP: 7, 10, 13,…

Here, a₁ = a = 7, a₂ = 10

Common difference, d = a₂ – a₁ = 10 – 7 = 3

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

a_n = 7 + (n – 1)3

= 7 + 3n – 3

= 3n + 4

To find the total terms of the A.P., put a_n = 43 as 43 is the last term of A.P.

3n + 4 = 43

3n = 43 – 4

3n = 39

n = 39/3

= 13

Hence, total 13 terms exist in the given A.P.

(ii) How many terms are there in the A.P. -1, -5/6, -2/3, -1/2, …, 10/3?

Solution:

Given:

AP: -1, -5/6, -2/3, -1/2, …

Here, a₁ = a = -1, a₂ = -5/6

Common difference, d = a₂ – a₁

= -5/6 – (-1)

= -5/6 + 1

= (-5+6)/6

= 1/6

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

a_n = -1 + (n – 1) 1/6

= -1 + 1/6n – 1/6

= (-6-1)/6 + 1/6n

= -7/6 + 1/6n

To find the total terms of the AP,

Put a_n = 10/3 [Since, 10/3 is the last term of AP]

a_n = -7/6 + 1/6n = 10/3

1/6n = 10/3 + 7/6

1/6n = (20+7)/6

1/6n = 27/6

n = 27

Hence, total 27 terms exist in the given A.P.

7. The first term of an A.P. is 5, the common difference is 3, and the last term is 80; find the number of terms.

Solution:

Given:

First term, a = 5; last term, l = a_n = 80

Common difference, d = 3

We know, a_n = a + (n – 1) d [where a is the  first term or a₁ and d is the common difference and n is any natural number]

a_n = 5 + (n – 1)3

= 5 + 3n – 3

= 3n + 2

To find the total terms of the A.P., put a_n = 80 as 80 is the last term of A.P.

3n + 2 = 80

3n = 80 – 2

3n = 78

n = 78/3

= 26

Hence, total 26 terms exist in the given A.P.

8. The 6^th and 17^th terms of an A.P. are 19 and 41, respectively. Find the 40^th term.

Solution:

Given:

6^th term of an A.P is 19, and 17^th term of an A.P. is 41

So, a₆ = 19 and a₁₇ = 41

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

When n = 6:

a₆ = a + (6 – 1) d

= a + 5d

Similarly, When n = 17:

a₁₇ = a + (17 – 1)d

= a + 16d

According to the question,

a₆ = 19 and a₁₇ = 41

a + 5d = 19 ……………… (i)

And a + 16d = 41………….. (ii)

Let us subtract equation (i) from (ii), and we get,

a + 16d – (a + 5d) = 41 – 19

a + 16d – a – 5d = 22

11d = 22

d = 22/11

= 2

Put the value of d in equation (I).

a + 5(2) = 19

a + 10 = 19

a = 19 – 10

= 9

As, a_n = a + (n – 1)d

a₄₀ = a + (40 – 1)d

= a + 39d

Now, put the value of a = 9 and d = 2 in a₄₀, and we get,

a₄₀ = 9 + 39(2)

= 9 + 78

= 87

Hence, the 40^th term of the given A.P. is 87.

9. If the 9^th term of an A.P. is Zero, prove that its 29^th term is double the 19^th term.

Solution:

Given:

9^th term of an A.P is 0

So, a₉ = 0

We need to prove: a₂₉ = 2a₁₉

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

When n = 9:

a₉ = a + (9 – 1)d

= a + 8d

According to the question,

a₉ = 0

a + 8d = 0

a = -8d

When n = 19:

a₁₉ = a + (19 – 1)d

= a + 18d

= -8d + 18d

= 10d

When n = 29:

a₂₉ = a + (29 – 1)d

= a + 28d

= -8d + 28d [Since, a = -8d]

= 20d

= 2×10d

a₂₉ = 2a₁₉ [Since, a₁₉ = 10d]

Hence proved.

10. If 10 times the 10^th term of an A.P. is equal to 15 times the 15^th term, show that the 25^th term of the A.P. is Zero.

Solution:

Given:

10 times the 10^th term of an A.P. is equal to 15 times the 15^th term

So, 10a₁₀ = 15a₁₅

We need to prove: a₂₅ = 0

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

When n = 10:

a₁₀ = a + (10 – 1)d

= a + 9d

When n = 15:

a₁₅ = a + (15 – 1)d

= a + 14d

When n = 25:

a₂₅ = a + (25 – 1)d

= a + 24d ………(i)

According to the question,

10a₁₀ = 15a₁₅

10(a + 9d) = 15(a + 14d)

10a + 90d = 15a + 210d

10a – 15a + 90d – 210d = 0

-5a – 120d = 0

-5(a + 24d) = 0

a + 24d = 0

a₂₅ = 0 [From (i)]

Hence Proved.

11. The 10^th and 18^th terms of an A.P. are 41 and 73, respectively. Find the 26^th term.

Solution:

Given:

The 10^th term of an A.P is 41, and the 18^th term of an A.P. is 73.

So, a₁₀ = 41 and a₁₈ = 73

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

When n = 10:

a₁₀ = a + (10 – 1)d

= a + 9d

When n = 18:

a₁₈ = a + (18 – 1)d

= a + 17d

According to the question,

a₁₀ = 41 and a₁₈ = 73

a + 9d = 41 ………………(i)

And a + 17d = 73…………..(ii)

Let us subtract equation (i) from (ii), and we get,

a + 17d – (a + 9d) = 73 – 41

a + 17d – a – 9d = 32

8d = 32

d = 32/8

d = 4

Put the value of d in equation (I), and we get,

a + 9(4) = 41

a + 36 = 41

a = 41 – 36

a = 5

We know, a_n = a + (n – 1)d

a₂₆ = a + (26 – 1)d

= a + 25d

Now, put the value of a = 5 and d = 4 in a₂₆

a₂₆ = 5 + 25(4)

= 5 + 100

= 105

Hence, the 26^th term of the given A.P. is 105.

12. In a certain A.P., the 24^th term is twice the 10^th term. Prove that the 72^nd term is twice the 34^th term.

Solution:

Given:

The 24^th term is twice the 10^th term.

So, a₂₄ = 2a₁₀

We need to prove: a₇₂ = 2a₃₄

We know, a_n = a + (n – 1) d [where a is the first term or a₁ and d is the common difference and n is any natural number]

When n = 10:

a₁₀ = a + (10 – 1)d

= a + 9d

When n = 24:

a₂₄ = a + (24 – 1)d

= a + 23d

When n = 34:

a₃₄ = a + (34 – 1)d

= a + 33d ………(i)

When n = 72:

a₇₂ = a + (72 – 1)d

= a + 71d

According to the question,

a₂₄ = 2a₁₀

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a – 2a + 23d – 18d = 0

-a + 5d = 0

a = 5d

Now, a₇₂ = a + 71d

a₇₂ = 5d + 71d

= 76d

= 10d + 66d

= 2(5d + 33d)

= 2(a + 33d) [since, a = 5d]

a₇₂ = 2a₃₄ (From (i))

Hence proved.

EXERCISE 19.3 PAGE NO: 19.15

1. The Sum of the three terms of an A.P. is 21, and the product of the first and the third terms exceed the second term by 6, find three terms.

Solution:

Given:

The sum of the first three terms is 21.

Let us assume the first three terms as a – d, a, a + d [where a is the first term and d is the common difference]

So, the sum of the first three terms is

a – d + a + a + d = 21

3a = 21

a = 7

It is also given that the product of the first and third terms exceeds the second by 6.

So, (a – d)(a + d) – a = 6

a² – d² – a = 6

Substituting the value of a = 7, we get

7² – d² – 7 = 6

d² = 36

d = 6 or d = – 6

Hence, the terms of AP are a – d, a, a + d which is 1, 7, 13.

2. Three numbers are in A.P. If the sum of these numbers is 27 and the product is 648, find the numbers

Solution:

Given:

The sum of the first three terms is 27.

Let us assume the first three terms as a – d, a, a + d [where a is the first term and d is the common difference]

So, the sum of the first three terms is

a – d + a + a + d = 27

3a = 27

a = 9

It is given that the product of three terms is 648.

So, a³ – ad² = 648

Substituting the value of a = 9, we get

9³ – 9d² = 648

729 – 9d² = 648

81 = 9d²

d = 3 or d = – 3

Hence, the given terms are a – d, a, a + d which are 6, 9, 12.

3. Find the four numbers in A.P. whose sum is 50 and in which the greatest number is 4 times the least.

Solution:

Given:

The sum of the four terms is 50.

Let us assume these four terms as a – 3d, a – d, a + d, a + 3d

It is given that the sum of these terms is 4a = 50

So, a = 50/4

= 25/2 … (i)

It is also given that the greatest number is 4 times the least.

a + 3d = 4(a – 3d)

Substitute the value of a = 25/2, and we get

(25+6d)/2 = 50 – 12d

30d = 75

d = 75/30

= 25/10

= 5/2 … (ii)

Hence, the terms of AP are a – 3d, a – d, a + d, a + 3d which is 5, 10, 15, 20

4. The sum of three numbers in A.P. is 12, and the sum of their cubes is 288. Find the numbers.

Solution:

Given:

The sum of three numbers is 12.

Let us assume the numbers in AP are a – d, a, a + d

So,

3a = 12

a = 4

It is also given that the sum of their cube is 288.

(a – d)³ + a³ + (a + d)³ = 288

a³ – d³ – 3ad(a – d) + a³ + a³ + d³ + 3ad(a + d) = 288

Substitute the value of a = 4, and we get

64 – d³ – 12d(4 – d) + 64 + 64 + d³ + 12d(4 + d) = 288

192 + 24d² = 288

d = 2 or d = – 2

Hence, the numbers are a – d, a, a + d which is 2, 4, 6 or 6, 4, 2

5. If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.

Solution:

Given:

The sum of the first three terms is 24.

Let us assume the first three terms are a – d, a, a + d [where a is the first term and d is the common difference]

So, the sum of the first three terms is a – d + a + a + d = 24

3a = 24

a = 8

It is given that the product of three terms is 440.

So a³ – ad² = 440

Substitute the value of a = 8, and we get

8³ – 8d² = 440

512 – 8d² = 440

72 = 8d²

d = 3 or d = – 3

Hence, the given terms are a – d, a, a + d which are 5, 8, 11.

6. The angles of a quadrilateral are in A.P., whose common difference is 10. Find the angles

Solution:

Given: d = 10

We know that the sum of all angles in a quadrilateral is 360.

Let us assume the angles are a – 3d, a – d, a + d, a + 3d

So, a – 2d + a – d + a + d + a + 2d = 360

4a = 360

a = 90… (i)

And,

(a – d) – (a – 3d) = 10

2d = 10

d = 10/2

= 5

Hence, the angles are a – 3d, a – d, a + d, a + 3d which is 75^o, 85^o, 95^o, 105^o

EXERCISE 19.4 PAGE NO: 19.30

1. Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, …. to 10 terms

(ii) 1, 3, 5, 7, … to 12 terms

(iii) 3, 9/2, 6, 15/2, … to 25 terms

(iv) 41, 36, 31, … to 12 terms

(v) a+b, a-b, a-3b, … to 22 terms

(vi) (x – y)², (x² + y²), (x + y)², … to n terms

(vii) (x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), … to n terms

Solution:

(i) 50, 46, 42, …. to 10 terms

n = 10

First term, a = a₁ = 50

Common difference, d = a₂ – a₁ = 46 – 50 = -4

By using the formula,

S = n/2 (2a + (n – 1) d)

Substitute the values of ‘a’ and ‘d’, we get

S = 10/2 (100 + (9) (-4))

= 5 (100 – 36)

= 5 (64)

= 320

∴ The sum of the given AP is 320.

(ii) 1, 3, 5, 7, … to 12 terms

n = 12

First term, a = a₁ = 1

Common difference, d = a₂ – a₁ = 3 – 1 = 2

By using the formula,

S = n/2 (2a + (n – 1) d)

Substitute the values of ‘a’ and ‘d’, we get

S = 12/2 (2(1) + (12-1) (2))

= 6 (2 + (11) (2))

= 6 (2 + 22)

= 6 (24)

= 144

∴ The sum of the given AP is 144.

(iii) 3, 9/2, 6, 15/2, … to 25 terms

n = 25

First term, a = a₁ = 3

Common difference, d = a₂ – a₁ = 9/2 – 3 = (9 – 6)/2 = 3/2

By using the formula,

S = n/2 (2a + (n – 1) d)

Substitute the values of ‘a’ and ‘d’, we get

S = 25/2 (2(3) + (25-1) (3/2))

= 25/2 (6 + (24) (3/2))

= 25/2 (6 + 36)

= 25/2 (42)

= 25 (21)

= 525

∴ The sum of the given AP is 525.

(iv) 41, 36, 31, … to 12 terms

n = 12

First term, a = a₁ = 41

Common difference, d = a₂ – a₁ = 36 – 41 = -5

By using the formula,

S = n/2 (2a + (n – 1) d)

Substitute the values of ‘a’ and ‘d’, we get

S = 12/2 (2(41) + (12-1) (-5))

= 6 (82 + (11) (-5))

= 6 (82 – 55)

= 6 (27)

= 162

∴ The sum of the given AP is 162.

(v) a+b, a-b, a-3b, … to 22 terms

n = 22

First term, a = a₁ = a+b

Common difference, d = a₂ – a₁ = (a-b) – (a+b) = a-b-a-b = -2b

By using the formula,

S = n/2 (2a + (n – 1) d)

Substitute the values of ‘a’ and ‘d’, we get

S = 22/2 (2(a+b) + (22-1) (-2b))

= 11 (2a + 2b + (21) (-2b))

= 11 (2a + 2b – 42b)

= 11 (2a – 40b)

= 22a – 440b

∴ The sum of the given AP is 22a – 440b.

(vi) (x – y)², (x² + y²), (x + y)², … to n terms

n = n

First term, a = a₁ = (x-y)²

Common difference, d = a₂ – a₁ = (x² + y²) – (x-y)² = 2xy

By using the formula,

S = n/2 (2a + (n – 1) d)

Substitute the values of ‘a’ and ‘d’, we get

S = n/2 (2(x-y)² + (n-1) (2xy))

= n/2 (2 (x² + y² – 2xy) + 2xyn – 2xy)

= n/2 × 2 ((x² + y² – 2xy) + xyn – xy)

= n (x² + y² – 3xy + xyn)

∴ The sum of the given AP is n (x² + y² – 3xy + xyn).

(vii) (x – y)/(x + y), (3x – 2y)/(x + y), (5x – 3y)/(x + y), … to n terms

n = n

First term, a = a₁ = (x-y)/(x+y)

Common difference, d = a₂ – a₁ = (3x – 2y)/(x + y) – (x-y)/(x+y) = (2x – y)/(x+y)

By using the formula,

S = n/2 (2a + (n – 1) d)

Substitute the values of ‘a’ and ‘d’, we get

S = n/2 (2((x-y)/(x+y)) + (n-1) ((2x – y)/(x+y)))

= n/2(x+y) {n (2x-y) – y}

∴ The sum of the given AP is n/2(x+y) {n (2x-y) – y}

2. Find the sum of the following series:
(i) 2 + 5 + 8 + … + 182

(ii) 101 + 99 + 97 + … + 47

(iii) (a – b)² + (a² + b²) + (a + b)² + s…. + [(a + b)² + 6ab]

Solution:

(i) 2 + 5 + 8 + … + 182

First term, a = a₁ = 2

Common difference, d = a₂ – a₁ = 5 – 2 = 3

a_n term of given AP is 182.

a_n = a + (n-1) d

182 = 2 + (n-1) 3

182 = 2 + 3n – 3

182 = 3n – 1

3n = 182 + 1

n = 183/3

= 61

Now,

By using the formula,

S = n/2 (a + l)

= 61/2 (2 + 182)

= 61/2 (184)

= 61 (92)

= 5612

∴ The sum of the series is 5612

(ii) 101 + 99 + 97 + … + 47

First term, a = a₁ = 101

Common difference, d = a₂ – a₁ = 99 – 101 = -2

a_n term of given AP is 47.

a_n = a + (n-1) d

47 = 101 + (n-1)(-2)

47 = 101 – 2n + 2

2n = 103 – 47

2n = 56

n = 56/2 = 28

Then,

S = n/2 (a + l)

= 28/2 (101 + 47)

= 28/2 (148)

= 14 (148)

= 2072

∴ The sum of the series is 2072

(iii) (a – b)² + (a² + b²) + (a + b)² + s…. + [(a + b)² + 6ab]

First term, a = a₁ = (a-b)²

Common difference, d = a₂ – a₁ = (a² + b²) – (a – b)² = 2ab

a_n term of given AP is [(a + b)² + 6ab]

a_n = a + (n-1) d

a² + b² + 2ab + 6ab = a² + b² – 2ab + 2abn – 2ab

a² + b² + 8ab – a² – b² + 2ab + 2ab = 2abn

12ab = 2abn

n = 12ab / 2ab

= 6

Then,

S = n/2 (a + l)

= 6/2 ((a-b)² + [(a + b)² + 6ab])

= 3 (a² + b² – 2ab + a² + b² + 2ab + 6ab)

= 3 (2a² + 2b² + 6ab)

= 3 × 2 (a² + b² + 3ab)

= 6 (a² + b² + 3ab)

∴ The sum of the series is 6 (a² + b² + 3ab)

3. Find the sum of first n natural numbers.

Solution:

Let AP be 1, 2, 3, 4, …, n

Here,

First term, a = a₁ = 1

Common difference, d = a₂ – a₁ = 2 – 1 = 1

l = n

So, the sum of n terms = S = n/2 [2a + (n-1) d]

= n/2 [2(1) + (n-1) 1]

= n/2 [2 + n – 1]

= n/2 [n – 1]

∴ The sum of the first n natural numbers is n(n-1)/2

4. Find the sum of all-natural numbers between 1 and 100, which are divisible by 2 or 5.

Solution:

The natural numbers which are divisible by 2 or 5 are:

2 + 4 + 5 + 6 + 8 + 10 + … + 100 = (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95)

Now, (2 + 4 + 6 +…+ 100) + (5 + 15 + 25 +…+95) are AP with a common difference of 2 and 10.

So, for the 1^st sequence => (2 + 4 + 6 +…+ 100)

a = 2, d = 4-2 = 2, a_n = 100

By using the formula,

a_n = a + (n-1)d

100 = 2 + (n-1)2

100 = 2 + 2n – 2

2n = 100

n = 100/2

= 50

So now, S = n/2 (2a + (n-1)d)

= 50/2 (2(2) + (50-1)2)

= 25 (4 + 49(2))

= 25 (4 + 98)

= 2550

Again, for the 2^nd sequence, (5 + 15 + 25 +…+95)

a = 5, d = 15-5 = 10, a_n = 95

By using the formula,

a_n = a + (n-1)d

95 = 5 + (n-1)10

95 = 5 + 10n – 10

10n = 95 +10 – 5

10n = 100

n = 100/10

= 10

So now, S = n/2 (2a + (n-1)d)

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

= 5 (10 + 9(10))

= 5 (10 + 90)

= 500

∴ The sum of the numbers divisible by 2 or 5 is: 2550 + 500 = 3050

5. Find the sum of the first n odd natural numbers.

Solution:

Given an AP of first n odd natural numbers whose first term a is 1, and common difference d is 3.

The sequence is 1, 3, 5, 7……n

a = 1, d = 3-1 = 2, n = n

By using the formula,

S = n/2 [2a + (n-1)d]

= n/2 [2(1) + (n-1)2]

= n/2 [2 + 2n – 2]

= n/2 [2n]

= n²

∴ The sum of the first n odd natural numbers is n².

6. Find the sum of all odd numbers between 100 and 200

Solution:

The series is 101, 103, 105, …, 199

Let the number of terms be n

So, a = 101, d = 103 – 101 = 2, a_n = 199

a_n = a + (n-1)d

199 = 101 + (n-1)2

199 = 101 + 2n – 2

2n = 199 – 101 + 2

2n = 100

n = 100/2

= 50

By using the formula,

The sum of n terms = S = n/2[a + l]

= 50/2 [101 + 199]

= 25 [300]

= 7500

∴ The sum of the odd numbers between 100 and 200 is 7500.

7. Show that the sum of all odd integers between 1 and 1000, which are divisible by 3, is 83667.

Solution:

The odd numbers between 1 and 1000 divisible by 3 are 3, 9, 15,…,999.

Let the number of terms be ‘n’, so the nth term is 999.

a = 3, d = 9-3 = 6, a_n = 999

a_n = a + (n-1)d

999 = 3 + (n-1)6

999 = 3 + 6n – 6

6n = 999 + 6 – 3

6n = 1002

n = 1002/6

= 167

By using the formula,

Sum of n terms, S = n/2 [a + l]

= 167/2 [3 + 999]

= 167/2 [1002]

= 167 [501]

= 83667

∴ The sum of all odd integers between 1 and 1000, which are divisible by 3, is 83667.

Hence proved.

8. Find the sum of all integers between 84 and 719, which are multiples of 5.

Solution:

The series is 85, 90, 95, …, 715

Let there be ‘n’ terms in the AP.

So, a = 85, d = 90-85 = 5, a_n = 715

a_n = a + (n-1)d

715 = 85 + (n-1)5

715 = 85 + 5n – 5

5n = 715 – 85 + 5

5n = 635

n = 635/5

= 127

By using the formula,

Sum of n terms, S = n/2 [a + l]

= 127/2 [85 + 715]

= 127/2 [800]

= 127 [400]

= 50800

∴ The sum of all integers between 84 and 719, which are multiples of 5, is 50800.

9. Find the sum of all integers between 50 and 500, which are divisible by 7.

Solution:

The series of integers divisible by 7 between 50 and 500 are 56, 63, 70, …, 497

Let the number of terms be ‘n’

So, a = 56, d = 63-56 = 7, a_n = 497

a_n = a + (n-1)d

497 = 56 + (n-1)7

497 = 56 + 7n – 7

7n = 497 – 56 + 7

7n = 448

n = 448/7

= 64

By using the formula,

Sum of n terms, S = n/2 [a + l]

= 64/2 [56 + 497]

= 32 [553]

= 17696

∴ The sum of all integers between 50 and 500, which are divisible by 7, is 17696.

10. Find the sum of all even integers between 101 and 999.

Solution:

We know that all even integers will have a common difference of 2.

So, AP is 102, 104, 106, …, 998

We know, a = 102, d = 104 – 102 = 2, a_n = 998

By using the formula,

a_n = a + (n-1)d

998 = 102 + (n-1)2

998 = 102 + 2n – 2

2n = 998 – 102 + 2

2n = 898

n = 898/2

= 449

By using the formula,

Sum of n terms, S = n/2 [a + l]

= 449/2 [102 + 998]

= 449/2 [1100]

= 449 [550]

= 246950

∴ The sum of all even integers between 101 and 999 is 246950.

EXERCISE 19.5 PAGE NO: 19.42

1. If 1/a, 1/b, 1/c are in A.P., prove that:

(i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P.

(ii) a(b + c), b(c + a), c(a + b) are in A.P.

Solution:

(i) (b+c)/a, (c+a)/b, (a+b)/c are in A.P.

We know that if a, b, c are in AP, then b – a = c – b

If, 1/a, 1/b, 1/c are in AP

Then, 1/b – 1/a = 1/c – 1/b

If (b+c)/a, (c+a)/b, (a+b)/c are in AP,

Then, (c+a)/b – (b+c)/a = (a+b)/c – (c+a)/b

Let us take LCM

Since, 1/a, 1/b, 1/c are in AP

1/b – 1/a = 1/c – 1/b

C (b – a) = a (b-c)

Hence, the given terms are in AP.

(ii) a(b + c), b(c + a), c(a + b) are in A.P.

We know that if, b(c + a) – a(b+c) = c(a+b) – b(c+a)

Consider LHS.

b(c + a) – a(b+c)

Upon simplification, we get,

b(c + a) – a(b+c) = bc + ba – ab – ac

= c (b-a)

Now,

c(a+b) – b(c+a) = ca + cb – bc – ba

= a (c-b)

We know,

1/a, 1/b, 1/c are in AP

So, 1/a – 1/b = 1/b – 1/c

Or c(b-a) = a(c-b)

Hence, the given terms are in AP.

2. If a², b², c² are in AP., prove that a/(b+c), b/(c+a), c/(a+b) are in AP.

Solution:

If a², b², c² are in AP then, b² – a² = c² – b²

If a/(b+c), b/(c+a), c/(a+b) are in AP then,

b/(c+a) – a/(b+c) = c/(a+b) – b/(c+a)

Let us take LCM on both the sides we get,

Since, b² – a² = c² – b²

Substituting b² – a²= c² – b² in above, we get

LHS = RHS

Hence, given terms are in AP.

3. If a, b, c are in A.P., then show that:
(i) a²(b + c), b²(c + a), c²(a + b) are also in A.P.
(ii) b + c – a, c + a – b, a + b – c are in A.P.
(iii) bc – a², ca – b², ab – c² are in A.P.

Solution:

(i) a²(b + c), b²(c + a), c²(a + b) are also in A.P.

If b²(c + a) – a²(b + c) = c²(a + b) – b²(c + a)

b²c + b²a – a²b – a²c = c²a + c²b – b²a – b²c

Given, b – a = c – b

And since a, b, c are in AP,

c(b² – a² ) + ab(b – a) = a(c² – b² ) + bc(c – b)

(b – a) (ab + bc + ca) = (c – b) (ab + bc + ca)

Upon cancelling, ab + bc + ca from both sides

b – a = c – b

2b = c + a [which is true]

Hence, given terms are in AP.

(ii) b + c – a, c + a – b, a + b – c are in A.P.

If (c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

Then, b + c – a, c + a – b, a + b – c are in A.P.

Let us consider LHS and RHS.

(c + a – b) – (b + c – a) = (a + b – c) – (c + a – b)

2a – 2b = 2b – 2c

b – a = c – b

And since a, b, c are in AP,

b – a = c – b

Hence, the given terms are in AP.

(iii) bc – a², ca – b², ab – c² are in A.P.

If (ca – b²) – (bc – a²) = (ab – c²) – (ca – b²)

Then, bc – a², ca – b², ab – c² are in A.P.

Let us consider LHS and RHS.

(ca – b²) – (bc – a²) = (ab – c²) – (ca – b²)

(a – b² – bc + a²) = (ab – c² – ca + b²)

(a – b) (a + b + c) = (b – c) (a + b + c)

a – b = b – c

And since a, b, c are in AP,

b – c = a – b

Hence, the given terms are in AP.

4. If (b+c)/a, (c+a)/b, (a+b)/c are in AP., prove that:

(i) 1/a, 1/b, 1/c are in AP

(ii) bc, ca, ab are in AP

Solution:

(i) 1/a, 1/b, 1/c are in AP

If 1/a, 1/b, 1/c are in AP then,

1/b – 1/a = 1/c – 1/b

Let us consider LHS.

1/b – 1/a = (a-b)/ab

= c(a-b)/abc [by multiplying with ‘c’ on both the numerator and denominator]

Let us consider RHS.

1/c – 1/b = (b-c)/bc

= a(b-c)/bc [by multiplying with ‘a’ on both the numerator and denominator]

Since, (b+c)/a, (c+a)/b, (a+b)/c are in AP.

(ii) bc, ca, ab are in AP

If bc, ca, ab are in AP, then,

ca – bc = ab – ca

c (a-b) = a (b-c)

If 1/a, 1/b, 1/c are in AP, then,

1/b – 1/a = 1/c – 1/b

c (a-b) = a (b-c)

Hence, the given terms are in AP.

5. If a, b, c are in A.P., prove that:
(i) (a – c)² = 4 (a – b) (b – c)

(ii) a² + c² + 4ac = 2 (ab + bc + ca)

(iii) a³ + c³ + 6abc = 8b³

Solution:

(i) (a – c)² = 4 (a – b) (b – c)

Let us expand the above expression

a² + c² – 2ac = 4(ab – ac – b² + bc)

a² + 4c²b² + 2ac – 4ab – 4bc = 0

(a + c – 2b)² = 0

a + c – 2b = 0

Since a, b, c are in AP,

b – a = c – b

a + c – 2b = 0

a + c = 2b

Hence, (a – c)² = 4 (a – b) (b – c)

(ii) a² + c² + 4ac = 2 (ab + bc + ca)

Let us expand the above expression.

a² + c² + 4ac = 2 (ab + bc + ca)

a² + c² + 2ac – 2ab – 2bc = 0

(a + c – b)² – b² = 0

a + c – b = b

a + c – 2b = 0

2b = a+c

b = (a+c)/2

Since a, b, c are in AP,

b – a = c – b

b = (a+c)/2

Hence, a² + c² + 4ac = 2 (ab + bc + ca)

(iii) a³ + c³ + 6abc = 8b³

Let us expand the above expression.

a³ + c³ + 6abc = 8b³

a³ + c³ – (2b)³ + 6abc = 0

a³ + (-2b)³ + c³ + 3a(-2b)c = 0

Since, if a + b + c = 0, a³ + b³ + c³ = 3abc

(a – 2b + c)³ = 0

a – 2b + c = 0

a + c = 2b

b = (a+c)/2

Since a, b, c are in AP,

a – b = c – b

b = (a+c)/2

Hence, a³ + c³ + 6abc = 8b³

6. If a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP., prove that a, b, c are in AP.

Solution:

Here, we know a(1/b + 1/c), b(1/c + 1/a), c(1/a + 1/b) are in AP

Also, a(1/b + 1/c) + 1, b(1/c + 1/a) + 1, c(1/a + 1/b) + 1 are in AP

Let us take LCM for each expression then we get,

(ac+ab+bc)/bc , (ab+bc+ac)/ac, (cb+ac+ab)/ab are in AP.

1/bc, 1/ac, 1/ab are in AP

Let us multiply the numerator with ‘abc’, and we get

abc/bc, abc/ac, abc/ab are in AP

∴ a, b, c are in AP.

Hence proved.

7. Show that x² + xy + y², z² + zx + x² and y² + yz + z² are in consecutive terms of an A.P., if x, y and z are in A.P.

Solution:

x, y, z are in AP

Given, x² + xy + y², z² + zx + x² and y² + yz + z² are in AP

(z² + zx + x²) – (x² + xy + y²) = (y² + yz + z²) – (z² + zx + x²)

Let d = common difference

So, Y = x + d and x = x + 2d

Let us consider the LHS.

(z² + zx + x²) – (x² + xy + y²)

z² + zx – xy – y²

(x + 2d)² + (x + 2d)x – x(x + d) – (x + d)²

x² + 4xd + 4d² + x² + 2xd – x² – xd – x² – 2xd – d²

3xd + 3d²

Now, let us consider RHS.

(y² + yz + z²) – (z² + zx + x²)

y² + yz – zx – x²

(x + d)² + (x + d)(x + 2d) – (x + 2d)x – x²

x² + 2dx + d² + x² + 2dx + xd + 2d² – x² – 2dx – x²

3xd + 3d²

LHS = RHS

∴ x² + xy + y², z² + zx + x² and y² + yz + z² are in consecutive terms of A.P.

Hence proved.

EXERCISE 19.6 PAGE NO: 19.46

1. Find the A.M. between:
(i) 7 and 13 (ii) 12 and – 8 (iii) (x – y) and (x + y)

Solution:

(i) Let A be the Arithmetic mean.

Then 7, A, 13 are in AP.

Now, let us solve

A-7 = 13-A

2A = 13 + 7

A = 10

(ii) Let A be the Arithmetic mean.

Then 12, A, – 8 are in AP.

Now, let us solve

A – 12 = – 8 – A

2A = 12 + 8

A = 2

(iii) Let A be the Arithmetic mean.

Then x – y, A, x + y are in AP.

Now, let us solve

A – (x – y) = (x + y) – A

2A = x + y + x – y

A = x

2. Insert 4 A.M.s between 4 and 19.

Solution:

Let A₁, A₂, A₃, A₄ be the 4 AM Between 4 and 19.

Then, 4, A₁, A₂, A₃, A₄, 19 are in AP.

By using the formula,

d = (b-a) / (n+1)

= (19 – 4) / (4 + 1)

= 15/5

= 3

So,

A₁ = a + d = 4 + 3 = 7

A₂ = A1 + d = 7 + 3 = 10

A₃ = A2 + d = 10 + 3 = 13

A₄ = A3 + d = 13 + 3 = 16

3. Insert 7 A.M.s between 2 and 17.

Solution:

Let A₁, A₂, A₃, A₄, A₅, A₆, A₇ be the 7 AMs between 2 and 17.

Then, 2, A₁, A₂, A₃, A₄, A₅, A₆, A₇, 17 are in AP.

By using the formula,

a_n = a + (n – 1)d

a_n = 17, a = 2, n = 9

so,

17 = 2 + (9 – 1)d

17 = 2 + 9d – d

17 = 2 + 8d

8d = 17 – 2

8d = 15

d = 15/8

So,

A₁ = a + d = 2 + 15/8 = 31/8

A₂ = A₁ + d = 31/8 + 15/8 = 46/8

A₃ = A₂ + d = 46/8 + 15/8 = 61/8

A₄ = A₃ + d = 61/8 + 15/8 = 76/8

A₅ = A₄ + d = 76/8 + 15/8 = 91/8

A₆ = A₅ + d = 91/8 + 15/8 = 106/8

A₇ = A₆ + d = 106/8 + 15/8 = 121/8

∴ the 7 AMs between 2 and 7 are 31/8, 46/8, 61/8, 76/8, 91/8, 106/8, 121/8

4. Insert six A.M.s between 15 and – 13.

Solution:

Let A₁, A₂, A₃, A₄, A₅, A₆ be the 7 AM between 15 and – 13.

Then, 15, A₁, A₂, A₃, A₄, A₅, A₆, – 13 are in AP.

By using the formula,

a_n = a + (n – 1)d

a_n = -13, a = 15, n = 8

so,

-13 = 15 + (8 – 1)d

-13 = 15 + 7d

7d = -13 – 15

7d = -28

d = -4

So,

A₁ = a + d = 15 – 4 = 11

A₂ = A1 + d = 11 – 4 = 7

A₃ = A2 + d = 7 – 4 = 3

A₄ = A3 + d = 3 – 4 = -1

A₅ = A4 + d = -1 – 4 = -5

A₆ = A5 + d = -5 – 4 = -9

5. There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n.

Solution:

Let the series be 3, A₁, A₂, A₃, …….., A_n, 17

Given, a_n/a₁ = 3/1

We know the total terms in AP are n + 2

So, 17 is the (n + 2)th term

By using the formula,

A_n = a + (n – 1)d

A_n = 17, a = 3

So, 17 = 3 + (n + 2 – 1)d

17 = 3 + (n + 1)d

17 – 3 = (n + 1)d

14 = (n + 1)d

d = 14/(n+1)

Now,

A_n = 3 + 14/(n+1) = (17n + 3) / (n+1)

A₁ = 3 + d = (3n+17)/(n+1)

Since,

a_n/a₁ = 3/1

(17n + 3)/ (3n+17) = 3/1

17n + 3 = 3(3n + 17)

17n + 3 = 9n + 51

17n – 9n = 51 – 3

8n = 48

n = 48/8

= 6

∴ There are 6 terms in the AP

6. Insert A.M.s between 7 and 71 in such a way that the 5^th A.M. is 27. Find the number of A.M.s.

Solution:

Let the series be 7, A₁, A₂, A₃, …….., A_n, 71

We know the total terms in AP are n + 2

So 71 is the (n + 2)th term

By using the formula,

A_n = a + (n – 1)d

A_n = 71, n = 6

A₆ = a + (6 – 1)d

a + 5d = 27 (5^th term)

d = 4

so,

71 = (n + 2)th term

71 = a + (n + 2 – 1)d

71 = 7 + n(4)

n = 15

∴ There are 15 terms in AP,

7. If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.

Solution:

Let a and b be the first and last terms.

The series be a, A₁, A₂, A₃, …….., A_n, b

We know, Mean = (a+b)/2

Mean of A₁ and A_n = (A₁ + A_n)/2

A₁ = a+d

A_n = a – d

So, AM = (a+d+b-d)/2

= (a+b)/2

AM between A₂ and A_n-1 = (a+2d+b-2d)/2

= (a+b)/2

Similarly, (a + b)/2 is constant for all such numbers.

Hence, AM = (a + b)/2

8. If x, y, z are in A.P., and A₁is the A.M. of x and y, and A₂ is the A.M. of y and z, then prove that the A.M. of A₁ and A₂ is y.

Solution:

Given that,

A₁ = AM of x and y

And A₂ = AM of y and z

So, A₁ = (x+y)/2

A₂ = (y+x)/2

AM of A₁ and A₂ = (A₁ + A₂)/2

= [(x+y)/2 + (y+z)/2]/2

= [x+y+y+z]/2

= [x+2y+z]/2

Since x, y, z are in AP, y = (x+z)/2

AM = [(x + z/2) + (2y/2)]/2

= (y + y)/2

= 2y/2

= y

Hence proved.

9. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P

Solution:

Let A₁, A₂, A₃, A₄, A₅ be the 5 numbers between 8 and 26.

Then, 8, A₁, A₂, A₃, A₄, A₅, 26 are in AP.

By using the formula,

A_n = a + (n – 1)d

A_n = 26, a = 8, n = 7

26 = 8 + (7 – 1)d

26 = 8 + 6d

6d = 26 – 8

6d = 18

d = 18/6

= 3

So,

A₁ = a + d = 8 + 3 = 11

A₂ = A₁ + d = 11 + 3 = 14

A₃ = A₂ + d = 14 + 3 = 17

A₄ = A₃ + d = 17 + 3 = 20

A₅ = A₄ + d = 20 + 3 = 23

EXERCISE 19.7 PAGE NO: 19.48

1. A man saved ₹ 16500 in ten years. In each year after the first, he saved ₹ 100/- more than he did in the preceding year. How much did he save in the first year?

Solution:

Given: A man saved ₹16500 in ten years

Let ₹ x be his savings in the first year

His savings increased by ₹ 100 every year.

So,

A.P will be x, 100 + x, 200 + x………………..

Where x is the first term and

Common difference, d = 100 + x – x = 100

We know S_n is the sum of n terms of an A.P

By using the formula,

S_n = n/2 [2a + (n – 1)d]

Where a is the first term, d is the common difference and n is the number of terms in an A.P.

Given:

S_n = 16500 and n = 10

S₁₀ = 10/2 [2x + (10 – 1)100]

16500 = 5{2x + 9(100)}

16500 = 5(2x + 900)

16500 = 10x + 4500

-10x = 4500 – 16500

–10x = –12000

x = 12000/10

= 1200

Hence, his saving in the first year is ₹ 1200.

2. A man saves ₹ 32 during the first year, ₹ 36 in the second year and in this way, he increases his savings by ₹ 4 every year. Find in what time his saving will be ₹ 200.

Solution:

Given:

First-year savings is ₹ 32

Second-year savings is ₹ 36

In this process, he increases his savings by ₹ 4 every year.

Then,

A.P. will be 32, 36, 40,………

Where 32 is the first term and common difference, d = 36 – 32 = 4

We know S_n is the sum of n terms of an A.P

By using the formula,

S_n = n/2 [2a + (n – 1)d]

Where a is the first term, d is the common difference and n is the number of terms in an A.P.

Given:

S_n = 200, a = 32, d = 4

S_n = n/2 [2a + (n – 1)d]

200 = n/2 [2(32) + (n-1)4]

200 = n/2 [64 + 4n – 4]

400 = n [60 + 4n]

400 = 4n [15 + n]

400/4 = n [15 + n]

100 = 15n + n²

n² + 15n – 100 = 0

n² + 20n – 5n – 100 = 0

n (n + 20) – 5 (n + 20) = 0

(n + 20) – 5 (n + 20) = 0

(n + 20) (n – 5) = 0

n = -20 or 5

n = 5 [Sincen is a positive integer]

Hence, the man requires 5 days to save ₹ 200

3. A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments, which form an arithmetic series. When 30 of the instalments are paid, he dies, leaving one-third of the debt unpaid. Find the value of the instalment.

Solution:

Given:

40 annual instalments, which form an arithmetic series.

Let the first instalment be ‘a’.

S₄₀ = 3600, n = 40

By using the formula,

S_n = n/2 [2a + (n – 1)d]

3600 = 40/2 [2a + (40-1)d]

3600 = 20 [2a + 39d]

3600/20 = 2a + 39d

2a + 39d – 180 = 0 ….. (i)

Given:

The sum of the first 30 terms is paid, and one-third of the debt is unpaid.

So, paid amount = 2/3 × 3600 = ₹ 2400

S_n = 2400, n = 30

By using the formula,

S_n = n/2 [2a + (n – 1)d]

2400 = 30/2 [2a + (30-1)d]

2400 = 15 [2a + 29d]

2400/15 = 2a + 29d

2a + 29d -160 = 0 …. (ii)

Now, let us solve equations (i) and (ii) by substitution method, and we get

2a + 39d = 180

2a = 180 – 39d … (iii)

Substitute the value of 2a in equation (ii).

2a + 29d – 160 = 0

180 – 39d + 29d – 160 = 0

20 – 10d = 0

10d = 20

d = 20/10

= 2

Substitute the value of d in equation (iii).

2a = 180 – 39d

2a = 180 – 39(2)

2a = 180 – 78

2a = 102

a = 102/2

= 51

Hence, the value of the first instalment ‘a’ is ₹ 51

4. A manufacturer of radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find
(і) the production in the first year,
(іі) the total product in the 7 years and
(ііі) the product in the 10^th year.

Solution:

Given:

600 and 700 radio sets units are produced in the third and the seventh year, respectively.

a₃ = 600 and a₇ = 700

(і) The production in the first year

We need to find the production in the first year.

Let first-year production be ‘a’

So the AP formed is, a, a+x, a+2x, ….

By using the formula,

a_n = a + (n-1)d

a₃ = a + (3-1)d

600 = a + 2d …. (i)

a₇ = a + (7-1)d

700 = a + 6d

a = 700 – 6d…. (ii)

Substitute value of a in (I), and we get,

600 = a + 2d

600 = 700 – 6d + 2d

700 – 600 = 4d

100 = 4d

d = 100/4

= 25

Now, substitute the value of d in (ii), and we get,

a = 700 – 6d

= 700 – 6(25)

= 700 – 150

= 550

∴ The production in the first year, ‘a’ is 550.

(іі) the total product in the 7 years

We need to find the total product in 7 years, i.e., is S₇

By using the formula,

S_n = n/2 [2a + (n-1)d]

n = 7, a = 550, d = 25

S₇ = 7/2 [2(550) + (7-1)25]

= 7/2 [1100 + 150]

= 7/2 [1250]

= 7 [625]

= 4375

∴ The total product in the 7 years is 4375.

(ііі) the product in the 10^th year.

We need to find the product in the 10^th year, i.e., a₁₀

By using the formula,

a_n = a + (n-1)d

n = 10, a = 550, d = 25

a₁₀ = 550 + (10-1)25

= 550 + (9)25

= 550 + 225

= 775

∴ The product in the 10^th year is 775.

5. There are 25 trees at equal distances of 5 meters in line with a well, the distance of the well from the nearest tree being 10 meters. A gardener waters all the trees separately, starting from the well, and returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Solution:

Given: Total number of trees is 25, and the equal distance between two adjacent trees is 5 meters.

We need to find the total distance the gardener will cover.

As the gardener is coming back to well after watering every tree,

The distance covered by the gardener to water 1^st tree and return back to the initial position is 10m + 10m = 20m

Now, the distance between adjacent trees is 5m.

The distance covered by him to water 2^nd tree and return back to the initial position is 15m + 15m = 30m

The distance covered by the gardener to water 3^rd tree return back to the initial position is 20m + 20m = 40m

Hence, the distance covered by the gardener to water the trees is in A.P.

A.P. is 20, 30, 40 ………upto 25 terms

Here, first term, a = 20, common difference, d = 30 – 20 = 10, n = 25

We need to find S₂₅ which will be the total distance covered by the gardener to water 25 trees.

So by using the formula,

S_n = n/2 [2a + (n – 1)d]

S₂₅ = 25/2 [2(20) + (25-1)10]

= 25/2 [40 + (24)10]

= 25/2 [40 + 240]

= 25/2 [280]

= 25 [140]

= 3500

∴ The total distance covered by the gardener to water trees all 25 trees is 3500m.

6. A man is employed to count ₹ 10710. He counts at the rate of ₹ 180 per minute for half an hour. After this, he counts at the rate of ₹ 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

Solution:

Given: The amount to be counted is ₹ 10710

We need to find the time taken by man to count the entire amount.

He counts at the rate of ₹ 180 per minute for half an hour or 30 minutes.

So, the amount to be counted in an hour = 180 × 30 = ₹ 5400

Amount left = 10710 – 5400 = ₹ 5310

S_n = 5310

After an hour, the rate of counting decreases to ₹ 3 per minute. This rate will form an A.P.

A.P. is 177, 174, 171,……

Here, a = 177 and d = 174 – 177 = –3

By using the formula,

S_n = n/2 [2a + (n – 1)d]

5310 = n/2 [2(177) + (n-1) (-3)]

5310 = n/2 [354 -3n + 3]

5310 × 2 = n [357 – 3n]

10620 = 357n – 3n²

10620 = 3n(119 – n)

10620/3 = n(119 – n)

3540 = 119n – n²

n² – 119n + 3540 = 0

n² – 59n – 60n + 3540 = 0

n(n – 59) – 60(n – 59) = 0

(n – 59) (n – 60) = 0

n = 59 or 60

We shall consider the value of n = 59. Since, at 60^th min he will count ₹ 0

∴ The total time taken by him to count the entire amount = 30 + 59 = 89 minutes.

7. A piece of equipment costs a certain factory ₹ 600,000. If it depreciates in value 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

Solution:

Given: A piece of equipment costs a certain factory is ₹ 600,000

We need to find the value of the equipment at the end of 10 years.

The price of equipment depreciates 15%, 13.5%, 12% in 1^st, 2^nd, 3^rd year and so on.

So the A.P. will be 15, 13.5, 12,…………… up to 10 terms

Here, a = 15, d = 13.5 – 15 = –1.5, n = 10

By using the formula,

S_n = n/2 [2a + (n – 1)d]

S₁₀ = 10/2 [2(15) + (10-1) (-1.5)]

= 5 [30 + 9(-1.5)]

= 5 [30 – 13.5]

= 5 [16.5]

= 82.5

The value of equipment at the end of 10 years is = [100 – Depreciation %]/100 × cost

= [100 – 82.5]/100 × 600000

= 175/10 × 6000

= 175 × 600

= 105000

∴ The value of the equipment at the end of 10 years is ₹ 105000.

8. A farmer buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount. How much did the tractor cost him?

Solution:

Given: The price of the tractor is ₹12000.

We need to find the total cost of the tractor if he buys it in instalments.

Total price = ₹ 12000

Paid amount = ₹ 6000

Unpaid amount = ₹ 12000 – 6000 = ₹ 6000

He pays the remaining ₹ 6000 in ‘n’ number of instalments of ₹ 500 each.

So, n = 6000/500 = 12

The cost incurred by him to pay the remaining 6000 is

The AP will be

(500 + 12% of 6000) + (500 + 12% of 5500) + … up to 12 terms

500 × 12 + 12% of (6000 + 5500 + … up to 12 terms)

By using the formula,

S_n = n/2 [2a + (n – 1)d]

n = 12, a = 6000, d = -500

S₁₂ = 500×12 + 12/100 × 12/2 [2(6000) + (12-1) (-500)]

= 6000 + 72/100 [12000 + 11 (-500)]

= 6000 + 72/100 [12000 – 5500]

= 6000 + 72/100 [6500]

= 6000 + 4680

= 10680

Total cost = 6000 + 10680

= 16680

∴ The total cost of the tractor, if he buys it in instalments, is ₹ 16680.

Also, access exercises of RD Sharma Solutions for Class 11 Maths Chapter 19 – Arithmetic Progressions

Exercise 19.1 Solutions

Exercise 19.2 Solutions

Exercise 19.3 Solutions

Exercise 19.4 Solutions

Exercise 19.5 Solutions

Exercise 19.6 Solutions

Exercise 19.7 Solutions