**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

Question 1.

Find the sum of the following A.P.’s

(i) 1, 3, 5, 7, …, upto 12 terms

(ii) 8, 3, -2, …, upto 22 terms

(iii) 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} = [2a + (n – 1)d]

S_{12} = [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 difference (d) = 3 – 8 = -5

sum of n terms

S_{n} = [2a + (n – 1)d]

= [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. upto 11 terms

First term (a) =

Common difference (d) = –

= = =

and number of term n = 11

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

Hence, sum of first 11 terms of series =

Question 2.

Find the sum of the following :

(i) 3 + 11 + 19 + … 803

(ii) 7 + 10 + 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 =

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 + 14 + …. + 84

Here, First term (a) = 7

common difference (d) = 10 – 7 = 3

=

If 84 is n^{th} term of given series, then

∴ n^{th} term a_{n} = 84

a + (n – 1)d = 84

⇒ 7 + (n – 1) = 84

⇒ 1 + = 12

⇒ = 12 – 1

⇒ = 11 ⇒ n – 1 = 22

n = 22 + 1 = 23

There are 23 terms in series

Sum of 23 terms

Hence, 7 + 10 + 14 + …… + 84 = 1046

Question 3.

Find the number of terms

(i) How many 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

⇒ [2a + (n – 1)d] = 636

⇒ [2 × 9+ (n – 1)8] = 636

⇒ [18 + 8n – 8] = 636

⇒ [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 –

∵ n cannot be negative

So, ignore n = –

∴ 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} = [2a + (n – 1)d]

⇒ 693 = [2 × 63 + (n – 1) (-3)]

⇒ 693 = [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 of given A.P. we will get sum 693.

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 is an A.P.

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

∵ S_{n} = [2a + (n – 1)d]

S_{25} = [2 × 7 + (25 – 1)4]

= [14 + 24 × 4]

= [14 + 96]

= × 110

= 25 × 55

= 1375

Thus S_{25} = 1375

(ii)Given : a_{n} = 7 – 3n

substituting various values 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} = [2a + (n – 1)d]

S_{n} = [2 x 4 + (25 – 1)(-3)

= [8 + 24 × -3]

= [8 – 72]

= × – 64

= -25 × 32

= -800

Hence S_{25} = – 800.

Question 5.

Find the sum of first 51 terms of AP. in 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} = [2a + (n – 1)d]

∴ S_{51} = [2 × 10 + (51 – 1)4]

= [20 + 50 × 4]

= [20 + 200]

= × 220 = 51 × 110 = 5610

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

Question 6.

The first and last term of an A.P. are 17 and 350 respectively. If 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 = = 37

∴ n = 37 + 1 = 38

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

∴ = (17 + 350)

= 19 × 367 = 6973

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

Question 7.

Find the sum of all odd numbers, 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 =

⇒ n = 167

∴ Required sum

S_{n} = (a + l)

S_{167} = (3 + 999)

= × 1002

= 167 × 501

= 83667

Hence, required sum = 83667.

Question 8.

The first term of A.P. is 8, n^{th} term is 33 and sum 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} = [2a + (n – 1)d]

⇒ 123 = [2 × 8 + 25] [From equation (i)]

⇒ 123 = (16 + 25)

⇒ 123 = × 41

⇒ n =

⇒ n = 6

Put the value of n in equation (i)

⇒ (6 – 1)d = 25

⇒ 5d = 25

⇒ d = 5

Thus, n = 6 and d = 5

Question 9.

A sum of ₹ 280 is to be used to give four 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} = [2a + (n – 1)d]

S_{4} = [2a + (4 – 1) × -20]

⇒ 280 = 2[2a + 3 × – 20]

⇒ 280 = 2[2a – 60]

⇒ 140 = 2a – 60

⇒ 2a = 140 + 60

⇒ a = = 100

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

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

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. 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 = = 25

Put the value of d in 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 of T.V. sets in 10^{th} year = 775 sets

(iii) Total production in 7 years

S_{n} = [2a + (n – 1)d]

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

= [1100 + 6 × 25]

= [1100 + 150]

= × 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.

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