Ex-9.6 Arithmetic Progressions (Part - 4), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 46. If the sum of first n terms of an A.P. is 1/2(3n² + 7n) , then find its n^th term. Hence write the 20^thterm.

Solution:  Let a be the first term and d be the common difference.

We know that, sum of first n terms is:

S_n = n/2(2a + (n – 1)d)

It is given that the sum of the first n terms of an A.P. is:

1/2(3n² + 7n)

Therefore, first term (a) = S₁ = 1/2(3(1)² + 7(1))

=  1/2(3X1 + 7)

=  1/2(10)

= 5

Sum of first two terms = S₂ = 1/2(3(2)² + 7(2))

=  1/2(3X4 + 14)

=  1/2(26)

= 13

Therefore, second term = S₂  – S₁

= 13 – 5 = 8

Common difference = d = second term – first term

= 8 – 5 = 3

Also, n^th term of the A.P. is : a + (n – 1)d

= 5 + (n – 1)3

= 5 + 3n – 3

= 3n + 2

Thus, n^th term of this A.P. is 3n + 2.

Now, we have to find the 20^th term, so

Putting n = 2o in the above equation, we get

a₂₀  = 3 (20) + 2

= 60 + 2

= 62

Thus, 20^th term of this A.P. is 62.

Question 47. In an A.P. the sum of first n terms is 3n²/2 + (13/2).n. Find its 25^th term.

Solution:  Here the sum of first n terms is given by the expression,

S_n = 3n²/2 + (13/2).n

We need to find the 25^th term of the A.P.

So, we know that the n^th term of an A.P. is given by,

a_n  = S_n  – S_n-1

So, a₂₅  = S₂₅  – S₂₄                                                               . . . . (1)

So, using the expression for the sum of n terms,

we find the sum of 25 terms (S₂₅) and the sum of 24 terms (S₂₅), we get,

=  2200/2

= 1100

Similarly,

=  2040/2

= 1020

Now, using the above values in (1),

a₂₁  = S₂₅  – S₂₄

= 1100 – 1020

= 80

Therefore, a₂₅  = 80

Question 48. Find the sum of all natural numbers between 1 and 100, which are divisible by 3.

Solution:  In this problem,

we need to find the sum of all the multiples of 3 lying between 1 and 100.

So, we know that the first multiple of 3 after 1 is 3

and the last multiple of 3 before 100 is 99.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (a) = 3

Last term (l) = 99

Common difference (d) = 3

So, here the first step is to find the total number of terms.

Let us take the number of terms as n.

Now, as we know,

a_n  = a + (n – 1)d

So, for the last term,

99 = 3 + (n – 1)3

⇒ 99 = 3 + 3n – 3

⇒ 99 = 3n

Further simplifying,

⇒ n = 33

Now, using the formula for the sum of n terms,

i.e. S_n = n/2[2a + (n – 1)d]

we get,

⇒  S₃₃ = 33/2[2(3) + (33 – 1)3]

=  S₃₃ = 33/2[6 + (32)3]

=  S₃₃ = 33/2[6 + 96]

=  S₃₃ = (33(102))/2

= 33 (51)

= 1683

Therefore, the sum of all the mulstiples of 3 lying between 1 and 100 is S_n  = 1683

Question 50. Find the sum of all odd numbers between (i) 0 and 50 (ii) 100 and 200.

Solution:

(i) In this problem, we need to find the sum of all odd numbers lying between 0 and 50.

So, we know that the first odd number after 0 is 1

and the last odd number before 50 is 49.

Also, all these terms will form an AP. with the common difference of 2.

So here,

First term (a) = 1

Last term (0 = 49

Common difference (d) = 2

So, here the first step is to find the total number of terms.

Let us take the number of terms as n. Now, as we know,

a_n  = a + (n -1)d

So, for the last term,

⇒ 49 = 1 + (n – 1)d

⇒ 49 = 1 + 2n – 2

⇒ 49 = 2n – 2

⇒ 49 + 1 = 2n

Further simplifying,

⇒ 50 = 2n

⇒ n = 25

Now, using the formula for the sum of n terms,

⇒  S_n = n/2[2a + (n – 1)d]

for n = 25, we get

⇒  S₂₅ = 25/2[2(1) + (25 – 1)2]

=  25/2[2 + 24×2]

= 25 X 25

= 625

Therefore, the sum of all the odd numbers lying between 0 and 50 is 625.

(ii) In this problem,

we need to find the sum of all odd numbers lying between 100 and 200.

So, we know that the first odd number after 0 is 101

and the last odd number before 200 is 199.

Also, all these terms will form an AR. with the common difference of 2.

So here,

First term (a) = 101

Last term (a_n) = 199

Common difference (d) = 2

So, here the first step is to find the total number of term.

Let us take the number of terms as n.

Now, as we know,

a_n  = a + (n – 1)d

So, for the last term,

⇒ 199 = 101 + (n – 1)2

⇒ 199 = 101 + 2n – 2

⇒ 199 = 99 + 2n

⇒ 199 – 99 = 2n

Further simplifying,

⇒ 100 = 2n

⇒ n = 50

Now, using the formula for the sum of n terms,

S_n = n/2[2(a) + (n – 1)d]

For n = 50, we get

⇒  S₅₀ = 50/2[2(101) + (50 – 1)2]

= 25 [ 202 + (49) 2 ]

= 25(202 + 98)

= 25 (300)

= 7500

Therefore, the sum of all the odd numbers lying between 100 and 200 is 7500

Question 52. Find the sum of all integers between 84 and 719, which are multiples of 5.

Solution: In this problem,

we need to find the sum of all the multiples of 5 lying between 84 and 719.

So, we know that the first multiple of 5 after 84 is 85

and the last multiple of 5 before 719 is 715.

Also, all these terms will form an A.P.

with the common difference of 5.

So here,

First term (a) = 85

Last term (l) = 715

Common difference (d) = 5

So, here the first step is to find the total number of terms.

Let us take the number of terms as n.

Now, as we know,

a = a + (n -1)d

So, for the last term.

715 = 85 + (n – 1)5

715 = 85 + 5n – 5

715 = 80 + 5n

715 – 80 = 5n

Further simplifying,

635 = 5n

n = 127

Now, using the formula for the sum of n terms,

S_n = n/2[2a + (n – 1)d]

For n = 127,

S₁₂₇ = 127/2[2(85) + (127 – 1)5]

=  127/2[170 + 630]

=  (127(800))/2

= 50800

Therefore, the sum of all the multiples of 5 lying between 84 and 719 is 50800.

Question 53. Find the sum of all integer between 50 and 500, which are divisible by 7.

Solution:  In this problem,

we need to find the sum of all the multiples of 7 lying between 50 and 500.

So, we know that the first multiple of 7 after 50 is 56

and the last multiple of 7 before 500 is 497.

Also, all these terms will form an A.P. with the common difference of 7.

So here,

First term (a) = 56

Last term (l) = 497

Common difference (d) = 7

So, here the first step is to find the total number of terms.

Let us take the number of terms as n.

Now, as we know,

a_n  = a   + (n – 1)d

So, for the last term.

497 = 56 + (n – 1)7

⇒ 497 = 56 + 7n -7

⇒ 497 = 49 + 7n

⇒ 497  – 49 = 7n

Further simplifying,

448 = 7n

n = 64

Now, using the formula for the sum of n terms,

S_n = n/2[2a + (n – 1)d]

for n = 64, we get

S₆₄ = 64/2[2(56) + (64 – 1)7]

= 32 [ 112 + (63)7]

= 32 [112 + 441]

= 32 (553)

= 17696

Therefore, the sum of all the multiples of 7 lying between 50 and 500 is 17696

Question 54. Find the sum of all even integers between 101 and 999.

Solution:  In this problem,

we need to find the sum of all the even numbers lying between 101 and 999.

So, we know that the first even number after 101 is 102

and the last even number before 999 is 998.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 102

Last term (l) = 998

Common difference (d) = 2

So, here the first step is to find the total number of terms.

Let us take the number of terms as n.

Now, as we know,

a_n  = a   + (n – 1)d

So, for the last term,

⇒ 998 = 102 + (n – 1)2

⇒ 998 = 102 + 2n – 2

⇒ 998 = 100 + 2n

⇒ 998 – 100 = 2n

Further simplifying,

⇒ 898 = 2n

⇒ n = 449

Now, using the formula for the sum of n terms,

S_n = n/2[2a + (n – 1)d]

For n = 449, we get

S₄₄₉ = 449/2[2(102) + (449 – 1)2]

=  449/2[204 + (448)2]

=  449/2[204 + 896]

=  449/2[1100]

= 449 (550)

= 246950

Therefore, the sum of all even numbers lying between 101 and 999 is  246950

Question 55.
  (i) Find the sum of all integers between 100 and 550, which are divisible by 9.

Solution:  In this problem,

we need to find the sum of all the multiples of 9 lying between 100 and 550.

So, we know that the first multiple of 9 after 100 is 108

and the last multiple of 9 before 550 is 549.

Also, all these terms will form an A.P. with the common difference of 9.

So here,

First term (a) = 108

Last term (l) = 549

Common difference (d) = 9

So, here the first step is to find the total number of terms.

Let us take the number of terms as n.

Now, as we know,

a_n  = a + (n – 1)d

So, for the last term.

⇒ 549 = 108 + (n – 1)d

⇒ 549 = 108 + 9n – 9

⇒ 549 = 99 + 9n

⇒ 549 – 99 = 9n

Further simplifying

⇒ 9n = 450

⇒ n = 50

Now, using the formula for the sum of n terms,

S_n = n/2[2a + (n – 1)d]

We get,

S_n = 50/2[2(108) + (50 – 1)9]

= 25 [ 216 + (49)9 ]

= 25 (216 + 441)

= 25 (657)

= 16425

Therefore, the sum of all the multiples of 9 lying between 100 and 550 is 16425

Question 56. Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_n denotes its n^th term and S_n the sum of first n terms, find.

(i) n and S_n, if a = 5 , d = 3 , and a_n  = 50.

(ii) n and a, if a_n  = 4 ,  d = 2 and S_n  = -14.

(iii) d, if a = 3, n = 8 and S_n  = 192.

(iv) a, if a_n  = 28, S_n  = 144 and n = 9.

(v) n and d, if a = 8, a_n  = 62 and S_{­n­ ­} = 120.

(vi) n and a_n , if a = 2, d = 8 and S_n  = 90.

Solution:

(i) Here, we have an A.P. whose n^th term (a_n), first term (a) and common difference (d) are given. We need to find the number of terms (n) and the sum of first n terms (S_n).

Here,

First term (a) = 5

Last term (a_n) = 50

Common difference (d) = 3

So here we will find the value of n using the formula, a_n  = a + (n – 1) d

So, substituting the values in the above mentioned formula

⇒ 50 = 5 + (n – 1) 3

⇒ 50 = 5 + 3n – 3

⇒ 50 = 2 + 3n

⇒ 3n = 50 – 2

Further simplifying for n,

3n = 48

n = 16

Now, here we can find the sum of the n terms of the given A.P., using the formula,

S_n = (n/2)(a + l)

Where, a = the first term

l = the last term

So, for the given A.P,

on substituting the values in the formula for the sum of n terms of an A.P., we get,

S₁₆ = (16/2)(5 + 50)

= 8 (55)

= 440

Therefore, for the given A.P. we have, n = 16 and S₁₆  = 440

 (ii) Here, we have an A.P. whose n^th term (a_n), sum of first n terms (S0) and common difference (d) are given. We need to find the number of terms (n) and the first term (a).

Here,

Last term (l) = 4

Common difference (d) = 2

Sum of n terms (S_n) = -14

So here we will find the value of n using the formula, a_n  = a + (n -1) d

So, substituting the values in the above mentioned formula

⇒ 4 = a + (n – 1) 2

⇒ 4 = a + 2n - 2

⇒ 4 + 2 = a + 2n

⇒ n =  (6 – a)/2                               . . . . (1)

Now, here the sum of the n terms is given by the formula,

S_n = (n/2)(a + l)

Where, a = the first term

l = the last term

So, for the given A.P,

on substituting the values in the formula for the sum of n terms of an A.P., we get,

⇒  −14 = n/2(a + 4)

⇒ 14 (2) = n (a + 4)

⇒  n = −28/(a + 4)                                      . . . . (2)

Equating (1) and (2), we get,

(6 – a)(a + 4) = -28(2)

6a – a²  + 24 – 4a = -56

-a²  + 2a + 24 + 56 = 0

So, we get the following quadratic equation,

-a ²  + 2a + 80 = 0

a²  – 2a – 80 = 0

Further solving it for a by splitting the middle term,

a²  – 2a – 80 = 0

a²  – 10a + 8a – 80 = 0

a(a – 10) + 8(a – 10) = 0

(a – 10)(a + 8) = 0

So, we get,

a – 10 = 0

a = 10

or,

a + 8 = 10

a = -8

Substituting, a = 10 in (1)

n = (6 – 10)/2

n = −4/2

n = -2

Here, we get n as negative, which is not possible. SO, we take a = -8

n = (6 – (−8))/2

n = (6 + 8)/2

n = 14/2

n = 7

Therefore, for the given A.P. n = 7 and a = -8

(iii) Here, we have an A.P. whose first term (a), sum of first n terms (S0) and the number of terms (n) are given. We need to find common difference (d).

Here,

First term (a) = 3

Sum of n terms (S_n) = 192

Number of terms (n) = 8

So here we will find the value of n using the formula, a_n  = a + (n -1) d

So, to find the common difference of this A.P.,

we use the following formula for the sum of n terms of an A.P

S_n = (n/2)[2a + (n – 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

S₈ = (8/2)[2(3) + (8 – 1)d]

192 = 4 [ 6 + 7d ]

192 = 24 + 28d

28d = 192 – 24

28d = 168

d = 6

Therefore, the common difference of the given A.P. is d = 6

(iv) Here, we have an A.P. whose n^th term (a_n), sum of first n terms (S_n) and the number of terms (n) are given. We need to find first term (a).

Here,

Last term (a₉) = 28

Sum of n terms (S_n) = 144

Number of terms (n) = 9

Now,

a₉  = a + 8d

28 = a + 8d                                                                . . . . (1)

Also, using the following formula for the sum of n terms of an AP

S_n = (n/2)[2a + (n – 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 9, we get,

S₉ = (9/2)[2a + (9 – 1)d]

144 (2) = 9 [2a + 8d]

288 = 18a + 72d                                                             . . . . (2)

Multiplying (1) by 9, we get

9a + 72d = 252                                                                . . . .  (3)

Further, subtracting (3) from (2), we get

9 a = 36

a = 4

Therefore, the first term of the given A.P. is a = 4

(v) Here, we have an A.P. whose n^th term (a_n), sum of first n terms (S_n) and first term (a) are given. We need to find the number of terms (n) and the common difference (d).

Here,

First term (a) = 8

Last term (a_n ) = 62

Sum of n terms (S_n) = 210

Now, here the sum of the n terms is given by the formula,

S_n = (n/2)(a + l)

Where, a = the first term

l = the last term

So, for the given A.P,

on substituting the values in the formula for the sum of n terms of an A.P., we get,

210 = n/2[8 + 62]

210 (2) = n (70)

n = 420/70

n = 7

Also, here we will find the value of d using the formula,

a_n  = a + (n – 1) d

So, substituting the values in the above mentioned formula

62 = 8 + (6 – 1)d

5d = 54

d = 54/5

Therefore, for the given A.P. n = 6 and d = 54/5

(vi) Here, we have an A.P. whose first term (a), common difference (d) and sum of first n terms are given. We need to find the number of terms (n) and the nth term (a_n).

Here,

First term (a) = 2

Sum of first nth terms (S_n) = 90

Common difference (d) = 8

So, to find the number of terms (n) of this A.P.,

we use the following formula for the sum of n terms of an A.P

S_n = (n/2)[2a + (n – 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for d = 8, we get,

S_n = (n/2)[2(2) + (n – 1)8]

90 = (n/2)[4 + 8n – 8]

90 (2) = n [ 8n – 4 ]

180 = 8n²  – 4n

Further solving the above quadratic equation,

8n² -4n -180 = 0

2n²  – n – 45 = 0

Further solving for n,

2n²  – 10n + 9n – 45 = 0

2n(n – 5) + 9(n – 5) = 0

(2n – 9)(n – 5) = 0

Now,

2n + 9 = 0

n = −9/2

Also,

n – 5 = 0

n = 5

Since, n cannot be a fraction.

Thus, n = 5

Also, we will find the value of n^th term (a_n), using the formula,

a_n  = a + (n – 1) d

So, substituting the values in the above formula,

a_n  = 2 + (5 – 1) 8

a_n  = 2 + 4 (8)

a_n  = 2 + 32

a_n  = 34

Therefore, for the given A.P., n = 5 and a_n  = 34