RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.3

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 (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]
S₁₂ = \(\frac { 12 }{ 2 }\) [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 = \(\frac { 33 }{ 20 }\)

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\(\frac { 1 }{ 2 }\)

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
⇒ 4n² + 5n = 636
⇒ 4n² + 5n – 636 = 0
⇒ 4n² + 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 – 3n²
⇒ 3n² – 129n + 1386 = 0
⇒ n² – 43n + 462 = 0
⇒ n² – 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.

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)
a₁ = 3 + 4(1) = 7
a₂ = 3 + 4(2) = 11
a₃ = 3 + 4(3) = 15, …
common difference (d) = a₂ – a₁ = 11 – 7 = 4
a₃ – a₂ = 15 – 11 = 4
∵ a₂ – a₁ = a₃ – a₂ = 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]
S₂₅ = \(\frac { 25 }{ 2 }\) [2 × 7 + (25 – 1)4]
= \(\frac { 25 }{ 2 }\) [14 + 24 × 4]
= \(\frac { 25 }{ 2 }\) [14 + 96]
= \(\frac { 25 }{ 2 }\) × 110
= 25 × 55
= 1375
Thus S₂₅ = 1375

(ii)Given : a_n = 7 – 3n
substituting various values (RBSESolutions.com) of n in equation (i)
a₁ = 7 – 3(1) = 4
a₂ = 7 – 3(2) = 1
a₃ = 7 – 3(3) = -2
common difference (d) = a₂ – a₁ = 1 – 4 = -3
and a₃ – a₂ = -2 – 1 = -3
∵ a₂ – a₁ = a₃ – a₂ = -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 S₂₅ = – 800.

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. a₂ = 14
and third term a₃ = 18
∴ Common difference d = a₃ – a₂
= 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]
∴ S₅₁ = \(\frac { 51 }{ 2 }\) [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

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

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)
S₁₆₇ = \(\frac { 167 }{ 2 }\) (3 + 999)
= \(\frac { 167 }{ 2 }\) × 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 (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

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 a₂ = ₹ (a – 20)
III^rd prize a₃ = ₹ [(a – 20) – 20]
a₃ = ₹ (a – 40)
IV^th prize a₄ = ₹ [(a – 40) – 20)]
a₄ = ₹ (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]
S₄ = \(\frac { 4 }{ 2 }\) [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

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
a₃ = 600
and in seventh year
a₇ = 100.
Now, a₃ = a + (3 – 1)d
600 = a + 2d …..(i)
and a₇ = a + (7 – 1)d
700 = a + 6d …(ii)
subtracting eqⁿ (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, S₇ = \(\frac { 7 }{ 2 }\) [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.

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