# RS Aggarwal Solutions Chapter 11 Arithmetic Progression Exercise 11D Class 10 Maths

 Chapter Name RS Aggarwal Chapter 11 Arithmetic Progression Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 11AExercise 11BExercise 11C Related Study NCERT Solutions for Class 10 Maths

### Exercise 11D Solutions

1.Find the sum of each of the following Aps:

(i) 2, 7, 12, 17, ...... to 19 terms.

(ii) 9, 7, 5, 3..... to 14 terms

(iii) – 37, -33, -29, ..... to 12 terms.

(iv) 1/15, 1/12, 1/10, ...... to 11 terms.

(v) 0.6, 1.7, 2.8, ..... to 100 terms

Solution

(i) The given AP is 2, 7, 12, 17, ......

Here, a = 2 and d = 7 – 2 = 5

Using the formula, Sn = n/2[2a + (n – 1)d], we have

Sn = 19/2[2 × 2 + (19 – 1) × 5]

= 19/2 × (4 + 90)

= 19/2 × 94

= 893

(ii) The given AP is 9, 7, 5, 3, .....

Here, a = 9 and d = 7 – 9 = - 2

Using the formula, Sn = n/2[2a + (n – 1)d], we have

S14 = 14/2 [2 × 9 + (14 – 1) × (-2)]

= 7 × (18 – 26)

= 7 × (-8)

= - 56.

(iii) The given AP is -37, -33, -29, ......

Here, a = - 37 and d = - 33 – (-37) = - 33 + 37 = 4

Using the formula, Sn = n/2[2a + (n – 1)d], we have

S12 = 12/2[2 × (-37) + (12 – 1) × 4]

= 6 × (-74 + 44)

= 6 × (-30)

= -180

(iv) The given AP is 1/15, 1/12, 1/10, .......

Here, a = 1/15 and d = 1/12 – 1/15

= (5 – 4)/60

= 1/60

Using the formula, Sn = n/2[2a + (n – 1)d], we have

S11 = 11/2[2 × (1/15) + (11 – 1) × 1/60]

= 11/2 × (2/15 + 10/60)

= 11/2 × (18/60)

= 33/20

(v) The given AP is 0.6, 1.7, 2.8, ........

Here, a = 0.6 and d = 1.7 – 0.6 = 1.1

Using formula, Sn = n/2[2a + (n – 1)d], we have

S100 = 100/2 [2 × 0.6 + (100 – 1) × 1.1]

= 50 × (1.2 + 108.9)

= 5505

2. Find the sum of each of the following arithmetic series:

(i) 7 + 10.1/2 + 14 + .......84

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

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

Solution

(i) The given arithmetic series is 7 + 10.1/2 + 14 + .... + 84.

Here, a = 7, d = 10.1/2 – 7 = 21/2 – 7

= (21 – 4)/2

= 7/2 and l = 84.

Let the given series contains n terms. Then,

an = 84

⇒ 7 + (n – 1) × 7/2 = 84 [an = a + (n – 1)d]

⇒ (7/2).n + 7/2 = 84

⇒ (7/2).n = 84 – 7/2 = 161/2

⇒ n = 161/7 = 23

∴ Required sum = 23/2 × (7 + 84) [Sn = n/2(a + 1)]

= 23/3 × 91

= 2030/2

= 1046.1/2

(ii) The given arithmetic series is 34 + 32 + 30 + .... +10.

Here, a = 34, d = 32 – 34 = -2 and l = 10.

Let the given series contain n terms. Then,

an = 10

⇒ 34 + (n – 1) × (-2) = 10 [an = a + (n – 1)d]

⇒ - 2n + 36 = 10

⇒ - 2n = 10 – 36 = - 26

⇒ n = 13

∴ Required sum = 13/2 × (34 + 10) [Sn = n/2(a + 1)]

= 13/2 × 44

= 286

(iii) The given arithmetic series is (-5) + (-8) + (-11) + ...... + (-230).

Here, a = - 5, d = - 8 – (-5)

= - 8 + 5

= - 3

and l = 230.

Let the given series contain n terms. Then,

an = - 230

⇒ - 5 + (n – 1) × (-3) = - 230 [an = a + (n – 1)d]

⇒ - 3n – 2 = - 230

⇒ - 3n = - 230 + 2 = - 228

⇒ n = 76

∴ Required sum = 76/2 × [(-5) + (-230)] [Sn = n/2(a + 1)]

= 76/2 × (-235)

= -8930

3. Find the sum of first n terms of an AP whose nth term is (5 – 6n). Hence, find its sum of its first 20 terms.

Solution

Let an be the nth term of the AP.

∴ an = 5 – 6n

Putting n = 1, we get

First term, a = a1 = 5 – 6 × 1 = - 1

Putting n = 2, we get

a2 = 5 – 6 × 2 = - 7

Let d be the common difference of the AP.

∴ d = a2 – a1

= -7 – (-1)

= - 7 + 1

= - 6

Sum of first n terms of the AP, Sn

= n/2[2 × (-1) + (n – 1) × (-6)] {Sn = n/2[2a + (n – 1)d}

= n/2(-2 – 6n + 6)

= n(2 – 3n)

= 2n – 3n2

Putting n = 20, we get

S20 = 2 × 20 – 3 × 202

= 40 – 1200

= -1160

4. The sum of the first n terms of an AP is (3n2 + 6n). Find the nth term and the 15th term of this AP.

Solution

Let Sn denotes the sum of first n terms of the AP.

∴ Sn = 3n2 + 6n

⇒ Sn-1 = 3(n – 1)2 + 6(n – 1)

= 3(n2 – 2n + 1) + 6(n – 1)

= 3n2 – 3

∴ nth terms of the AP, an

Sn – Sn- 1

= (3n2 + 6n) – (3n2 – 3)

= 6n + 3

Putting n = 15, we get

a15 = 6 × 15 + 3

= 90 + 3

= 93

Hence, the nth term is (6n + 3) and 15th term is 93.

5. The sum of the first n terms of an AP is given by Sn = (3n2 – n). Find its

(i) nth term,

(ii) first term and

(ii) common difference

Solution

Given:

Sn = (3n2 – n) ...(i)

Replacing n by (n – 1) in (i), we get:

Sn-1 = 3(n – 1)2 - (n – 1)

= 3(n2 – 2n + 1)- n + 1

= 3n2 – 7n + 4

(i) Now, Tn = (Sn – Sn-1)

= (3n2 – n) – (3n2 – 7n + 4) = 6n – 4

∴ nth term, Tn = (6n – 4) .....(ii)

(ii) Putting n = 1 in (ii), we get:

T1 = (6 × 1) – 4 = 2

(iii) Putting n = 2 in (ii), we get:

T2 = (6 × 2) – 4 = 8

∴ Common difference, d = T2 – T1

= 8 - 2

= 6

6. The sum of the first n terms of an AP is (5n2/2 + 3n/2). Find its nth term and the 20th term of this AP.

Solution

Sn = (5n2/2 + 3n/2)

= 1/2 (5n2 + 3n) .....(i)

Replacing n by (n – 1) in (i), we get:

Sn-1 = 1/2 × [5(n – 1)2 + 3(n – 1)]

= 1/2 × [5n2 – 10n + 5 + 3n – 3]

= 1/2 × [5n2 – 7n + 2]

∴ Tn = Sn – Sn-1

= 1/2(5n2 + 3n) – 1/2 × [5n2 – 7n + 2]

= 1/2(10n – 2)

= 5n – 1 .....(ii)

Putting n = 20 in (ii), we get

T20 = (5 × 20) – 1 = 99

Hence, the 20th term is 99.

7. The sum of the first n term sofa an AP is (3n2/2 + 5n/2). Find its nth term and the 25th term.

Solution

Let Sn denotes the sum of first n terms of the AP.

∴ Sn = 3n2/2 + 5n/2

⇒ Sn-1 = 3(n – 1)2/2 + {5(n – 1)}/2

= 3(n2 – 2n + 1)/2 + 5(n – 1)/2

= (3n2 – n – 2)/2

∴ nth term of the AP, an

= Sn – Sn-1

= (3n2 + 5n)/2 – (3n2 – n – 2)/2

= (6n + 2)/2

= 3n + 2

Putting n = 25, we get

a25 = 3 × 25 + 1 = 75 + 1 = 76

Hence, the nth term is (3n + 1) and 25th term is 76.

8. How many terms of the AP 21, 18, 15, .... must be added to get the sum 0?

Solution

The given AP is 21, 18, 15, .....

Here, a = 21 and d = 18 – 21 = - 3

Let the required -number of terms be n. Then,

Sn = 0

⇒ n/2[2 × 21 + (n – 1) × (-3)] = 0 {Sn = n/2 [2a + (n – 1)]}

⇒ n/2(42 – 3n + 3) = 0

⇒ n(45 – 3n) = 0

⇒ n = 0 or 45 – 3n = 0

⇒ n = 0 or n = 15

∴ n = 15 (Number of terms cannot be zero)

Hence, the required number of terms is 15.

9. How many terms of the AP 9, 17, 25, ..... must be taken so that their sum is 636?

Solution

The given AP is 9, 17, 25, ........

Here, a = 9 and d = 17 – 9 = 8

Let the required’ number of terms be n. Then,

Sn = 636

⇒ n/2[2 × 9 + (n – 1) × 8] = 636 {Sn = n/2[2a + (n – 1)d]}

⇒ n/2(18 + 8n – 8) = 636

⇒ n/2(10 + 8n) = 636

⇒ n(5 + 4n) = 636

⇒ 4n2 + 5n – 636 = 0

⇒ 4n2 – 48n + 53n – 636 = 0

⇒ 4n(n – 12) + 53(n – 12) = 0

⇒ (n – 12)(4n + 53) = 0

⇒ n – 12 = 0 or 4n + 53 = 0

⇒ n = 12 or n = -(53/4)

∴ n = 12 (Number of terms cannot negative)

Hence, the required number of terms is 12.

10. How many terms of AP 63, 60, 57, 54, ...... must be taken so that their sum is 693? Explain the double answer:

Solution

The given AP is 63, 60, 57, 54, .......

Here, a = 63 and d = 60 – 63 = - 3

Let the required number terms be n. Then,

Sn = 693

⇒ n/2[2 × 63 + (n – 1) × (-3)] = 693 {Sn = n/2[2a + (n – 1)d]}

⇒ n/2(126 – 3n + 3) = 693

⇒ n(129 – 3n) = 1386

⇒ 3n2 - 129n + 1386 = 0

⇒ 3n2 – 66n – 63n + 1386 = 0

⇒ 3n(n – 22) – 63(n – 22) = 0

⇒ (n – 22)(3n – 63) = 0

⇒ n – 22 = 0 or 3n – 63 = 0

⇒ n = 22 or n = 21

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of the AP is 0.

a22 = 63 + (22 – 1) × (-3)

= 63 – 63

= 0

Hence, the required number of terms is 21 or 22.

11. How many terms of the AP 20, 19.1/3, 18.2/3, ..... must be taken so that their sum is 300? Explain the double answer.

Solution

The given AP is 20, 19.1/3, 18.2/3, ......

Here, a = 20 and d = 19.1/3 – 20 = 58/3 – 20

= (58 – 60)/3

= -(2/3)

Let the required number of terms ne n. Then,

Sn = 300

⇒ n/2[2 × 20 + (n – 1) × -(2/3)] = 300 {S= n/2[2a + (n – 1)d]}

⇒ n/2(40 – (2/3)n + 2/3) = 300

⇒ n/2 × (122 – 2n)/3 = 300

⇒ 122n – 2n2 = 1800

⇒ 2n2 – 122n + 1800 = 0

⇒ 2n2 – 50n – 72n + 1800 = 0

⇒ 2n(n – 25) – 72(n – 25) = 0

⇒ (n – 25)(2n – 72) = 0

⇒ n – 25 = 0 or 2n – 72 = 0

⇒ n = 25 or n = 36

So, the sum of first 25 terms as well as that of first 36 terms is 300. This because the sum of all terms from 26th to 36th is 0.

12. Find the sum of all odd numbers between 0 and 50.

Solution

All odd numbers between 0 and 50 are 1,

This is an in which a = 1, d = (3 – 1) = 2 and l = 49.

This is an AP in which a = 1, d = (3 – 1) = 2 and l = 49.

Let the number of terms be n.

Then, Tn = 49

⇒ a + (n – 1)d = 49

⇒ 1 + (n – 1) × 2 = 49

⇒ 2n = 50

⇒ n = 25

∴ Required sum = n/2(a + 1)

= 25/2[l + 49]

= 25 × 25

= 625

Hence, the required sum is 625.

13. Find the sum of all natural numbers between 200 and 400 which are divisible by 7.

Solution

Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, .... 399.

This is an AP with a = 203, d = 7 and l = 399.

Suppose there are n terms in the AP. Then,

an = 399

⇒ 203 + (n – 1) × 7 = 399 [an = a + (n – 1)d]

⇒ 7n + 196 = 399

⇒ 7n = 399 – 196 = 203

⇒ 7n = 399 – 196 = 203

⇒ n = 29

∴ Required sum = 29/2(203 + 399) [Sn = n/2(a + l)]

= 29/2 × 602

= 8729

Hence, the required sum is 8729.

14. Find the sum of first positive integers divisible by 6.

Solution

The positive integers divisible by 6 are 6, 12, 18, ......

This an AP with a = 6 and d = 6.

Also, n = 40 (Given)

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S40 = 40/2 [2×6 + (40 – 1)×6]

= 20(12 + 234)

= 20 × 246

= 4920

Hence, the required sum is 4920.

15. Find the sum of first 15 multiples of 8.

Solution

The first 15 multiples of 8 are 8, 16, 24, 32, ......

This is an AP in which a = 8, d = (16 – 8) = 8 and n = 15.

Thus, we have:

l = a + (n – 1)d

= 8 + (15 – 1)8

= 120

∴ Required sum = n/2(a + l)

= 15/2[8 + 120]

= 15 × 64

= 960

Hence, the required sum is 960.

16. Find the sum of all multiples of 9 lying between 300 and 700.

Solution

The multiples of 9 lying between 300 and 700 are 306, 315, ......., 693.

This is AP with a = 306, d = 9 and l = 693.

Suppose these are n terms are n terms in the AP. Then,

an = 693

⇒ 306 + (n – 1) × 9 = 693 [an = a + (n – 1)d]

⇒ 9n + 297 = 693

⇒ 9n = 693 – 297 = 396

⇒ n = 44

∴ Required sum = 44/2(306 = 693) [Sn = n/2(a + l)]

= 22 × 999

= 21978

Hence, the required sum is 21978.

17. Find the sum of all three-digits natural numbers which are divisible by 13.

Solution

All three-digit numbers which are divisible by 13 are 104, 117, 130, 143, ...... 938.

This is an AP in which a = 104, d = (117 – 104) = 13 and l = 938

Let the number of terms be n.

Then, Tn = 938

⇒ a + (n – 1)d = 988

⇒ 104 + (n – 1) × 13 = 988

⇒ 13n = 897

⇒ n = 69

∴ Required sum = n/2(a + l)

= 69/2[104 + 988]

= 69 × 546

= 37674

Hence, the required sum is 37674.

18. Find the sum of first 100 even number which are divisible by 5.

Solution

The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...

This is an AP in which a = 10, d = (20 – 10) = 10 and n = 100

The sum of n terms of an AP is given by

Sn = n/2[2a + (n – 1)d]

(100/2) × [2 × 10 + (100 – 1) × 10] [∵ a = 10, d = 10 and n =100]

= 50 × [20 + 990]

= 50 × 1010

= 50500

Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.

19. Find the sum of the following:

(1 – 1/n) + (1 – 2/n) + (1 – 3/n) + ...... upto n terms.

Solution

On simplifying the given series, we get:

(1 – 1/n) + (1 – 2/n) + (1 – 3/n) + ..... n terms

= (1 + 1 + 1 + ..... n terms) – (1/n + 2/n + 3/n + ..... + n/n)

= n – (1/n + 2/n + 3/n + ..... + n/n)

Here, (1/n + 2/n + 3/n + ...... + n/n) is an AP whose first term is 1/n and the common difference is (2/n – 1/n)

= 1/n.

The sum of terms of an AP is given by

Sn = n/2[2a + (n – 1)d]

= n – [n/2{2 × (1/n) + (n – 1) × (1/n)}]

= n – [n/2 [(2/n) + (n – 1)/n]]

= n – {n/2(n + 1)/n}

= n – (n + 1)/2

= (n – 1)/2

20. In an AP. It is given that S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms.

Solution

Let a be the first term and d be the common difference of three AP. Then,

S5 + S7 = 167

⇒ 5/2(2a + 4d) + 7/2(2a + 6d) = 167 {Sn = n/2[2a + (n – 1)d]}

⇒ 5a + 10d + 7a + 21d = 167

⇒ 12a + 31d = 167 ...(1)

Also,

Sn = 235

⇒ 10/2(2a + 9d) = 235

⇒ 5(2a + 9d) = 235

⇒ 2a + 9d = 47

Multiplying both sides by 6, we get

12a + 54d = 282 ....(2)

Subtracting (1) from (2), we get

12a + 54d – 12a – 31d = 282 – 167

⇒ 23d = 115

⇒ d = 5

Putting d = 5 in (1), we get

12a + 31 × 5 = 167

⇒ 12a + 155 = 167

⇒ 12a = 167 – 155 = 12

⇒ a = 1

Hence, the AP is 1, 6, 11, 16, ....

21. In an AP, the first term is 2, the last term is 29 and the sum of all terms is 155. Find the common difference.

Solution

Here, a = 2, l = 29 and Sn = 155

Let d be the common difference of the given AP and n be the total number of terms.

Then, Tn = 29

⇒ a + (n – 1)d = 29

⇒ 2 + (n – 1)d = 29 ...(i)

The sum of n terms of an AP is given by

Sn = n/2[a + l] = 155

⇒ n/2[2 + 29] = (n/2) × 31 = 155

⇒ n = 10

Putting the value of n in (i), we get:

⇒ 2 + 9d = 29

⇒ 9d = 27

⇒ d = 3

Thus, the common difference of the given AP is 3.

22. In an AP, the first term is -4, the las term is 29 and the sum of all its terms is 150. Find its common difference.

Solution

Suppose there are n terms in the AP.

Here, a = - 4, l = 29 and Sn = 150

Sn = 150

⇒ n/2 (-4 + 29) = 150 [Sn = n/2(a + l)]

⇒ n = (150 × 2)/25 = 12

Thus, the AP contains 12 terms.

Let d be the common difference of the AP.

∴ a12 = 29

⇒ -4 + (12 – 1) × d = 29 [an = a + (n – 1)d]

⇒ 11d = 29 + 4 = 33

⇒ d = 3

Hence, the common difference of the AP is 3.

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

Solution

Suppose there are n terms in the AP.

Here, a = 17, d = 9 and l = 350.

∴ an = 350

⇒ 17 + (n – 1) × 9 = 350 [an = a + (n – 1)d]

⇒ 9n + 8 = 350

⇒ 9n = 350 – 8 = 342

⇒ n = 38

Thus, there are 38 terms in the AP.

∴ S38 = 28/2 (17 + 350) [Sn = n/2(a + l)]

= 19 × 367

= 6973

Hence, the required sum is 6973.

24. The first and last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms.

Solution

Suppose there are n term in the AP.

Here, a = 5, l = 45 and Sn = 400

Sn = 400

⇒ n/2(5 + 45) = 400 [Sn = n/2(a + l)]

⇒ n/2 × 50 = 400

⇒ n = (400 × 2)/50

= 16

Thus, there are 16 terms in the AP.

Let d be the common difference of the AP.

∴ a16 = 45

⇒ 5 + (16 – 1) × d = 45 [an = a + (n – 1)d]

⇒ 15d = 45 – 5 = 40

⇒ d = 40/15 = 8/3

Hence, the common difference of the AP is 8/3.

25. In an AP, the first term is 22, nth terms is -11 and sum of first n terms is 66. Find the n and hence find the 4 common difference.

Solution

Here, a = 22, Tn = -11 and Sn = 66

Let d be the common difference of the given AP.

Then, Tn = 11

⇒ a + (n – 1)d = 22 + (n – 1)d = -11

⇒ (n – 1)d = - 33 ....(i)

The sum of n terms of an AP is given by

Sn = n/2[2a + (n – 1)d] = 66 [Substituting the value of (n – 1)d from (i)]

⇒ n/2[2 × 22 + (-33)]

= (n/2) × 11

= 66

⇒ n = 12

Putting the value of n in (i), we get:

11d = -33

⇒ d = - 3

Thus, n = 12 and d = -3

26. The 12th term of an AP is -13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.

Solution

Let a be the first term and d be the common difference of the AP. Then,

a12 = - 13

⇒ a + 11d = - 13 ...(1) [an = a + (n – 1)d]

Also,

S4 = 24

⇒ 4/2(2a + 3d) = 24 {Sn = n/2[2a + (n – 1)d]}

⇒ 2a + 3d = 12 ...(2)

Solving (1) and (2), we get

2(-13 – 11d) + 3d = 12

⇒ - 26 – 22d + 3d = 12

⇒ - 19d = 12 + 26 = 38

⇒ d = - 2

Putting d = - 2 in (1), we get

a + 11 × (-2) = - 13

⇒ a = - 13 + 22 = 9

∴ Sum oof its first 20 terms, S10

= 10/2[2 × 9 + (10 – 1) × (-2)]

= 5 × (18 – 18)

= 5 × 0

= 0

Hence, the required sum s is 0.

27. The sum of the first 7 terms of an Ap is 182. If its 4th and 17th terms are in the ratio 1:5, find the AP.

Solution

Let a be the first term and d be the common difference of the AP.

∴ S7 = 182

⇒ 7/2(2a + 6d) = 182 {Sn = n/2[2a + (n – 1))d]}

⇒ a + 3d = 26 ....(1)

Also,

a4 : a17 = 1 : 5 (Given)

⇒ (a + 3d)/(a + 16d) = 1/5 [an = a + (n – 1)d]

⇒ 5a + 155d = a + 16d

⇒ d = 4a ....(2)

Solving (1) and (2), we get

a + 3 × 4a = 26

⇒ 13a = 26

⇒ a = 2

Putting a = 2 in in (2), we get

d = 4 × 2 = 8

Hence, the required AP is 2, 10, 18, 26, ......

28. The sum of first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and common difference of the AP.

Solution

Here, a = 4, d =7 and l = 81

Let the nth term be 81.

Then Tn = 81

⇒ a + (n – 1)d = 4(n – 1)7 = 81

⇒ (n – 1)7 = 77

⇒ (n – 1) = 11

⇒ n = 12

Thus, there are 12 terms in AP.

The sum of n terms of an AP is given by

Sn = n/2[a + l]

∴ S12 = 12/2 [4 + 81]

= 6 × 85

= 510

Thus, the required sum is 510.

29. The sum of the first 7 terms of an AP is 49 and the sum of its first 17 term is 289. Find the sum of its first n terms.

Solution

Let a be the first term and d be the common difference of the given AP.

Then, we have:

Sn = n/2[2a + (n – 1)d]

S7 = 7/2 [2a + 6d] = 7[a + 3d]

S17 = 17/2 [2a + 16d] = 17[a + 8d]

However, S7 = 49 and S17 = 289

Now, 7(a + 3d) = 49

⇒ a + 3d = 7 ...(i)

Also, 17[a + 8d] = 289

⇒ a + 8d = 17 ...(ii)

Subtracting (i) from (ii), we get:

5d = 10

⇒ d = 2

Putting d = 2 in (i), we get

a + 6 = 7

⇒ a = 1

Thus, a = 1 and d = 2

∴ Sum of n terms of AP = n/2[2 × 1 + (n – 1) × 2]

= n[1 + (n – 1)]

= n2

30. Two Aps have the same common difference. If the first terms of these Aps be 3 and 8 respectively. Find the difference between the sums of their first 50 terms.

Solution

Let a1 and a2 be the first terms of the two APs.

Here, a1 = 8 and a2 = 3

Suppose d be the common difference of the two Aps

Let S50 and S’50 = 50/2[2a1 + (50 – 1)d] – 50/2[2a2 + (50 – 1)d]

= 25(2 × 8 × 49d) – 25(2 × 3 + 49d)

= 25 × (16 – 6)

= 250

Hence, the required difference between the two sums is 250.

31. The sum first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.

Solution

Let a be the first term and d be the common difference of the AP. Then,

S10 = - 150 (Given)

⇒ 10/2 (2a + 9d) = -150 {Sn = n/2[2a + (n – 1)d}

⇒ 5(2a + 9d) = - 150

⇒ 2a + 9d = - 30 ....(1)

It is given that the sum of its next 1o terms is – 550.

Now,

S20 = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = - 150 + (-550)

= -700

∴ S20 = -700

⇒ 20/2(2a + 19d) = - 700

⇒ 10(2a + 19d) = - 700

⇒ 2a + 19d = - 70 ...(2)

Subtracting (1) from (2), we get

(2a + 19d) – (2a + 9d) = - 70 – (-30)

⇒ 10d = - 40

⇒ d = - 4

Putting d = - 4 in (1), we get

2a + 9 × (-4) = - 30

⇒ 2a = - 30 + 36 = 6

⇒ a = 3

Hence, the required AP is 3, -1, -5, -9, ....

32. The 13th terms of an AP is 4 times its 3rd term. If its 5th term is 16. Find the sum of its first 10 terms.

Solution

Let a be the first term and d be the common difference of the AP. Then,

a13 = 4 × a3 (Given)

⇒ a + 12d = 4(2 + 2d) [an = a + (n – 1)d]

⇒ a + 12d = 4a + 8d

⇒ 3a = 4d ...(1)

Also,

a5 = 16 (Given)

⇒ a + 4d = 16 ...(2)

Solving (1) and (2), we get

a + 3a = 16

⇒ 4a = 16

⇒ a = 4

Putting a = 4 in (1), we get

4d = 3 × 4 = 12

⇒ d = 3

Using the formula, S4 = n/2[2a + (n – 1)d], we get

S10 = 10/2[2 × 4 + (10 – 1) × 3]

= 5 × (8 + 27)

= 5 × 35

= 175

Hence, the required sum is 175.

33. The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.

Solution

Let a be the first term and d be the common difference of the AP. Then,

a16 = 5 × a3 (Given)

Let a1 and a2 be the first terms of the two APs.

Here, a1 = 8 and a2 = 3

Suppose d be the common difference of the two Aps

Let S50 and S’50 = 50/2[2a1 + (50 – 1)d] – 50/2[2a2 + (50 – 1)d]

= 25(2 × 8 × 49d) – 25(2 × 3 + 49d)

= 25 × (16 – 6)

= 250

Hence, the required difference between the two sums is 250.

31. The sum first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.

Solution

Let a be the first term and d be the common difference of the AP. Then,

S10 = - 150 (Given)

⇒ 10/2 (2a + 9d) = - 150 {Sn = n/2[2a + (n – 1)d}]

⇒ 5(2a + 9d) = - 150

⇒ 2a + 9d = - 30 ...(1)

It is given that the sum of its next 1o terms is – 550.

Now,

S20 = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = - 150 + (-550)

= -700

∴ S20 = -700

⇒ 20/2(2a + 19d) = - 700

⇒ 10(2a + 19d) = - 700

⇒ 2a + 19d = - 70 ...(2)

Subtracting (1) from (2), we get

(2a + 19d) – (2a + 9d) = - 70 – (-30)

⇒ 10d = - 40

⇒ d = - 4

Putting d = - 4 in (1), we get

2a + 9 × (-4) = -30

⇒ 2a = - 30 + 36 = 6

⇒ a = 3

Hence, the required AP is 3, -1, -5, -9, ....

32. The 13th terms of an AP is 4 times its 3rd term. If its 5th term is 16. Find the sum of its first 10 terms.

Solution

Let a be the first term and d be the common difference of the AP. Then,

a13 = 4 × a3 (Given)

⇒ a + 12d = 4(2 + 2d) [an = a + (n – 1)d]

⇒ a + 12d = 4a + 8d

⇒ 3a = 4d ...(1)

Also,

a5 = 16 (Given)

⇒ a + 4d = 16 ...(2)

Solving (1) and (2), we get

a + 3a = 16

⇒ 4a = 16

⇒ a = 4

Putting a = 4 in (1), we get

4d = 3 × 4 = 12

⇒ d = 3

Using the formula, S4 = n/2[2a + (n – 1)d], we get

S10 = 10/2[2 × 4 + (10 – 1) × 3]

= 5 × (8 + 27)

= 5 × 35

= 175

Hence, the required sum is 175.

33. The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.

Solution

Let a be the first term and d be the common difference of the AP. Then,

a16 = 5 × a3 (Given)

⇒ a + 15d = 5(a + 2d) [an = a + (n – 1)d]

⇒ a + 15d = 5a + 10d

⇒ 4a = 5d ...(1)

Also,

a10 = 41 (Given)

⇒ a + 9d – 41 ...(2)

Solving (1) and (2), we get

a + 9 × 4a/5 = 41

⇒ (5a + 36a)/5 = 41

⇒ 41a/5 = 41

⇒ a = 5

Putting a = 5 in (1), we get

5d = 4 × 5 = 20

⇒ d = 4

Using the formula, Sn = n/2[2a + (n – 1)d], we get

Sn = 15/2[2 × 5 + (15 – 1) × 4]

= 15/2 × (10 + 56)

= 15/2 × 66

= 495

Hence, the required sum is 495.

34. An AP 5, 12, 19, ..... has 50 term. Find its last term. Hence, find the sum of its last 15 terms.

Solution

The given AP is 5, 12, 19, .....

Here, a = 5, d = 12 – 5 = 7 and n = 50.

Since, there are 50 terms in the AP, so the last term of the AP is a50.

l = a50 = 5+ (50 – 1) × 7 [an = a + (n – 1)d]

= 5 + 343

= 348

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 terms of the AP.

= S50 – S35

= 50/2[2 × 5 + (50 – 1) × 7] – 35/2[2 × 5 + (35 – 1) × 7]

{Sn = n/2[2a + (n – 1)d]}

= 50/2 × (10 + 343) – 35/2 × (10 + 238)

= 50/2 × 353 – 35/2 × 248

= (17650 – 8680)/2

= 8970/2

= 4485

Hence, the require sum is 4485.

35. An AP 8, 10, 12, ...has 60 terms. Find its last term. Hence, find the sum of its terms.

Solution

The given AP is 8, 10, 12, ......

Here, a = 8, d = 10 – 8 = 2 and n = 60

Since, there are 60 terms in the AP, so the last term of the AP is a60.

l = a60 = 8 + (60 – 1)×2 [an = a + (n – 1)d]

= 8 + 118

= 126

Thus, the last term of the AP is 126.

Now,

Sum of the last 10 terms of AP

= S60 – S50

= 60/2[2 × 8 + (60 – 1) × 2] – 50/2[2 × 8 + (50 – 1) × 2]

{Sn = n/2[2a + (n – 1)d]}

= 30 × (16 + 118) – 25 × (16 + 98)

= 30 × 134 – 25 × 114

= 4020 – 2850

= 1170

Hence, the required sum is 1170.

36. The sum of the 4th and 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44.

Find the sum of its first terms.

Solution

Let a be the first and d be the common difference of the AP.

∴ a4 + a8 = 24 (Given)

⇒ (a + 3d) + (a + 7d) = 24 [an = a + (n – 1)d]

⇒ 2a + 10d = 24

⇒ a + 5d = 12 ...(1)

Also,

∴ a6 + a10 = 44 (Given)

⇒ (a + 5d) + (a + 9d) = 44 [an = a + (n – 1)d]

⇒ 2a + 14d = 44

⇒ a + 7d = 22 ...(2)

Subtracting (1) from (2), we get

(a + 7d) – (a + 5d) = 22 – 12

⇒ 2d = 10

⇒ d = 5

Putting d = 5 in (1), we get

a + 5 × 5 = 12

⇒ a = 12 – 25 = -13

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S10 = 10/2[2 × (-13) + (10 – 1) × 5]

= 5 × (-26 + 45)

= 5 × 19

= 95

Hence, the required sum is 95.

37. The sum of first m terms of an AP is (4m2 – m). If its nth term is 107, find the value of n. Also, find the 21st term of this AP.

Solution

Let Sm denotes the sum of the first m terms of the AP. Then,

Sm = 4m2 – m

⇒ Sm-1 = 4(m – 1)2 – (m – 1)

= 4(m2 – 2m + 1) – (m – 1)

= 4m2 – 9m + 5

Suppose am denote the mth term of the AP.

∴ am = Sm – Sm-1

= (4m2 – m) – (4m2 – 9m + 5)

= 8m – 5 ...(1)

Now,

an = 107 (Given)

⇒ 8n – 5 = 107 [From (1)]

⇒ 8n = 107 + 5 = 112

⇒ n = 14

Thus, the value of n is 14.

Putting m = 21 in (1), we get

a21 = 8 × 21 – 5 = 168 – 5 = 163

Hence, the 21st term of the AP is 163.

38. The sum of first q terms of an AP is (63q – 3q2). If its pth term is -60, find the value of p. Also, find the 11th term of its AP.

Solution

Let Sq denote the sum of the first q terms of the AP. Then,

Sq = 63q – 3q2

⇒ Sq-1 = 63(q – 1) – 3(q – 1)2 = 63q – 63 – 3(q2 – 2q + 1)

= - 3q2 + 69q – 66

Suppose aq denote the qth term of the AP.

∴ aq = Sq – Sq-1

= (63q – 3q2) – (-3q2 + 69q – 66)

= -6q + 66 ...(1)

Now,

ap = - 60 (Given)

⇒ - 6p + 66 = -60 [From (1)]

⇒ -6p = -60 – 66 = -126

⇒ p = 21

Thus, the value of p is 21.

Putting q = 11 in (1), we get

a11 = -6×11 + 66 = - 66 + 66 = 0

Hence, the 11th term of the AP is 0.

39. Find the number of terms of the AP -12, -9, -6, ...., 21. If 1 is added to each term of this AP then the sum of all terms of the AP thus obtained.

Solution

The given AP is -12, -9, -6, ....., 21.

Here, a = - 12d, d = -9 – (-12)

= -9 + 12 = 3 and l = 2l

Suppose there are n terms in the AP.

∴ l = an = 21

⇒ - 12 + (n – 1) × 3 = 21 [an = a + (n – 1)d]

⇒ 3n – 15 = 21

⇒ 3n = 21 + 15 = 36

⇒ n= 12

Thus, there are 12 terms in the AP.

If 1 is added to each term of the AP, then the new AP so obtained is -11, -8, -5, ....., 22.

Here, first term, A = - 11; last term, L = 22 and n = 12

∴ Sum of the terms of this AP

= 12/2 (-11 + 22) [Sn = n/2(a + 1)]

= 6 × 11

= 66

Hence, the required sum is 56.

40. Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term.

Solution

Here, a = 10 and n = 14

Now,

S14 = 1505 (Given)

⇒ 14/2 [2 × 10 + (14 – 1) × d] = 1505 {Sn = n/2[2a + (n – 1)d]}

⇒ 7(20 + 13d) = 1505

⇒ 20 + 13d = 215

⇒ 13d = 215 – 20 = 195

⇒ d = 15

∴ 25th term of the AP, a25

= (10 + (25 – 1) × 15 [an = a + (n – 1)d]

= 10 + 360

= 370

Hence, the required term is 370.

41. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution

Let a be the first term and d be the common difference of the AP. Then,

d = a3 – a2 = 18 – 14 = 4

Now,

a2 = 14 (Given)

⇒ a + d = 14 [an = a + (n – 1)d]

⇒ a + 4 = 14

⇒ a = 14 – 4 = 10

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S51 = 51/2[2 × 10 + (51 – 1) × 4]

= 51/2(20 + 200)

= 51/2 × 220

= 5610

Hence, the required sum is 5610.

42. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class how two section, find how many trees were planted by student. Which value is shown in the question?

Solution

Number of trees planted by the students of each section of class 1 = 2

There are two section of class 1.

∴ Number of trees planted by the students of class l = 2 × 2 = 4

Number of trees planted by the students of each section of class 2 = 4

There are two sections of class 2.

∴ Number of trees planted by the students of class 2 = 2 × 4 = 8

Similarly,

Number of trees planted by the students of class 3 = 2 × 6 = 12

So, the number of trees planted by the students of different classes are 4, 8, 12, .....

∴ Total number of trees planted by the students = 4 + 8 + 12 + ...... up to 12 terms

This series is an arithmetic series.

Here, a = 4, d = 8 – 4 = 4 and n = 12

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S12 = 12/2[2 × 4 + (12 – 1) × 4]

= 6 × (8 + 44)

= 6 × 52

= 312

Hence, the total number of trees planted by the students is 312. The values shown in the question are social responsibility and awareness for conserving nature.

43. In a potato race, a bucket is placed at the starting point, which is 5 m from the potato, and the other potatoes are placed 3 m apart in a straight line. There are 10 potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and he continues in the same way until all the potatoes are in the bucket. what is the total distance the competitor has to run?

Solution

Distance covered by the competitor to pick and drop the first potato = 2 × 5m

= 10 m

Distance covered by the competitor to pick and drop the second potato = 2 × (5 + 3)m = 2 × 8 m = 16 m

Distance covered by the competitor to pick and drop the third potato = 2 × (5 + 3 + 3)m

= 2 × 11m

= 22m and so on.

∴ Total distance covered by the competitor = 10m + 16m + 22m + ....... upto 10 terms

This is an arithmetic series.

Here, a = 10, d = 16 – 10 = 6 and n = 10

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S10 = 10/2[2 × 10 + (10 – 1) × 6]

= 5 × (20 + 54)

= 5 × 74

= 370

Hence, the total distance the competitor has to run is 370 m.

44. There ere 25 trees at equal distance of 5 m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m. A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next. Find the total distance covered by the gardener in order to water all the trees.

Solution

Distance covered by the gardener to water the first tress and return to the water tank = 10 m + 10 m

= 20 m

Distance covered by the gardener to water the second tree and return to the water tank = 15m + 15m = 30 m

Distance covered by the gardener to water the third tree and return to the water tank = 20 m + 20m

= 40 m and so on.

∴ Total distance covered by the gardener to water all the trees = 20m + 30 m + 40 m + .... up to 25 terms.

This series is an arithmetic series.

Here, a = 20, d = 30 – 20 = 10 and n = 25

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S25 = 25/2[2 × 20 + (25 – 1) × 10]

= 25/2(40 + 240)

= 25/2

= 280

= 3500

Hence, the total distance covered by the gardener to water all the trees 3500 m.

45. A sum of 700 is to be give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each prize.

Solution

Let the value of the first prize be a.

Since the value of each prize is 20 less than its preceding prize, so the values of the prizes are in AP with common difference = ₹20.

⇒ 40/2[2a + (40 – 1)d] = 36000

∴ d = -₹

⇒ 20(2a + 39d) = 36000

⇒ 2a + 39d = 1800 ......(2)

Number of cash prizes to be given to the students, n = 7

Total sum of the prizes, S7 = ₹700

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S7 = 7/2[2a + (7 – 1) × (-20)] = 700

⇒ 7/2(2a – 120) = 700

⇒ 7a – 420 = 700

⇒ 7a = 700 + 420 = 1120

⇒ a = 160

Thus, the value of the first prize is ₹160.

Hence, the value of each prize is ₹160, ₹140, ₹120, ₹100, ₹80, ₹ 60 and ₹40.

46. A man saved ₹33000 in 10 months. In each month after the first, he saved ₹ 100 more than he did in the preceding month. How much did he save in the first month?

Solution

Let the money saved by the man in the first month be ₹a

It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹ 100.

∴ d = ₹100

Number of months, n = 10

Sum of money saved in 10 months, S10 = ₹ 330,000

Using the formula, Sn = n/2[2a + (n – 1)d], we get

S10 = 10/2[2a + (10 – 1) × 100] = 33000

⇒ 5(2a + 900) = 33000

⇒ 2a + 900 = 6600

⇒ 2a = 6600 – 900 = 5700

⇒ a = 2850

Hence, the money saved by the man in the first month is ₹2,850.

47. A man arranges to pay off debt of ₹ 36000 by 40 monthly instalments which form an arithmetic series. When 30 of the installment are paid, he dies leaving on-third of the debt unpaid. Find the value of the first instalment.

Solution

Let the value of the first installment be ₹a.

Since the monthly installments form an arithmetic series, so let us suppose the man increases the value of each installment be ₹ d every month.

∴ Common difference of the arithmetic series = ₹d

Amount paid in 30 installments = ₹36,000 – 1/3 × ₹ 36,000

= ₹36,000 - ₹12,000

= ₹ 24,000

Let Sn denote the total amount of money paid in the n installments. Then,

S30 = 24,000

⇒ 30/2 [2a + (30 – 1)d] = 24000 {Sn = n/2[2a + (n – 1)d]}

⇒ 15(2a + 29d) = 24000

⇒ 2a + 29d = 1600 ....(1)

Also,

S40 = ₹36,000

⇒ 40/2[2a + (40 – 1)d] = 36000

⇒ 20(2a + 39d) = 36000

⇒ 2a + 39d = 1800 ...(2)

Subtracting (1) from (2), we get

(2a + 39d) – (2a + 29d) = 1800 – 1600

⇒ 10d = 200

⇒ d = 20

Putting d = 20 in (1), we get

2a + 29 × 20 = 1600

⇒ 2a + 580 = 1600

⇒ 2a = 1600 – 580 = 1020

⇒ a = 150

Thus, the value of the first installment is ₹510.

48. A contact on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 just for the first day, ₹250 for the second day, ₹300 for the third day, etc. the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution

It is given that the penalty for each succeeding day is 50 more than for the preceding day, so the amount of penalties are in AP with common difference ₹50.

Number of days in the delay of the work = 30

The amount of penalties are ₹200, ₹ 250, ₹300, .... upto 30 terms.

∴ Total amount of money paid by the contractor as penalty,

S30 = ₹ 200 + ₹250 + ₹ 300 + .... upto 30 terms

Here. a = ₹ 200, d = ₹ 50 and n = 30

Using the formula, Sn = n/2[2a + (n – 1)d)], we get

S30 = 30/2[2 × 200 + (30 – 1) × 50]

= 15(400 – 1450)

= 15 × 1850

= 27750

Hence, the contractor has to pay ₹27,750 as penalty.