Ex-9.6 Arithmetic Progressions (Part - 2), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 12. Find the sum of

(i) First 15 multiples of 8

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(iii) all 3 – digit natural numbers which are divisible by 13.

Solution:  In the given problem,

we need to find the sum of terms for different arithmetic progressions.

So, here 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

(i) First 15 multiples of 8.

So, we know that the first multiple of 8 is 8 and the last multiple of 8 is 120.

Also, all these terms will form an A.P. with the common difference of 8.

So here,

First term (a) = 8

Number of terms (n) = 15

Common difference (d) = 8

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

S_n = 15/2[2(8) + (15 – 1)8]

=  15/2[16 + (14)8]

=  15/2[16 + 12]

=  15/2[128]

= 960

Therefore, the sum of the first 15 multiples of 8 is 960

(ii) (a) First 40 positive integers divisible by 3

So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.

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

So here,

First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

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

S_n = 40/2[2(3) + (40 – 1)3]

= 20 [ 6 + (39)3 ]

= 20 (6 + 117)

= 20 (123)

= 2460

Therefore, the sum of first 40 multiples of 3 is 2460

(b) First 40 positive integers divisible by 5

So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.

Also, all these terms will form an A.P. with the common difference of 5.

So here,

First term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

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

S_n = 40/2[2(5) + (40 – 1)5]

= 20 [ 10 + (39)5 ]

= 20 (10 + 195)

= 20 (205)

= 4100

Therefore, the sum of first 40 multiples of 5 is 4100

(c) First 40 positive integers divisible by 6

So, we know that the first multiple of 6 is 6 and the last multiple of 6 is 240.

Also, all these terms will form an A.P. with the common difference of 6.

So here,

First term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

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

S_n = 40/2[2(6) + (40 – 1)6]

= 20[12 + (39) 6]

= 20 (12 + 234)

= 20(246)

= 4920

Therefore, the sum of first 40 multiples of 6 is 4920

(ii) All 3 digit natural number which are divisible by 13

So, we know that the first 3 digit multiple of 13 is 104

and the last 3 digit multiple of 13 is 988.

Also, all these terms will form an AR with the common difference of 13.

So here,

First term (a) = 104

Last term (l) = 988

Common difference (d) = 13

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,

an = a + (n – 1)d

So, for the last term,

988 = 104 + (n – 1)13

⇒ 988 = 104 + 13n – 13

⇒ 988 = 91 + 13n

⇒ 13n = 897

⇒ n = 69

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

S_n = 69/2[2(104) + (69 – 1)13]

=  69/2[208 + 884]

=  69/2[1092]

= 69 (546)

= 37674
Therefore, the sum of all 3 digit multiples of 13 is 37674

Question 13. Find the sum:

(i) 2 + 4 + 6 + . . . + 200

(ii) 3 + 11 + 19 + . . . + 803

(iii) (-5) + (-8) + (-11) + . . . + (-230)

(iv) 1 + 3 + 5 + 7 + . . . + 199

(v) 7 + (10)/12 + 14 + ... + 84

(vi) 34 + 32 + 30 + . . . + 10

(vii) 25 + 28 + 31 + . . . + 100

Solution:  In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

S_n = n/2[2a + (n – 1)d]

or

S_n = n/2[a + l]

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

d = common difference of the given A.P.

l = last term

n = number of terms

(i) 2 + 4 + 6 + … + 200

Common difference of the A.P. (d) = a₂  – a₁

= 6 – 4

= 2

So here,

First term (a) = 2

Last term (l) = 200

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,

200 = 2 + (n – 1)2

200 = 2 + 2n – 2

200 = 2n

Further simplifying,

n = 100

Now using the formula for sum of n terms,

S₁₀₀ = 100/2[a + l]

= 50 [ 2 + 200 ]

= 50 X 202

= 10100

Therefore, the sum of the A.P is 10100

(ii) 3 + 11 + 19 + . . . + 803

Common difference of the A.P. (d) = a₂  – a₁

= 19 – 11

= 8

So here,

First term (a) = 3

Last term (l) = 803

Common difference (d) = 8

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, Further simplifying,

803 = 3 + (n – 1)8

⇒ 803 = 3 + 8n – 8

⇒ 803 + 5 = 8n

⇒ 808 = 8n

⇒ n = 101

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

S₁₀₁ = 101/2[a + l]

=  101/2[3 + 803]

=  101/2[806]

= 101 (403)

= 40703

Therefore, the sum of the A.P. is 40703

(iii) (-5) + (-8) + (-11) + . . . + (-230)

Common difference of the A.P. (d) = a₂  – a₂

= -8 – (-5)

= -8 + 5

= -3

So here,

First term (a) = -5

Last term (l) = -230

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,

-230 = -5 + (n -1)(-3)

⇒ -230 = -5 -3n + 3

⇒ -230 + 2 = -3n

⇒ -228 = -3n

⇒ n = 76

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

S₇₆ = 76/2[a + l]

= 38[ (- 5) + (-230)]

= 38(-235)

= -8930

Therefore, the sum of the A.P. is -8930

(iv) 1 + 3 + 5 + 7 + . . . + 199

Common difference of the A.P. (d) = a₂  – a₁

= 3 – 1

= 2

So here,

First term (a) = 1

Last term (l) = 199

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,

199 = 1 + (n – 1)2

⇒ 199 = 1 + 2n – 2

⇒ 199 + 1 = 2n

⇒ n = 100

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

S₁₀₀ = 100/2[a + l]

= 50 [ 1 + 199 ]

= 50 (200)

= 10000

Therefore, the sum of the A.P. is 10000

(v) 7 + 1012 + 14 + ... + 84

Common difference of the A.P. (d) = a₂  – a₁

=  (10)1/2 – 7

=  21/2 – 7

=  (21 – 14)/2

=  7/2

So here,

First term (a) = 7

Last term (l) = 184

Common difference (d) = 7/2

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,

84 = 7 + (n – 1)7/2

84 = 7 + 7n/2 – 7/2

84 = 14 – 72 + 7n/2

84 (2) = 7 + 7n

7n = 161

n = 23

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

S_n = 23/2[2(7) + (23 – 1)72]

=  23/2[14 + (22)72]

=  23/2[14 + 77]

=  23/2[91]

=  2093/2

Therefore, the sum of the A.P. is 2093/2

(vi) 34 + 32 + 30 + . . . + 10

Common difference of the A.P. (d) = a₂  – a₁

= 32 -34

= -2

So here,

First term (a) = 34

Last term (l) = 10

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,

⇒ 10 = 34 + (n – 1)(-2)

⇒ 10 = 34 – 2n + 2

⇒ 10 = 36 – 2n

⇒ 10 – 36 = -2n

Further solving for n,

⇒ -2n = -26

⇒ n = 13

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

S_n = 13/2[a + l]

=  13/2[34 + 10]

=  13/2[44]

= 12 (22)

= 286

Therefore, the sum of the A.P. is 286

(vii) 25 + 28 + 31 + . . . + 100

Common difference of the A.P. (d) = a₂  – a₁

= 28 – 25

= 3

So here,

First term (a) = 25

Last term (l) = 100

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,

100 = 25 + (n – 1)(3)

100 = 25 + 3n – 3

100 = 22 + 3n

100 – 22 = 3n

Further solving for n,

78 = 3n

n = 26

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

S_n = 26/2[a + l]

= 13 [ 25 + 100 ]

= 13 (125)

= 1625

Therefore, the sum of the given A.P. is 1625

Question 14. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution :-  In the given problem, we have the first and the last term of an A.P.

along with the common difference of the AP Here,

we need to find the number of terms of the AP and the sum of all the terms.

Here,

The first term of the A.P (a) = 17

The last term of the A.P (l) = 350

The common difference of the A.P. = 9

Let the number of terms be n.

So, as we know that,

l = a + (n – 1)d

we get,

350 = 17 + (n- 1) 9

⇒ 350 = 17 + 9n – 9

⇒ 350 = 8 + 9n

⇒ 350 – 8 = 9n

Further solving this,

n = 38

Using the above values in the formula,

S_n = n/2[a + l]

⇒  38/2(17 + 350)

⇒ 19 X 367

⇒ 6973

Therefore, the number of terms is (n) 38 and the sum (S_n) is 6973

Question 15. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

Solution:  In the given problem, let us take the first term as a

and the common difference as d.

Here, we are given that,

a₃ = 7                                                                                                   . . . .(1)

a₇ = 3a₃  + 2                                                                                        . . . .(2)

So, using (1) in (2), we get,

a₇ = 3(7) + 2

= 21 + 2

= 23                                                                                                     . . . .(3)

Also, we know,

a_n = a + (n – 1)d

For the 3th term (n = 3),

a₃ = a + (3 – 1)d

⇒ 7 = a + 2d                                                   (Using 1)

⇒ a = 7 -2d                                                                                          . . . . (4)

Similarly, for the 7th term (n = 7),

a₇ = a + (7 – I) d

24 = a + 6d                                                      (Using 3)

a = 24 – 6d                                                                                           . . . . (5)

Subtracting (4) from (5), we get,

a – a = (23 – 6d) – (7 – 2d)

⇒ 0 = 23 – 6d – 7 + 2d

⇒ 0 = 16 – 4d

⇒ 4d = 16

⇒ d = 4

Now, to find a, we substitute the value of d in  (4),

a = 7 – 2(4)

⇒ a = 7 – 8

a = -1

So, for the given A.P, we have d = 4 and a = -1

So, to find the sum of first 20 terms of this A.P.,

we use 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 = 20, we get,

S₂₀ = 20/2[2(−1) + (20 – 1)(4)]

= (10)[-2 + (19) (4)]

= (10) [-2 + 76]

= (10) [74]

= 740

Therefore, the sum of first 20 terms for the given A.P. is S₂₀ = 740

Question 16. The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

Solution:  In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P.

Here, we need to find the common difference of the A.P.

Here,

The first term of the A.P (a) = 2

The last term of the A.P (I) = 50

Sum of all the terms S„ = 442

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

442 = (n/2)(2 + 50)

⇒  442 = (n/2)(52)

⇒ 26 n = 442

⇒ n = 17

Now, to find the common difference of the A.P. we use the following formula,

l = a + (n – 1)d

We get,

50 = 2 + (17 – 1)d

⇒ 50 = 2 + 16d

⇒ 16d = 48

⇒ d = 3

Therefore, the common difference of the A.P. is d = 3

Question 17. If 12^th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Solution:  In the given problem, we need to find the sum of first 10 terms of an A.P.

Let us take the first term a

and the common difference as d

Here, we are given that,

a₁₂ = -13

S₄ = 24

Also, we know,

a_n = a + (n – 1)d

For the 12th term (n = 12)

a₁₂ = a + (12 – 1)d

-13 = a + 11d

a = -13 – 11d                                                                                        . . . .(1)

So, as we know the formula for the sum of n terms of an A.P. is given by,

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 = 4, we get,

S₄ = 4/2[2(a) + (4 – 1)d]

⇒ 24 = (2) [ 2a + (3)(d) ]

⇒ 24 = 4a + 6d

⇒ 4a = 24 – 6d

⇒  a = 6 – 64d                                                  . . . .(2)

Subtracting (1) from (2), we get.

⇒  a – a = (6 – 6/4.d) – (−13 – 11d)

⇒  0 = 6 – 6/4.d + 13d + 11d

⇒  0 = 19 + ((44d – 6d)/4).s

On further simplifying for d, we get,

⇒  0 = 19 + 38/4.d

⇒ -19 = 19/2.d

⇒ – 19 X 2 = 19 d

⇒ d = -2

Now, we have to substitute the value of d in (1),

a = -13 – 11 (-2)

a = -13 + 22

a = 9

Now, using the formula for the sum of n terms of an A.P., for n = 10

we have,

S₁₀ = 10/2[2(9) + (10 – 1)(−2)]

= (5)[ 19 + (9)(-2) ]

= (5)(18 – 18)

= 0

Therefore, the sum of first 10 terms for the given A.P. is S₁₀ = 0.

Question 18. Find the sum of first 22 terms of an A.P. in which d = 22 and a₂₂ = 149.

Solution:  In the given problem, we need to find the sum of first 22 terms of an A.P.

Let us take the first term as a.

Here, we are given that,

a₂₂ = 149                                                                                  . . . .(1)

d = 22                                                                                     . . . .(2)

Also, we know,

a_n = a + (n – 1) d

For the 22^nd term (n = 22),

a₂₂ = a + (22 – 1) d

149 = a + (21) (22)                                                      (Using 1 and 2)

a = 149 – 462

a = – 313                                                                                              . . . .(3)

So, as we know the formula for the sum of n terms of an A.P. is given by,

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 = 22, we get,

S₂₂ = 22/2[2(−313) + (22 – 1)(22)]

= (11) [ -626 + 462 ]

= (11) [ -164 ]

= -1804

Therefore, The sum of first 22 terms for the given A.P. is S₂₂ = -1804

Question 19. In an A.P., if the first tern is 22, the common difference is -4 and the sum to n terms is 64, find n.

Solution:  In the given problem,

we need to find the number of terms of an A.P.

Let us take the number of terms as n.

Here, we are given that,

a = 22

d = -4

S„ = 64

So, as we know the formula for the sum of n terms of an A.P. is given by,

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 we get,

⇒  S_n = n/2[2(22) + (n – 1)(−4)]

⇒  64 = n/2[2(22) + (n – 1)(−4)]

⇒ 64(2) = n(48 – 4n)

⇒ 128 = 48n – 4n²

Further rearranging the terms, we get a quadratic equation,

4n²  – 48n + 128 = 0

On taking 4 common, we get,

n²  – 12n + 32 = 0

Further, on solving the equation for n by splitting the middle term, we get,

n²  – 12n + 32 = 0

n²  – 8n – 4n + 32 = 0

n (n – 8) – 4 (n – 8) = 0

(n – 8) (n – 4) = 0

So, we get

n – 8 = 0

⇒ n = 8

Also,

n – 4 = 0

⇒ n = 4

Therefore, n = 4 or 8.