RD Sharma Class 10 Ex 9.5 Solutions Chapter 9 Arithmetic Progressions

In this chapter, we provide RD Sharma Class 10 Ex 9.5 Solutions Chapter 9 Arithmetic Progressions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 9.5 Solutions Chapter 9 Arithmetic Progressions pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 9
Chapter NameArithmetic Progressions
Exercise9.5
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 9 Arithmetic Progressions Ex 9.5 Download PDF

Chapter 9: Arithmetic Progressions Exercise 9.5

Question: 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 22 tams

(viii) – 26, – 24, – 22, …. to 36 terms

Solution:

In an A.P let first term = a, common difference = d, and there are n terms.

Then, sum of n terms is,

(i) Given progression is,

50, 46, 42 to 10 term.

First term (a) = 50

Common difference (d) = 46 – 50 = – 4

nth term = 10

= 5{100 – 9.4}

= 5{100 – 36}

= S × 64

∴ S10 = 320

(ii) Given progression is, 1, 3, 5, 7, …..to 12 terms

First term difference (d) = 3 – 1 = 2 nth term = 12

= 6 × {2 + 22} = 6.24

∴ S12 = 144.

(iii) Given expression is 3, 9/2, 6, 15/2, … to 25 terms

First term (a) = 3

Common difference (d) = 9/2 – 3 = 3/2

Sum of nth terms Sn, given n = 25

(iv) Given expression is, 41, 36, 31 to 12 terms.

First term (a) = 41

Common difference (d) = 36 – 41 = – 5

Sum of nth terms Sn, given n = 12

 (v) a + b, a – b, a – 3b to 22 terms

First term (a) = a + b

Common difference (d) = a – b – a – b = -2b

Sum of nth terms Sn = n/2{2a(n – 1). d}

Here n = 22

S22 = 22/2{2.(a + b) + (22 – 1). -2b}

= 11{2(a + b) – 22b)

= 11{2a – 20b}

= 22a – 440b

∴S22 = 22a – 440b

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

First term (a) = (x – y)2

Common difference (d) = x2 + y2 – (x – y)2

= x2 + y2 – (x2 + y2 – 2xy)

= x2 + y2 – x2 + y2 + 2xy

= 2xy

Sum of nth terms Sn = n/2{2a(n – 1). d}

= n/2{2(x – y)2 + (n – 1). 2xy}

= n{(x – y)2 + (n – 1)xy}

∴ Sn = n{(x — y)2 + (n — 1). xy)

 (viii) Given expression -26, – 24. -22, to 36 terms

First term (a) = -26

Common difference (d) = -24 – (-26)

= -24 + 26 = 2

Sum of n tams Sn = n/2{2a + (n – 1)d)

Sum of n tams Sn = 36/2{2. -26 + (36 – 1)2}

= 18[-52 + 70]

= 18.18

= 324

∴ Sn = 324

Question: 2

Find the sum to n tam of the A.P. 5, 2, –1, – 4, –7, …

Solution:

Given AP is 5, 2, -1, -4, -7, …..

a = 5, d = 2 – 5 = -3

Sn = n/2{2a + (n – 1)d}

= n/2{2.5 + (n – 1) – 3}

= n/2{10 – 3(n – 1)}

= n/2{13 – 3n)

∴ Sn = n/2(13 – 3n)

Question: 3

Find the sum of n terms of an A.P. whose the terms is given by an = 5 – 6n.

Solution:

Given nth term an = 5 – 6n

Put n = 1, a1 = 5 – 6.1 = -1

We know, first term (a1) = -1

Last term (an) = 5 – 6n = 1

Then Sn = n/2(-1 + 5 – 6n)

= n/2(4 – 6n) = n/2(2 – 3n)

Question: 4

If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, … is 116. Find the last term.

Solution:

Given AP is 25, 22, 19, ……….

First term (a) = 25, d = 22 – 25 = -3.

Given, Sn = n/2(2a + (n – 1)d)

116 = n/2(2 × 25 + (n – 1) – 3)

232 = n(50 – 3(n – 1))

232 = n(53 – 3n)

232 = 53n – 3n2

3n2 – 53n + 232 = 0

(3n – 29)(n – 8) = 0

⟹ ag = 25 + (8 – 1) – 3

∴ n = 8, ag = 4

= 25 – 21 = 4

Question: 5

(i) How many terms of the sequence 18, 16, 14, … should be taken so that their

(ii) How many terms are there in the A P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?

(iii) How many terms of the A.P. 9, 17, 25, . must be taken so that their sum is 636?

(iv) How many terms of the A P. 63, 60, 57, … must be taken so that their sum is 693?

Solution:

(i) Given sequence, 18, 16, 14, …

a = 18, d = 16 – 18 = -2.

Let, sum of n terms in the sequence is zero

Sn = 0

n/2(2a + (n. – 1)d) = 0

n/2(2.18 + (n – 1) – 2) = 0

n(18 – (n – 1)) = 0

n(19 – n) = 0

n = 0 or n = 19

(ii) ∵ n = 0 is not possible.

Therefore, sum of 19 numbers in the sequence is zero.

Given, a = -14, a 5 = 2

a+ (5 – 1)d = 2

-14 + 4d = 2

4d = 16 ⟹ d = 4

Sequence is -14, -10, -6, -2, 2, ……….. 

Given Sn = 40

40 = n/2{2(-14) + (n – 1)4}

80 = n(-28 + 4n – 4)

80 = n(-32 + 4n)

4(20) = 4n(-8 + n)

n2 – 8n – 20 = 0

(n – 10)(n + 2) = 0

n = 10 or n = -2

Sum of 10 numbers is 40 (Since -2 is not a natural number)

(iii) Given AP 9. 17, 25, ……………… 

a = 9, d = 17 – 9 = 8, and Sn = 636

636 = n/2(2.9 + (n – 1)8)

1272 = n(18 – 8 + 8n)

1272 = n(10 + 8n)

2 × 636 = 2n(5 + 4n)

636 = 5n + 4n2

4n2 + 5n – 636 = 0

(4n + 53)(n – 12) = 0

∴ n = 12 (Since n (-53)/4 is not a natural number)

Therefore, value of n is 12.

(iv) Given AP, 63, 60, 57, …………. 

a = 63, d = 60 – 63 = -3

Sn  = 693

Sn = n/2(2a + (n – 1)d)

693 = n/2(2.63 + (n. – 1) – 3)

1386 = n(126 – 3n + 3)

1386 = (129 – 3n)n

3n2 – 129n + 1386 = 0

n2 – 43n + 462 = 0

n = 21, 22

∴ Sum of 71 or 77 tern is 693

All Chapter RD Sharma Solutions For Class10 Maths

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