**RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression** Ex 5.3 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.3.

## Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.3

Ex 5.3 Class 10 RBSE Question 1.

Find the sum of the (RBSESolutions.com) following A.P.’s

(i) 1, 3, 5, 7, …, upto 12 terms

(ii) 8, 3, -2, …, upto 22 terms

(iii) \(\frac { 1 }{ 15 } ,\frac { 1 }{ 22 } ,\frac { 1 }{ 10 } ,….\) upto 11 terms

Solution :

(i) Given A.P. 1, 3, 5, 7 ….. upto 12 terms

First term (a) = 1

Common difference (d) = 3 – 1 ⇒ 2

Sum of n terms

S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

S_{12} = \(\frac { 12 }{ 2 }\) [2 × 1 + (12 – 1)2]

= 6[2 + 11 × 2]

= 6[2 + 22]

= 6 × 24

= 144

Hence, sum of first 12 terms = 144.

(ii) Given. A.P. 8, 3, -2, …, upto 22 terms

First term (a) = 8

common (RBSESolutions.com) difference (d) = 3 – 8 = -5

sum of n terms

S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

= \(\frac { 22 }{ 2 }\) [2 × 8 + (22 – 1)(-5)]

= 11[16 + 21 × -5]

= 11[16 – 105]

= 11 [- 89]

= -979

Hence, sum of first 22 terms is -979.

(iii) GivenA.P. \(\frac { 1 }{ 15 } ,\frac { 1 }{ 12 } ,\frac { 1 }{ 10 } ,….\) upto 11 terms

First term (a) = \(\frac { 1 }{ 15 }\)

Common difference (d) = \(\frac { 1 }{ 12 }\) – \(\frac { 1 }{ 15 }\)

= \(\frac { 15-12 }{ 180 }\) = \(\frac { 3 }{ 180 }\) = \(\frac { 1 }{ 60 }\)

and number (RBSESolutions.com) of term n = 11

∵ Sum of n terms S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

Hence, sum of first 11 terms of series = \(\frac { 33 }{ 20 }\)

Arithmetic Progression Class 10 RBSE Question 2.

Find the sum of the (RBSESolutions.com) following :

(i) 3 + 11 + 19 + … 803

(ii) 7 + 10\(\frac { 1 }{ 2 }\) + 14 + …… + 84

Solution :

(i) Given :

3 + 11 + 19 + … + 803

Here first term a = 3

Common difference d = 11 – 3 = 8

If 803 is n^{th} term of given A.P, then

n^{th} term, a_{n} = 803

a + (n – 1)d = 803

3 + (n – 1)8 = 803

(n – 1)8 = 803 – 3

n – 1 = \(\frac { 800 }{ 8 }\)

n – 1 = 100

n = 100 + 1

n = 101

Hence, A.P. has 101 terms.

∴ Sum of n terms

Thus, 3 + 11+ 15 + … + 803 = 40703.

(ii) Given

7 + 10\(\frac { 1 }{ 2 }\) + 14 + …. + 84

Here, First term (a) = 7

common difference (d) = 10\(\frac { 1 }{ 2 }\) – 7 = 3\(\frac { 1 }{ 2 }\)

= \(\frac { 7 }{ 2 }\)

If 84 is n^{th} term of given (RBSESolutions.com) series, then

∴ n^{th} term a_{n} = 84

a + (n – 1)d = 84

⇒ 7 + (n – 1) \(\frac { 7 }{ 2 }\) = 84

⇒ 1 + \(\frac { n-1 }{ 2 }\) = 12

⇒ \(\frac { n-1 }{ 2 }\) = 12 – 1

⇒ \(\frac { n-1 }{ 2 }\) = 11 ⇒ n – 1 = 22

n = 22 + 1 = 23

There are 23 terms in series

Sum of 23 terms

Hence, 7 + 10\(\frac { 1 }{ 2 }\) + 14 + …… + 84 = 1046\(\frac { 1 }{ 2 }\)

RBSE Solutions For Class 10 Maths Chapter 5 Question 3.

Find the number of terms

(i) How many (RBSESolutions.com) terms of the A.P. : 9, 17, 25 ….. must be taken to give a sum of 636?

(ii) How many terms the A.P. 63, 60, 57, …… must be taken to give a sum of 693?

Solution :

(i) Given A.P. : 9, 17, 25, …

First term (a) = 9, common difference (d) = 17 – 9 = 8

Let no. of terms be n

S_{n} = 636

⇒ \(\frac { n }{ 2 }\) [2a + (n – 1)d] = 636

⇒ \(\frac { n }{ 2 }\) [2 × 9+ (n – 1)8] = 636

⇒ \(\frac { n }{ 2 }\) [18 + 8n – 8] = 636

⇒ \(\frac { n }{ 2 }\) [8n + 10] = 636

⇒ n(4n + 5) = 636

⇒ 4n^{2} + 5n = 636

⇒ 4n^{2} + 5n – 636 = 0

⇒ 4n^{2} + 53 n – 48n – 636 = 0

⇒ n(4n + 53) – 12(4n + 53) = 0

⇒ (4n + 53)(n – 12) = 0

⇒ n – 12 = 0 or 4n + 53 = 0

⇒ n = 12 or – \(\frac { 53 }{ 4 }\)

∵ n cannot be (RBSESolutions.com) negative

So, ignore n = – \(\frac { 53 }{ 4 }\)

∴ n = 12

Hence, sum of 12 terms of given A.P. is 636.

(ii) Given A.P. 63, 60, 57 …..

First term(a) = 63

Common difference (d) = 60 – 63 = -3

Let number of terms be n

S_{n} = 693

We know that

S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

⇒ 693 = \(\frac { n }{ 2 }\) [2 × 63 + (n – 1) (-3)]

⇒ 693 = \(\frac { n }{ 2 }\) [126 – 3n + 3]

⇒ 1386 = n(129 – 3n)

⇒ 1386 = 129n – 3n^{2}

⇒ 3n^{2} – 129n + 1386 = 0

⇒ n^{2} – 43n + 462 = 0

⇒ n^{2} – 21n – 22n + 462 = 0

⇒ = n(n – 21) – 22(n – 21) = 0

⇒ (n – 21)(n – 22) = 0

⇒ n – 21 = 0 or n – 22 = 0

n = 21 or n = 22

By taking 21 or 22 terms (RBSESolutions.com) of given A.P. we will get sum 693.

RBSE Class 10 Maths Chapter 5 Question 4.

Find the sum of first 25 terms of following series whose th term is given :

(i) a_{n} = 3 + 4n

(ii) a_{n} = 7 – 3n

Solution :

(i) Given a_{n} = 3 + 4n …..(i)

Substituting various values of n in equation (i)

a_{1} = 3 + 4(1) = 7

a_{2} = 3 + 4(2) = 11

a_{3} = 3 + 4(3) = 15, …

common difference (d) = a_{2} – a_{1} = 11 – 7 = 4

a_{3} – a_{2} = 15 – 11 = 4

∵ a_{2} – a_{1} = a_{3} – a_{2} = 4

Thus series is 7, 11, 15 ….

and given series (RBSESolutions.com) is an A.P.

Here, a = 7, d= 4 and n = 25

∵ S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

S_{25} = \(\frac { 25 }{ 2 }\) [2 × 7 + (25 – 1)4]

= \(\frac { 25 }{ 2 }\) [14 + 24 × 4]

= \(\frac { 25 }{ 2 }\) [14 + 96]

= \(\frac { 25 }{ 2 }\) × 110

= 25 × 55

= 1375

Thus S_{25} = 1375

(ii)Given : a_{n} = 7 – 3n

substituting various values (RBSESolutions.com) of n in equation (i)

a_{1} = 7 – 3(1) = 4

a_{2} = 7 – 3(2) = 1

a_{3} = 7 – 3(3) = -2

common difference (d) = a_{2} – a_{1} = 1 – 4 = -3

and a_{3} – a_{2} = -2 – 1 = -3

∵ a_{2} – a_{1} = a_{3} – a_{2} = -2 – 1 = -3

Thus A.P. is 4, 1, -2,…

Here a = 4 d = -3 and n = 25

We have to find sum of first 25 terms

S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

S_{n} = \(\frac { 25 }{ 2 }\) [2 x 4 + (25 – 1)(-3)

= \(\frac { 25 }{ 2 }\) [8 + 24 × -3]

= \(\frac { 25 }{ 2 }\) [8 – 72]

= \(\frac { 25 }{ 2 }\) × – 64

= -25 × 32

= -800

Hence S_{25} = – 800.

RBSE Solutions For Class 10 Maths Chapter 5.3 Question 5.

Find the sum of first 51 terms of AP. in (RBSESolutions.com) which II^{nd} and III^{rd} term are 14 and 18 respectively.

Solution :

Second term of A.P. a_{2} = 14

and third term a_{3} = 18

∴ Common difference d = a_{3} – a_{2}

= 18 – 14 = 4

Again, ∵ II^{nd} term = 14

∴ a + d = 14

⇒ a + 4 = 14

⇒ a = 14 – 4

⇒ a = 10

∵ a = 10, d = 4

Then, sum of n terms S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

∴ S_{51} = \(\frac { 51 }{ 2 }\) [2 × 10 + (51 – 1)4]

= \(\frac { 51 }{ 2 }\) [20 + 50 × 4]

= \(\frac { 51 }{ 2 }\) [20 + 200]

= \(\frac { 51 }{ 2 }\) × 220 = 51 × 110 = 5610

Hence, sum of 5 terms of given A.P. = 5610

Exercise 5.3 Class 10 RBSE Question 6.

The first and last term of an A.P. are 17 and 350 respectively. If (RBSESolutions.com) common difference is 9 then find number of terms in A.P. and their sum.

Solution :

Given

First term (a) = 17

Last term (l) = a_{n} = 350

and common difference (d) = 9

∵ a_{n} = 350

a + (n – 1)d = 350

⇒ 17 + (n – 1)9 = 350

⇒ 9(n – 1) = 350 – 17 = 333

⇒ n – 1 = \(\frac { 333 }{ 9 }\) = 37

∴ n = 37 + 1 = 38

Now, S_{n} = \(\frac { n }{ 2 }\)(a + l)

∴ = \(\frac { 38 }{ 2 }\) (17 + 350)

= 19 × 367 = 6973

Hence, n = 38 and sum of term (S_{n}) = 6973

RBSE Solutions For Class 10 Maths Chapter 5 Miscellaneous Question 7.

Find the sum of all odd numbers, (RBSESolutions.com) divisible by 3 between 1 and 1000.

Solution :

Odd numbers divisible by 3, between 1 and 1000 are 3, 9, 15, 21 …….. 999.

Clearly series 3, 9, 15, 15,21 …… 999 is A.P.

whose first term (a) = 3 and common difference (d) = 6.

Let us assume that this series contains n terms.

∴ a_{n} = 999

⇒ a + (n – 1)d = 999

⇒ 3 + (n – 1) × 6 = 999

⇒ 6n – 3 = 999

⇒ 6n = 1002

⇒ n = \(\frac { 1002 }{ 6 }\)

⇒ n = 167

∴ Required sum

S_{n} = \(\frac { n }{ 2 }\) (a + l)

S_{167} = \(\frac { 167 }{ 2 }\) (3 + 999)

= \(\frac { 167 }{ 2 }\) × 1002

= 167 × 501

= 83667

Hence, required sum = 83667.

RBSE Class 10 Maths Chapter 5.3 Question 8.

The first term of A.P. is 8, n^{th} term is 33 and sum (RBSESolutions.com) of first n terms is 123, then find n and common difference d.

Solution :

Given

First term (a) = 8

n^{th} term (a_{n}) = 33

sum of n terms (S_{n}) = 123

∵ n^{th} term a_{n} = a + (n – 1)d

⇒ 33 = 8 + (n – 1)d

⇒ (n – 1)d = 33 – 8

⇒ (n – 1)d = 25 ……(i)

Now, sum of n terms

S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

⇒ 123 = \(\frac { n }{ 2 }\) [2 × 8 + 25] [From equation (i)]

⇒ 123 = \(\frac { n }{ 2 }\) (16 + 25)

⇒ 123 = \(\frac { n }{ 2 }\) × 41

⇒ n = \(\frac { 123\times 2 }{ 41 }\)

⇒ n = 6

Put the value of n in equation (i)

⇒ (6 – 1)d = 25

⇒ 5d = 25

⇒ d = 5

Thus, n = 6 and d = 5

RBSE Solutions For Class 10 Maths Chapter 5.3 Question 9.

A sum of ₹ 280 is to be used to give four (RBSESolutions.com) cash prize. If each prize is ₹ 20 less than its preceding prize. Find the value of each of the prizes.

Solution :

Let first prize is ₹ a

∴ II^{nd} prize a_{2} = ₹ (a – 20)

III^{rd} prize a_{3} = ₹ [(a – 20) – 20]

a_{3} = ₹ (a – 40)

IV^{th} prize a_{4} = ₹ [(a – 40) – 20)]

a_{4} = ₹ (a – 60)

Here first term = a,

Common difference d = (a – 20) – a

d = -20

Number of terms n = 4

Sum of terms S_{n} = 280

By Formula, S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

S_{4} = \(\frac { 4 }{ 2 }\) [2a + (4 – 1) × -20]

⇒ 280 = 2[2a + 3 × – 20]

⇒ 280 = 2[2a – 60]

⇒ 140 = 2a – 60

⇒ 2a = 140 + 60

⇒ a = \(\frac { 200 }{ 2 }\) = 100

Hence I^{st} prize = 100, remaining (RBSESolutions.com) prize are ₹ (100 – 20), ₹ (100 – 20 – 20) and ₹(100 – 20 – 20 – 20).

Hence price are ₹ 100, ₹ 80, ₹ 60, and 40

Class 10 Maths RBSE Solution Chapter 5 Question 10.

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find

(i) the production in the I^{st} year.

(ii) The production in the 10^{th} year.

(iii) The total production in first 7 years.

Solution :

(i) Let production of T.V. sets in first years is a. (RBSESolutions.com) Given that production of T.V. sets in third year

a_{3} = 600

and in seventh year

a_{7} = 100.

Now, a_{3} = a + (3 – 1)d

600 = a + 2d …..(i)

and a_{7} = a + (7 – 1)d

700 = a + 6d …(ii)

subtracting eq^{n} (i) from (ii),

100 = 4d

d = \(\frac { 100 }{ 4 }\) = 25

Put the value of d in (RBSESolutions.com) equation (i)

⇒ 600 = a + 2 × 25

⇒ 600 = a + 50

⇒ a = 600 – 50

⇒ a = 550

Production of T.V. set in I^{st} year = 550

(ii) Production of TV. sets in 10^{th} year

Formula, a_{n} = a + (n – 1)d

a_{n} = 550 + (10 – 1)25

= 550 + 9 × 25

= 550 + 225 = 775

Thus, production (RBSESolutions.com) of T.V. sets in 10^{th} year = 775 sets

(iii) Total production in 7 years

S_{n} = \(\frac { 7 }{ 2 }\) [2a + (n – 1)d]

By formula, S_{7} = \(\frac { 7 }{ 2 }\) [2 × 550 + (7 – 1)25]

= \(\frac { 7 }{ 2 }\) [1100 + 6 × 25]

= \(\frac { 7 }{ 2 }\) [1100 + 150]

= \(\frac { 7 }{ 2 }\) × 1250 = 7 × 625

= 4375

Thus, total production in 7 years in 4375 etc.

We hope the given RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.3, drop a comment below and we will get back to you at the earliest.