CBSE NCERT Solutions for Class 10 Mathematics Chapter 5 · 2020-05-25 · CBSE NCERT Solutions for...

Class- X-CBSE-Mathematics Arithmetic Progressions

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CBSE NCERT Solutions for Class 10 Mathematics Chapter 5 Back of Chapter Questions

1. In which of the following situations, does the list of numbers involved make anarithmetic progression, and why?

(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes 14 of

the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.

(iv) The amount of money in the account every year, when ₹ 10000 isdeposited at compound interest at 8 % per annum.

Solution:

(i) From the given data we see that

Taxi fare for 1st km = 15

Taxi fare for first 2 kms = 15 + 8 = 23



Clearly 15, 23, 31, 39 … … forms an AP since every term is 8 more than the preceding term.

i.e., (23 − 15) = (31 − 23) = (39 − 31) = 8

(ii) Let the initial volume of air in a cylinder be 𝑉𝑉. In each stroke, the vacuumpump removes 1

4 of air remaining in the cylinder at a time. In other words,

after every stroke, only 1 − 14

= 34 the part of air will remain.

Air left in the cylinder after first removal = �34� 𝑉𝑉

Air again removed = 14�34�𝑉𝑉

Air left in the cylinder after second removal = �34� 𝑉𝑉 − 1

4�34� 𝑉𝑉 = �3

4�2𝑉𝑉

Air again removed =14�34�2𝑉𝑉

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Air left in the cylinder after third removal = �34�2𝑉𝑉 − 1

4�34�2𝑉𝑉 = �3

4�3𝑉𝑉

Therefore, volume of air left in cylinder follows the sequence:

𝑉𝑉, �34� 𝑉𝑉, �3

4�2𝑉𝑉, �3

4�3𝑉𝑉…

Clearly, the consecutive terms of this sequence do not have constant different between them, Thus, this is not an AP

(iii) Cost of digging for first metre = ₹ 150

Cost of digging for 2m metre = 150 + 50 = ₹ 200



Clearly, 150, 200, 250, 300 …. forms an AP because every term is 50 more than its preceding term.

(iv) We know that if Principal amount (P) of ₹ 10000 is deposited at rate (r) at 8 % compound interest per annum for 𝑛𝑛 years, our money (in ₹) will be

10000 �1 +8

100� , 10000 �1 +

8100

�2

, 10000 �1 +8

100�3

,

10000 �1 +8

100�4

, . ..

Clearly, adjacent terms of this sequence, do not have the same difference between them. Therefore, this is not an AP.

2. Write first four terms of the AP, when the first term 𝑎𝑎 and the common difference 𝑑𝑑 are given as follows:

(i) 𝑎𝑎 = 10,𝑑𝑑 = 10

(ii) 𝑎𝑎 = −2, 𝑑𝑑 = 0

(iii) 𝑎𝑎 = 4,𝑑𝑑 = −3

(iv) 𝑎𝑎 = −1, 𝑑𝑑 = 12

(v) 𝑎𝑎 = −1.25,𝑑𝑑 = −0.25

Solution:

(i) Given: 𝑎𝑎 = 10, 𝑑𝑑 = 10

First term = 𝑎𝑎1 = 𝑎𝑎 = 10

Second term = 𝑎𝑎2 = 𝑎𝑎1 + 𝑑𝑑 = 10 + 10 = 20





Third term = 𝑎𝑎3 = 𝑎𝑎2 + 𝑑𝑑 = 20 + 10 = 30

Fourth term = 𝑎𝑎4 = 𝑎𝑎3 + 𝑑𝑑 = 30 + 10 = 40

Hence, the first four terms of this AP are 10, 20, 30, and 40.

(ii) Given: 𝑎𝑎 = −2,𝑑𝑑 = 0

First term = 𝑎𝑎1 = 𝑎𝑎 = −2

Second term = 𝑎𝑎2 = 𝑎𝑎1 + 𝑑𝑑 = −2 + 0 = −2

Third term = 𝑎𝑎3 = 𝑎𝑎2 + 𝑑𝑑 = −2 + 0 = −2

Fourth term = 𝑎𝑎4 = 𝑎𝑎3 + 𝑑𝑑 = −2 + 0 = −2

Hence, the first four terms of this AP will be −2,−2,−2, and −2.

(iii) Given: 𝑎𝑎 = 4,𝑑𝑑 = −3

First term = 𝑎𝑎1 = 𝑎𝑎 = 4

Second term = 𝑎𝑎2 = 𝑎𝑎1 + 𝑑𝑑 = 4 − 3 = 1

Third term = 𝑎𝑎3 = 𝑎𝑎2 + 𝑑𝑑 = 1 − 3 = −2

Fourth term = 𝑎𝑎4 = 𝑎𝑎3 + 𝑑𝑑 = −2 − 3 = −5

Hence, the first four terms of this AP will be 4, 1,−2, and −5.

(iv) Given: 𝑎𝑎 = −1,𝑑𝑑 = 12

First term = 𝑎𝑎1 = 𝑎𝑎 = −1

Second term = 𝑎𝑎2 = 𝑎𝑎1 + 𝑑𝑑 = −1 + 12

= −12

Third term = 𝑎𝑎3 = 𝑎𝑎2 + 𝑑𝑑 = −12

+ 12

= 0

Fourth term = 𝑎𝑎4 = 𝑎𝑎3 + 𝑑𝑑 = 0 + 12

= 12

Hence, the first four terms of this AP will be −1,−12

, 0, and 12.

(v) Given: 𝑎𝑎 = −1.25,𝑑𝑑 = −0.25

First term = 𝑎𝑎1 = 𝑎𝑎 = −1.25

Second term = 𝑎𝑎2 = 𝑎𝑎1 + 𝑑𝑑 = −1.25 − 0.25 = −1.50

Third term = 𝑎𝑎3 = 𝑎𝑎2 + 𝑑𝑑 = −1.50 − 0.25 = −1.75

Fourth term = 𝑎𝑎4 = 𝑎𝑎3 + 𝑑𝑑 = −1.75 − 0.25 = −2.00

Hence, the first four terms of this AP will be −1.25,−1.50,−1.75, and −2.00.





3. For the following APs, write the first term and the common difference:

(i) 3, 1,−1,−3, …

(ii) −5,−1, 3, 7, …

(iii) 13

, 53

, 93

, 133

, …

(iv) 0.6, 1.7, 2.8, 3.9, …

Solution:

(i) 3, 1,−1,−3 …

Here, first term, 𝑎𝑎 = 3

Common difference, 𝑑𝑑 = Second term−First term

= 1 − 3 = −2

Thus, 𝑎𝑎 = 3 𝑎𝑎𝑛𝑛𝑑𝑑 𝑑𝑑 = −2

(ii) −5,−1, 3, 7 …

Here, first term, 𝑎𝑎 = −5


= (−1) − (−5) = −1 + 5 = 4

Thus, 𝑎𝑎 = −5 𝑎𝑎𝑛𝑛𝑑𝑑 𝑑𝑑 = 4

(iii) 13

, 53

, 93

, 133

, …

Here, first term, 𝑎𝑎 = 13


=53−

13

=43

Thus, 𝑎𝑎 = 13

𝑎𝑎𝑛𝑛𝑑𝑑 𝑑𝑑 = 43

(iv) 0.6, 1.7, 2.8, 3.9 …

Here, first term, 𝑎𝑎 = 0.6


= 1.7 − 0.6

= 1.1

Thus, 𝑎𝑎 = 0.6 𝑎𝑎𝑛𝑛𝑑𝑑 𝑑𝑑 = 1.1





4. Which of the following are APs ? If they form an AP, find the common difference 𝑑𝑑 and write three more terms.

(i) 2, 4, 8, 16, …

(ii) 2, 52

, 3, 72

, …

(iii) −1.2,−3.2,−5.2,−7.2, . ..

(iv) −10,−6,−2, 2, . . .

(v) 3, 3 + √2 , 3 + 2√2 , 3 + 3√2 , …

(vi) 0.2, 0.22, 0.222, 0.2222, . . .

(vii) 0,−4,−8,−12, . ..

(viii) −12

,−12

,−12

,−12

, …

(ix) 1, 3, 9, 27, . ..

(x) 𝑎𝑎, 2𝑎𝑎, 3𝑎𝑎, 4𝑎𝑎, . . .

(xi) 𝑎𝑎,𝑎𝑎2,𝑎𝑎3,𝑎𝑎4, . ..

(xii) √2,√8,√18 ,√32, . . .

(xiii) √3,√6,√9 ,√12 , . ..

(xiv) 12, 32, 52, 72, . ..

(xv) 12, 52, 72, 73, . . .

Solution:

(i) 2, 4, 8, 16 …

Here,

𝑎𝑎2 − 𝑎𝑎1 = 4 − 2 = 2

𝑎𝑎3 − 𝑎𝑎2 = 8 − 4 = 4

Since 2 ≠ 4

Hence, the given terms are not in AP.

(ii) 2, 52

, 3, 72

. ..

Here,

𝑎𝑎2 − 𝑎𝑎1 =52− 2 =

12





𝑎𝑎3 − 𝑎𝑎2 = 3 −52

=12

𝑎𝑎4 − 𝑎𝑎3 =72− 3 =

12

𝑖𝑖. 𝑒𝑒. , difference is the same every time.

Hence, 𝑑𝑑 = 12 and the given terms are in AP.

The next three more terms are

𝑎𝑎5 =72

+12

= 4

𝑎𝑎6 = 4 +12

=92

𝑎𝑎7 =92

+12

= 5

(iii) −1.2,−3.2,−5.2,−7.2 …

Here,

𝑎𝑎2 − 𝑎𝑎1 = (−3.2) − (−1.2) = −2

𝑎𝑎3 − 𝑎𝑎2 = (−5.2) − (−3.2) = −2

𝑎𝑎4 − 𝑎𝑎3 = (−7.2) − (−5.2) = −2


Hence, 𝑑𝑑 = −2 and the given terms are in AP.


𝑎𝑎5 = −7.2 − 2 = −9.2

𝑎𝑎6 = −9.2 − 2 = −11.2

𝑎𝑎7 = −11.2 − 2 = −13.2

(iv) −10,−6,−2, 2 …

Here,

𝑎𝑎2 − 𝑎𝑎1 = (−6) − (−10) = 4

𝑎𝑎3 − 𝑎𝑎2 = (−2) − (−6) = 4

𝑎𝑎4 − 𝑎𝑎3 = (2) − (−2) = 4








𝑎𝑎5 = 2 + 4 = 6

𝑎𝑎6 = 6 + 4 = 10

𝑎𝑎7 = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 …

Here,

𝑎𝑎2 − 𝑎𝑎1 = 3 + √2 − 3 = √2

𝑎𝑎3 − 𝑎𝑎2 = 3 + 2√2 − 3 − √2 = √2

𝑎𝑎4 − 𝑎𝑎3 = 3 + 3√2 − 3 − 2√2 = √2


Hence, 𝑑𝑑 = √2 and the given terms are in AP.


𝑎𝑎5 = 3 + 3√2 + √2 = 3 + 4√2

𝑎𝑎6 = 3 + 4√2 + √2 = 3 + 5√2

𝑎𝑎7 = 3 + 5√2 + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222....

Here,

𝑎𝑎2 − 𝑎𝑎1 = 0.22 − 0.2 = 0.02

𝑎𝑎3 − 𝑎𝑎2 = 0.222 − 0.22 = 0.002

𝑎𝑎4 − 𝑎𝑎3 = 0.2222 − 0.222 = 0.0002

Since, 0.02 ≠ 0.002

Hence, the given terms are not in AP

(vii) 0,−4,−8,−12. ..

Here,

𝑎𝑎2 − 𝑎𝑎1 = (−4) − 0 = −4

𝑎𝑎3 − 𝑎𝑎2 = (−8) − (−4) = −4

𝑎𝑎4 − 𝑎𝑎3 = (−12) − (−8) = −4


Hence, 𝑑𝑑 = −4 and the given terms are in AP.






𝑎𝑎5 = −12 − 4 = −16

𝑎𝑎6 = −16 − 4 = −20

𝑎𝑎7 = −20 − 4 = −24

(viii) −12

,  − 12

,−12

,−12

. . ..

Here,

𝑎𝑎2 − 𝑎𝑎1 = �−12� − �−

12� = 0

𝑎𝑎3 − 𝑎𝑎2 = �−12� − �−

12� = 0

𝑎𝑎4 − 𝑎𝑎3 = �−12� − �−

12� = 0




𝑎𝑎5 = −12− 0 = −

12

𝑎𝑎6 = −12− 0 = −

12

𝑎𝑎7 = −12− 0 = −

12

(ix) 1, 3, 9, 27

Here,

𝑎𝑎2 − 𝑎𝑎1 = 3 − 1 = 2

𝑎𝑎3 − 𝑎𝑎2 = 9 − 3 = 6

Since 2 ≠ 6


(x) 𝑎𝑎, 2𝑎𝑎, 3𝑎𝑎, 4𝑎𝑎, ….

Here,

𝑎𝑎2 − 𝑎𝑎1 = 2𝑎𝑎 − 𝑎𝑎 = 𝑎𝑎

𝑎𝑎3 − 𝑎𝑎2 = 3𝑎𝑎 − 2𝑎𝑎 = 𝑎𝑎





𝑎𝑎4 − 𝑎𝑎3 = 4𝑎𝑎 − 3𝑎𝑎 = 𝑎𝑎


Hence, 𝑑𝑑 = 𝑎𝑎 and the given terms are in AP.


𝑎𝑎5 = 4𝑎𝑎 + 𝑎𝑎 = 5𝑎𝑎



(xi) 𝑎𝑎,𝑎𝑎2,𝑎𝑎3,𝑎𝑎4. ..

𝑎𝑎2 − 𝑎𝑎1 = 𝑎𝑎2 − 𝑎𝑎 = 𝑎𝑎(𝑎𝑎 − 1)

𝑎𝑎3 − 𝑎𝑎2 = 𝑎𝑎3 − 𝑎𝑎2 = 𝑎𝑎2(𝑎𝑎 − 1)

𝑎𝑎4 − 𝑎𝑎3 = 𝑎𝑎4 − 𝑎𝑎3 = 𝑎𝑎3(𝑎𝑎 − 1)

𝑖𝑖. 𝑒𝑒. ,𝑎𝑎𝑥𝑥+1 − 𝑎𝑎𝑥𝑥 is not the same every time.


(xii) √2,√8,√18,√32. ..

Here,

𝑎𝑎2 − 𝑎𝑎1 = √8 − √2 = 2√2 − √2 = √2

𝑎𝑎3 − 𝑎𝑎2 = √18 − √8 = 3√2 − 2√2 = √2

𝑎𝑎4 − 𝑎𝑎3 = √32 − √18 = 4√2 − 3√2 = √2


Hence, 𝑑𝑑 = √2 and the given terms are in AP.


𝑎𝑎5 = √32 + √2 = 4√2 + √2 = 5√2 = √50

𝑎𝑎6 = 5√2 + √2 = 6√2 = √72

𝑎𝑎7 = 6√2 + √2 = 7√2 = √98

(xiii) √3,√6,√9,√12. ..

Here,

𝑎𝑎2 − 𝑎𝑎1 = √6 − √3 = √3 × 2 − √3 = √3 �√2 − 1�

𝑎𝑎3 − 𝑎𝑎2 = √9 − √6 = 3 − √6 = √3 �√3 − √2�





𝑎𝑎4 − 𝑎𝑎3 = √12 − √9 = 2√3 − √3 × 3 = √3 �2 − √3�

𝑖𝑖. 𝑒𝑒. ,𝑎𝑎𝑥𝑥+1 − 𝑎𝑎𝑥𝑥 is not the same every time.


(xiv) 12, 32, 52, 72. ..

Or, 1, 9, 25, 49 …

Here,

𝑎𝑎2 − 𝑎𝑎1 = 9 − 1 = 8

𝑎𝑎3 − 𝑎𝑎2 = 25 − 9 = 16

Since, 8 ≠ 16


(xv) 12, 52, 72, 73 . ..

Or, 1, 25, 49, 73 …

Here,

𝑎𝑎2 − 𝑎𝑎1 = 25 − 1 = 24

𝑎𝑎3 − 𝑎𝑎2 = 49 − 25 = 24

𝑎𝑎4 − 𝑎𝑎3 = 73 − 49 = 24




𝑎𝑎5 = 73 + 24 = 97

𝑎𝑎6 = 97 + 24 = 121

𝑎𝑎7 = 121 + 24 = 145

♦ ♦ ♦

EXERCISE 5.2

1. Fill in the blanks in the following table, given that 𝑎𝑎 is the first term, 𝑑𝑑 the common difference and 𝑎𝑎𝑛𝑛 the 𝑛𝑛th term of the AP:

𝑎𝑎 𝑑𝑑 𝑛𝑛 𝑎𝑎𝑛𝑛





(i) 7 3 8 ⋯

(ii) −18 ⋯ 10 0

(iii) ⋯ −3 18 −5

(iv) −18.9 2.5 ⋯ 3.6

(v) 3.5 0 105 ⋯

Solution:

(i) Given: 𝑎𝑎 = 7,𝑑𝑑 = 3,𝑛𝑛 = 8,𝑎𝑎𝑛𝑛 =?

We Know that,

For an AP ∶ 𝑎𝑎𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑

Substituting the values,

= 7 + (8 − 1) × 3

= 7 + (7) × 3

= 7 + 21 = 28

Thus, 𝑎𝑎𝑛𝑛 = 28

(ii) Given:

𝑎𝑎 = −18,𝑛𝑛 = 10,𝑎𝑎𝑛𝑛 = 0,𝑑𝑑 =?

We Know that,

𝑎𝑎𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑


⇒ 0 = −18 + (10 − 1) × 𝑑𝑑

⇒ 18 = 9 × 𝑑𝑑

⇒ 𝑑𝑑 =189

= 2

Thus, common difference, 𝑑𝑑 = 2

(iii) Given:

𝑑𝑑 = −3, 𝑛𝑛 = 18,𝑎𝑎𝑛𝑛 = −5,𝑎𝑎 =?

We Know that,







⇒ −5 = 𝑎𝑎 + (18 − 1) × (−3)

⇒ −5 = 𝑎𝑎 + (17) × (−3)

⇒ −5 = 𝑎𝑎 − 51

⇒ 𝑎𝑎 = 51 − 5 = 46

Thus, 𝑎𝑎 = 46

(iv) Given: 𝑎𝑎 = −18.9,𝑑𝑑 = 2.5,𝑎𝑎𝑛𝑛 = 3.6,𝑛𝑛 =?

We Know that,



⇒ 3.6 = −18.9 + (𝑛𝑛 − 1) × 2.5

⇒ 3.6 + 18.9 = (𝑛𝑛 − 1) × 2.5

⇒ 22.5 = (𝑛𝑛 − 1) × 2.5

⇒ (𝑛𝑛 − 1) =22.52.5

⇒ 𝑛𝑛 − 1 = 9

⇒ 𝑛𝑛 = 10

Thus, 𝑛𝑛 = 10

(v) Given: 𝑎𝑎 = 3.5,𝑑𝑑 = 0,𝑛𝑛 = 105, 𝑎𝑎𝑛𝑛 =?

We Know that,


⇒ 𝑎𝑎𝑛𝑛 = 3.5 + (105 − 1)0

⇒ 𝑎𝑎𝑛𝑛 = 3.5 + 104 × 0

⇒ 𝑎𝑎𝑛𝑛 = 3.5

Thus, 𝑎𝑎𝑛𝑛 = 3.5

2. Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, . . ., is

(A) 97

(B) 77





(C) – 77

(D) – 87

(ii) 11th term of the AP: −3,−12

, 2, … is

(A) 28

(B) 22

(C) −38

(D) −48 12

Solution:

(i) Given:

AP: 10, 7, 4 …

First term, 𝑎𝑎 = 10

Common difference, 𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 7 − 10 = −3

We know that, 𝑎𝑎𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑

Substituting 𝑛𝑛 = 30,𝑎𝑎 = 10 𝑎𝑎𝑛𝑛𝑑𝑑 𝑑𝑑 = −3, we get

𝑎𝑎30 = 10 + (30 − 1) × (−3)

⇒ 𝑎𝑎30 = 10 + (29) × (−3)

⇒ 𝑎𝑎30 = 10 − 87 = −77

Therefore, the correct answer is C.

(ii) Given:

AP: −3,−12

, 2 …

First term, 𝑎𝑎 = −3

Common difference, 𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1

= −12− (−3)

= −12

+ 3 =52


Substituting 𝑛𝑛 = 11,𝑎𝑎 = −3 𝑎𝑎𝑛𝑛𝑑𝑑 𝑑𝑑 = 52, we get





𝑎𝑎11 = −3 + (11 − 1) × �52�

⇒ 𝑎𝑎11 = −3 + (10) × �52�

⇒ 𝑎𝑎11 = −3 + 25

⇒ 𝑎𝑎11 = 22

Therefore, the correct answer is B.

3. In the following APs, find the missing terms in the boxes:

(i) 2, , 26

(ii) , 13, , 3

(iii) 5, , , 9 12

(iv) – 4, , , , , 6

(v) , 38, , , , – 22

Solution:

(i) 2, , 26

For the given AP,

𝑎𝑎 = 2, 𝑎𝑎3 = 26



𝑎𝑎3 = 2 + (3 − 1)𝑑𝑑

⇒ 26 = 2 + 2𝑑𝑑

⇒ 24 = 2𝑑𝑑

⇒ 𝑑𝑑 = 12

𝑎𝑎2 = 2 + (2 − 1)12 = 14

Hence, the missing term is 14.

(ii) , 13, , 3

For the given AP,

𝑎𝑎2 = 13 and 𝑎𝑎4 = 3







𝑎𝑎2 = 𝑎𝑎 + (2 − 1)𝑑𝑑

⇒ 13 = 𝑎𝑎 + 𝑑𝑑 … (i)

𝑎𝑎4 = 𝑎𝑎 + (4 − 1)𝑑𝑑

⇒ 3 = 𝑎𝑎 + 3𝑑𝑑 … (ii)

On subtracting (ii) from (i), we obtain

13 − 3 = (−2)𝑑𝑑

⇒ 10 = −2𝑑𝑑

⇒ 𝑑𝑑 = −5

Substituting the value of 𝑑𝑑 in (i),

𝑎𝑎 = 18

𝑎𝑎3 = 18 + (3 − 1)(−5)

= 18 + 2(−5) = 18 − 10 = 8

Hence, the missing terms are 18 and 8 respectively.

(iii) 5, , , 9 12

For the given AP,

𝑎𝑎 = 5,𝑎𝑎4 = 912

=192

We know that,



𝑎𝑎4 = 𝑎𝑎 + (4 − 1)𝑑𝑑

⇒192

= 5 + 3𝑑𝑑

⇒192− 5 = 3𝑑𝑑

⇒92

= 3𝑑𝑑

⇒ 𝑑𝑑 =32





𝑎𝑎2 = 𝑎𝑎 + 𝑑𝑑 = 5 +32

=132

𝑎𝑎3 = 𝑎𝑎 + 2𝑑𝑑 = 5 + 2 �32� = 8

Hence, the missing terms are 132

and 8 respectively.

(iv) – 4, , , , , 6

For the given AP,

𝑎𝑎 = −4 and 𝑎𝑎6 = 6

We know that,



𝑎𝑎6 = 𝑎𝑎 + (6 − 1)𝑑𝑑

⇒ 6 = −4 + 5𝑑𝑑

⇒ 10 = 5𝑑𝑑

⇒ 𝑑𝑑 = 2.

𝑎𝑎2 = 𝑎𝑎 + 𝑑𝑑 = −4 + 2 = −2

𝑎𝑎3 = 𝑎𝑎 + 2𝑑𝑑 = −4 + 2(2) = 0

𝑎𝑎4 = 𝑎𝑎 + 3𝑑𝑑 = −4 + 3(2) = 2

𝑎𝑎5 = 𝑎𝑎 + 4𝑑𝑑 = −4 + 4(2) = 4

Therefore, the missing terms are −2, 0, 2 and 4 respectively.

(v) , 38, , , , -22

For the given AP,

𝑎𝑎2 = 38

𝑎𝑎6 = −22

⇒ 38 = 𝑎𝑎 + 𝑑𝑑 … (1)

𝑎𝑎6 = 𝑎𝑎 + (6 − 1)𝑑𝑑

⇒ −22 = 𝑎𝑎 + 5𝑑𝑑 … (2)

On subtracting equation (1) from (2), we obtain

−22 − 38 = 4𝑑𝑑





⇒ −60 = 4𝑑𝑑

⇒ 𝑑𝑑 = −15

𝑎𝑎 = 𝑎𝑎2 − 𝑑𝑑 = 38 − (−15) [From (1)]

⇒ 𝑎𝑎 = 53

𝑎𝑎3 = 𝑎𝑎 + 2𝑑𝑑 = 53 + 2(−15) = 23

𝑎𝑎4 = 𝑎𝑎 + 3𝑑𝑑 = 53 + 3(−15) = 8

𝑎𝑎5 = 𝑎𝑎 + 4𝑑𝑑 = 53 + 4(−15) = −7

Therefore, the missing terms are 53, 23, 8 and −7 respectively.

4. Which term of the AP : 3, 8, 13, 18, . . ., is 78?

Solution:

3, 8, 13, 18, . ..

For the given AP,

𝑎𝑎 = 3

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 8 − 3 = 5

Let 𝑛𝑛thterm of this AP be 78.



78 = 3 + (𝑛𝑛 − 1)5

⇒ 75 = (𝑛𝑛 − 1)5

⇒ (𝑛𝑛 − 1) = 15

⇒ 𝑛𝑛 = 16

Therefore, 16th term of the given AP is 78.

5. Find the number of terms in each of the following APs:

(i) 7, 13, 19, . . . , 205

(ii) 18, 15 12

, 13, … ,−47

Solution:

(i) 7, 13, 19, . . . , 205





For the given AP,

𝑎𝑎 = 7

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 13 − 7 = 6

Let there be 𝑛𝑛 number of terms in this AP

𝑎𝑎𝑛𝑛 = 205


So, 205 = 7 + (𝑛𝑛 − 1)6

⇒ 198 = (𝑛𝑛 − 1)6

⇒ 33 = (𝑛𝑛 − 1)

⇒ 𝑛𝑛 = 34

Hence, this given AP has 34 terms in it.

(ii) 18, 15 12

, 13, … ,−47

For the given AP,

𝑎𝑎 = 18

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 1512− 18

𝑑𝑑 =31 − 36

2= −

52

Let there be 𝑛𝑛 number of terms in this AP

𝑎𝑎𝑛𝑛 = −47


So, −47 = 18 + (𝑛𝑛 − 1) �− 52�

⇒ −47 − 18 = (𝑛𝑛 − 1) �−52�

⇒ −65 = (𝑛𝑛 − 1) �−52�

⇒ (𝑛𝑛 − 1) =−130−5

⇒ (𝑛𝑛 − 1) = 26

⇒ 𝑛𝑛 = 27

Hence, this given AP has 27 terms in it.





6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . ..

Solution:

For the given AP,

𝑎𝑎 = 11

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 8 − 11 = −3

Let −150 be the 𝑛𝑛th term of this AP

We know that,


⇒ −150 = 11 + (𝑛𝑛 − 1)(−3)

⇒ −150 = 11 − 3𝑛𝑛 + 3

⇒ −164 = −3n

⇒ 𝑛𝑛 =164

3

Clearly, 𝑛𝑛 is not an integer.

Hence, −150 is not a term of this given AP.

Note: n is the number of terms in a sequence and can take only integral value.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

According to the question, it is given that,

𝑎𝑎11 = 38

𝑎𝑎16 = 73

We know that,



𝑎𝑎11 = 𝑎𝑎 + (11 − 1)𝑑𝑑

⇒ 38 = 𝑎𝑎 + 10𝑑𝑑 … (1)

Similarly,






𝑎𝑎16 = 𝑎𝑎 + (16 − 1)𝑑𝑑

⇒ 73 = 𝑎𝑎 + 15𝑑𝑑 … (2)

On subtracting (1) from (2), we obtain

35 = 5𝑑𝑑

⇒ 𝑑𝑑 = 7

From equation (1),

38 = 𝑎𝑎 + 10 × (7)

⇒ 38 − 70 = 𝑎𝑎

⇒ 𝑎𝑎 = −32

𝑎𝑎31 = 𝑎𝑎 + (31 − 1)𝑑𝑑

= −32 + 30 × (7)

= −32 + 210

= 178

Therefore, the 31st term in given AP is 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

According to the question it is given that,

a3 = 12

a50 = 106

We know that,

an = a + (n − 1)d


a3 = a + (3 − 1)d

⇒ 12 = a + 2d …(I)

Similarly, a50 = a + (50 − 1)d

⇒ 106 = a + 49d … (II)

On subtracting (I) from (II), we get





94 = 47d

⇒ d = 2

From equation (I), we get

12 = a + 2(2)

⇒ a = 12 − 4 = 8

a29 = a + (29 − 1)d

⇒ a29 = 8 + (28)2

⇒ a29 = 8 + 56 = 64

Hence, the 29th term in given AP is 64.

9. If the 3rd and the 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero?

Solution:

As per the question, it is given that,

a3 = 4

a9 = −8

We know that,

an = a(n − 1)d


a3 = a + (3 − 1)d

⇒ 4 = a + 2d … (I)

a9 = a + (9 − 1)d

⇒ −8 = a + 8d … (II)

On subtracting equation (I) from (II), we get

−12 = 6d

⇒ d = −2

From equation (I), we get

4 = a + 2(−2)

⇒ 4 = a − 4





⇒ a = 8

Let nth term of the given AP be zero,

an = a + (n − 1)d


0 = 8 + (n − 1)(−2)

⇒ 0 = 8 − 2n + 2

⇒ 2n = 10

⇒ n = 5

Therefore, the 5th term of the given AP is 0.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solutions:

For the given AP, we know that: an = a + (n − 1)d

a17 = a + (17 − 1)d

⇒ a17 = a + 16d

Similarly,a10 = a + 9d

But according to the question, it is given that

a17 − a10 = 7

⇒ (a + 16d) − (a + 9d) = 7

⇒ 7𝑑𝑑 = 7

⇒ d = 1

Hence, the common difference is 1.

11. Which term of the AP: 3, 15, 27, 39, . .. will be 132 more than its 54th term?

Solution:

Given: AP is 3,15,27,39, . . ..

a = 3

d = a2 − a1 = 15 − 3 = 12

a54 = a + (54 − 1)d





= 3 + (53)(12)

= 3 + 636 = 639

Required term = 132 + 639 = 771 (Given that term to be found is 132 more than 𝑎𝑎54)

Now, we need to determine the term of the given AP i.e., 771.

Let nth term be 771.

an = a + (n − 1)d


771 = 3 + (n − 1)12

⇒ 768 = (n − 1)12

⇒ (n − 1) = 64

⇒ n = 65

Hence, the 65th term will be 132 more than 54th term.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

Let the first term of the given two APs be a1 and a2 respectively and the common difference of these APs be d.

For first AP,

a100 = a1 + (100 − 1)d

= a1 + 99d

a1000 = a1 + (1000 − 1)d

= a1000 = a1 + 999d

For Second AP,

a′100 = a2 + (100 − 1)d

= a2 + 99d

a′1000 = a2 + (1000 − 1)d

= a2 + 999d

According to the question, it is given that the difference between





100th term of these APs = 100

Therefore, (a1 + 99d) − (a2 + 99d) = 100

a1 − a2 = 100 … (1)

Difference between 1000th terms of these APs

(a1 + 999d) − (a2 + 999d) = a1 − a2

From equation (1),

This difference, a1 − a2 = 100

Hence difference between 1000th terms of these AP will be 100.

13. How many three-digit numbers are divisible by 7?

Solution:

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

So, 105,112,119, . . . ..

All are 3-digit numbers that are divisible by 7 and therefore, all these are terms of an AP having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5.

Clearly, 999 − 5 = 994 is the maximum possible 3-digit number that is divisible by 7.

The AP is as follows.

105, 112, 119, . . . . . ,994

Let 994 be the nth term of this AP

a = 105

d = 7

an = 994

n =?

an = a + (n − 1)d

⇒ 994 = 105 + (n − 1)7

⇒ 889 = (n − 1)7





⇒ (n − 1) = 127

⇒ n = 128

Hence, there are 128 three-digit numbers that are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

Solution:

First multiple of 4 that is greater than 10 is 12. Next will be 16.

So, 12,16,20,24, . . . ..

All these are divisible by 4 and therefore, all these are terms of an AP with first term as 12 and common difference as 4.

when we divide 250 by 4, the remainder will be 2.

Hence, 250 − 2 = 248 is divisible by 4

The AP is as follows.

12, 16, 20, 24, . . . . . .248

Let 248 be the nth term of this AP

a = 12

d = 4

an = 248

an = a + (n − 1)d

⇒ 248 = 12 + (n − 1)4

⇒236

4= 𝑛𝑛 − 1

⇒ n = 60

So, there are 60 multiples of 4 lying between 10 and 250.

15. For what value of 𝑛𝑛, are the 𝑛𝑛th terms of two APs: 63, 65, 67, . .. and 3, 10, 17, . .. equal?

Solution:

Let us consider the first AP:

63, 65, 67, . . . ..





a = 63

d = a2 − a1 = 65 − 63 = 2

nth term of this AP = a + (n − 1)d

⇒ an = 63 + (n − 1)2 = 63 + 2n − 2

⇒ an = 61 + 2n … (1)

Now, let us consider the second AP:

3, 10, 17, . . . ..

a = 3

d = a2 − a1 = 10 − 3 = 7

nth term of this AP = 3 + (n − 1)7

an = 7n − 4 … (2)

In question, it is given that, nth term of the two APs are equal to each other.

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

61 + 2n = 7n − 4

⇒ 61 + 4 = 5n

⇒ 5n = 65

⇒ n = 13

Hence, 13th terms of two APs are equal.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

a3 = 16

Recall formula for nth term of an AP: an = a + (n − 1)d

Substituting the value of 𝑛𝑛 = 3, we get

a + (3 − 1)d = 16

⇒ a + 2d = 16 … (1)

a7 − a5 = 12

⇒ [a + (7 − 1)d] − [a + (5 − 1)d] = 12





⇒ (a + 6d) − (a + 4d) = 12

⇒ 2d = 12

⇒ d = 6

From equation (1), we get

a + 2(6) = 16

⇒ a + 12 = 16

⇒ a = 4

So, AP will be

4, 10, 16, 22, . . . . . . . . .,

17. Find the 20th term from the last term of the AP : 3, 8, 13, . . . , 253.

Solution:

According to the question, given AP is

3, 8, 13, . . . . . . . . . . , 253

Common difference for this AP is 5.

So, this AP can be written in reverse order as

253, 248, 243, . . . . . , 13, 8, 5

For this AP,

a = 253

d = 248 − 253 = −5

n = 20

a20 = a + (20 − 1)d

⇒ a20 = 253 + (19)(−5)

⇒ a20 = 253 − 95

a20 = 158

Hence, 20th term from the last term of the given AP is 158.

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.





Solution:

We know that,

an = a + (n − 1)d

⇒ a4 = a + (4 − 1)d

⇒ a4 = a + 3d

Similarly,

a8 = a + 7d

a6 = a + 5d

a10 = a + 9d

According to the question, it is given that, a4 + a8 = 24

⇒ (a + 3d) + (a + 7d) = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 … (1)

It is also given in the question that, a6 + a10 = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22 … (2)

On subtracting equation (1) from (2), we get

2d = 22 − 12

⇒ 2d = 10 ⇒ d = 5

From equation (1), we get

a + 5d = 12

⇒ a + 5(5) = 12

⇒ a + 25 = 12

⇒ a = −13

a2 = a + d = −13 + 5 = −8

a3 = a2 + d = −8 + 5 = −3

Hence, the first three terms of the given AP are −13,−8, and −3.





19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Solution:

We can observe that the income Subba Rao obtained in various years are in AP as every year, his salary is increased by ₹ 200.

So, the salaries of each year from 1995 are

5000, 5200, 5400, . . . . . . ..

Here, a = 5000

d = 200

Let after nth year, his salary be ₹ 7000.

Therefore, an = a + (n − 1)d

⇒ 7000 = 5000 + (n − 1)200

⇒ 200(n − 1) = 2000

⇒ (n − 1) = 10

⇒ n = 11

Hence, in 11th year, Subba Rao’s salary will become ₹ 7000.

20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the 𝑛𝑛th week, her weekly savings become ₹ 20.75, find 𝑛𝑛.

Solution:


Savings made by Ramkali in first week = a = 5

Increment in savings = d = 1.75

Let her savings become ₹ 20.75 in 𝑛𝑛𝑛𝑛ℎ week, i.e., an = 20.75

n =?

an = a + (n − 1)d

Substituting the values

20.75 = 5 + (n − 1)1.75

⇒ 15.75 = (n − 1)1.75

⇒ (n − 1) =15.751.75

=1575175





=637

= 9

⇒ n − 1 = 9

⇒ n = 10

So, her savings become ₹ 20.75 in10th week.

♦ ♦ ♦

EXERCISE 5.3

1. Find the sum of the following APs:

(i) 2, 7, 12, . . ., to 10 terms.

(ii) −37,−33,−29, . . ., to 12 terms.

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

(iv) 115

, 112

, 110

, …to 11 terms.

Solution:

(i) 2, 7 ,12, . . . . . , to 10 terms

For the given AP,

a = 2

d = a2 − a1 = 7 − 2 = 5

n = 10

We know that,

Sn =n2

[2a + (n − 1)d]


S10 =102

[2(2) + (10 − 1)5]

= 5 × [4 + (9) × (5)]

= 5 × 49 = 245

(ii) −37,−33,−29, . . . . . . .,to 12 terms





For the given AP,

a = −37

d = a2 − a1 = (−33) − (−37)

= −33 + 37 = 4

n = 12

We know that,

Sn =n2

[2a + (n − 1)d]


S12 =122

[2(−37) + (12 − 1)4]

= 6[−74 + 11 × 4]

= 6[−74 + 44]

= 6(−30) = −180

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

For the given AP,

a = 0.6

d = a2 − a1 = 1.7 − 0.6 = 1.1

n = 100

We know that,

Sn =n2

[2a + (n − 1)d]


S100 =100

2[2(0.6) + (100 − 1)1.1]

= 50[1.2 + (99) × (1.1)]

= 50[1.2 + 108.9]

= 50[110.1]

= 5505

(iv) 115

, 112

, 110

, . . . . . . ., to 11 terms

For the given AP,





a =1

15

n = 11

d = a2 − a1 =1

12−

115

=5 − 4

60=

160

We know that,

Sn =n2

[2a + (n − 1)d]


S11 =112�2 �

115� + (11 − 1)

160�

=112�

215

+1060�

= �112� �

930� =

3320

2. Find the sums given below:

(i) 7 + 10 12

+ 14 + ⋯+ 84

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

(iii) −5 + (−8) + (−11)+. . . +(−230)

Solution:

(i) 7 + 10 12

+ 14+. . . . . . . +84

For the given AP,

a = 7

Last term = 𝑙𝑙 = 84

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 1012− 7 =

212− 7 =

72

Let 84 be the nth term of the given AP

We know that,

an = a + (n − 1)d






84 = 7 + (𝑛𝑛 − 1)72

⇒ 77 = (𝑛𝑛 − 1)72

⇒ 22 = n − 1

So, n = 23

We know that,

𝑆𝑆𝑛𝑛 =𝑛𝑛2

(𝑎𝑎 + 𝑙𝑙)


𝑆𝑆23 =232

[7 + 84]

=23 × 91

2=

20932

= 104612

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

For the given AP,

a = 34

d = a2 − a1 = 32 − 34 = −2

Last term = 𝑙𝑙 = 10

Let 10 be the nth term of the given AP

We know that,

an = a + (n − 1)d

⇒ 10 = 34 + (n − 1)(−2)

⇒ −24 = (n − 1)(−2)

⇒ 12 = n − 1

So, n = 13

We know that,








𝑆𝑆13 =132

(34 + 10)

=13 × 44

2= 13 × 22

= 286

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

For the given AP,

a = −5

last term = 𝑙𝑙 = −230

d = a2 − a1 = (−8) − (−5)

= −8 + 5 = −3

Let −230 be the nth term of the given AP

We know that,

an = a + (n − 1)d

⇒ −230 = −5 + (n − 1)(−3)

⇒ −225 = (n − 1)(−3)

⇒ (n − 1) = 75

So, n = 76

And, 𝑆𝑆𝑛𝑛 = 𝑛𝑛2



𝑆𝑆76 = 762

[(−5) + (−230)]

= 38(−235)

= −8930

3. In an AP:

(i) given 𝑎𝑎 = 5,𝑑𝑑 = 3,𝑎𝑎𝑛𝑛 = 50, find 𝑛𝑛 and 𝑆𝑆𝑛𝑛.

(ii) given 𝑎𝑎 = 7,𝑎𝑎13 = 35, find 𝑑𝑑 and 𝑆𝑆13.

(iii) given 𝑎𝑎12 = 37,𝑑𝑑 = 3, find 𝑎𝑎 and 𝑆𝑆12.

(iv) given 𝑎𝑎3 = 15, 𝑆𝑆10 = 125, find 𝑑𝑑 and 𝑎𝑎10.





(v) given 𝑑𝑑 = 5, 𝑆𝑆9 = 75, find 𝑎𝑎 and 𝑎𝑎9.

(vi) given 𝑎𝑎 = 2,𝑑𝑑 = 8, 𝑆𝑆𝑛𝑛 = 90, find 𝑛𝑛 and 𝑎𝑎𝑛𝑛.

(vii) given 𝑎𝑎 = 8,𝑎𝑎𝑛𝑛 = 62, 𝑆𝑆𝑛𝑛 = 210, find 𝑛𝑛 and 𝑑𝑑.

(viii) given 𝑎𝑎𝑛𝑛 = 4,𝑑𝑑 = 2, 𝑆𝑆𝑛𝑛 = −14, find 𝑛𝑛 and 𝑎𝑎.

(ix) given 𝑎𝑎 = 3,𝑛𝑛 = 8, 𝑆𝑆 = 192, find 𝑑𝑑.

(x) given 𝑙𝑙 = 28, 𝑆𝑆 = 144, and there are total 9 terms. Find 𝑎𝑎.

Solution:

(i) Given that, 𝑎𝑎 = 5,𝑑𝑑 = 3,𝑎𝑎𝑛𝑛 = 50

We know that, 𝑎𝑎𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑,


∴ 50 = 5 + (𝑛𝑛 − 1)3

⇒ 45 = (𝑛𝑛 − 1)3

⇒ 15 = 𝑛𝑛 − 1

So, 𝑛𝑛 = 16

We also know that,


[𝑎𝑎 + 𝑎𝑎𝑛𝑛]


𝑆𝑆16 =162

[5 + 50]

= 8 × 55

= 440

(ii) Given that, 𝑎𝑎 = 7,𝑎𝑎13 = 35

We Know that, 𝑎𝑎𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑,

∴ 𝑎𝑎13 = 𝑎𝑎 + (13 − 1)𝑑𝑑

⇒ 35 = 7 + 12𝑑𝑑

⇒ 28 = 12𝑑𝑑

⇒ 𝑑𝑑 =73

We also know that,







𝑆𝑆13 =𝑛𝑛2

[𝑎𝑎 + 𝑎𝑎13]


=132

[7 + 35]

=13 × 42

2= 13 × 21

= 273

(iii) Given that, 𝑎𝑎12 = 37,𝑑𝑑 = 3



𝑎𝑎12 = 𝑎𝑎 + (12 − 1)3

⇒ 37 = 𝑎𝑎 + 33

So, 𝑎𝑎 = 4

We also know that,




𝑆𝑆12 =122

[4 + 37]

⇒ 𝑆𝑆12 = 6(41)

⇒ 𝑆𝑆12 = 246

(iv) Given that, 𝑎𝑎3 = 15, 𝑆𝑆10 = 125



𝑎𝑎3 = 𝑎𝑎 + (3 − 1)𝑑𝑑

⇒ 15 = 𝑎𝑎 + 2𝑑𝑑 … (i)

We also know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]






𝑆𝑆10 =102

[2𝑎𝑎 + (10 − 1)𝑑𝑑]

⇒ 125 = 5(2𝑎𝑎 + 9𝑑𝑑)

⇒ 25 = 2𝑎𝑎 + 9𝑑𝑑 … (ii)

By multiplying equation (1) by 2, we get 30 = 2𝑎𝑎 + 4𝑑𝑑 … (iii)

By subtracting equation (iii) from (ii), we get

−5 = 5𝑑𝑑

So, 𝑑𝑑 = −1

From equation (i),

15 = 𝑎𝑎 + 2(−1)

⇒ 15 = 𝑎𝑎 − 2

⇒ 𝑎𝑎 = 17



𝑎𝑎10 = 𝑎𝑎 + (10 − 1)𝑑𝑑

⇒ 𝑎𝑎10 = 17 + (9)(−1)

So, 𝑎𝑎10 = 17 − 9 = 8

(v) Given that, 𝑑𝑑 = 5, 𝑆𝑆9 = 75

We know that, 𝑆𝑆𝑛𝑛 = 𝑛𝑛2

[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑],


𝑆𝑆9 =92

[2𝑎𝑎 + (9 − 1)5]

⇒ 75 =92

(2𝑎𝑎 + 40)

⇒ 25 = 3(𝑎𝑎 + 20)

⇒ 25 = 3𝑎𝑎 + 60

⇒ 3𝑎𝑎 = 25 − 60

⇒ 𝑎𝑎 = −35

3

We also know that,







𝑎𝑎9 = 𝑎𝑎 + (9 − 1)(5)

=−35

3+ 8(5)

=−35

3+ 40

=−35 + 120

3=

853

(vi) Given that, 𝑎𝑎 = 2,𝑑𝑑 = 8, 𝑆𝑆𝑛𝑛 = 90

We know that,

𝑆𝑆𝑛𝑛 = 𝑛𝑛2

[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑],


90 =𝑛𝑛2

[4 + (𝑛𝑛 − 1)8]

⇒ 90 = 𝑛𝑛[2 + (𝑛𝑛 − 1)4]

⇒ 90 = 𝑛𝑛[2 + 4𝑛𝑛 − 4]

⇒ 90 = 𝑛𝑛(4𝑛𝑛 − 2) = 4𝑛𝑛2 − 2𝑛𝑛

⇒ 4𝑛𝑛2 − 2𝑛𝑛 − 90 = 0

⇒ 4𝑛𝑛2 − 20𝑛𝑛 + 18𝑛𝑛 − 90 = 0 (Factorisation by splitting the middle term)

⇒ 4𝑛𝑛(𝑛𝑛 − 5) + 18(𝑛𝑛 − 5) = 0

⇒ (𝑛𝑛 − 5)(4𝑛𝑛 + 18) = 0

Either 𝑛𝑛 − 5 = 0 or 4𝑛𝑛 + 18 = 0

⇒ 𝑛𝑛 = 5 or 𝑛𝑛 = −184

= −92

But, 𝑛𝑛 can neither be negative nor fractional.

Hence, 𝑛𝑛 = 5

We also know that,


⇒ 𝑎𝑎5 = 2 + (5 − 1)8





= 2 + (4)(8)

= 2 + 32 = 34

So, 𝑛𝑛 = 5 𝑎𝑎𝑛𝑛𝑑𝑑 𝑎𝑎5 = 34.

(vii) Given that, 𝑎𝑎 = 8,𝑎𝑎𝑛𝑛 = 62, 𝑆𝑆𝑛𝑛 = 210

We know that,




210 =𝑛𝑛2

[8 + 62]

⇒ 210 =𝑛𝑛2

(70)

So, 𝑛𝑛 = 6

We also know that,



62 = 8 + (6 − 1)𝑑𝑑

⇒ 62 − 8 = 5𝑑𝑑

Hence, 𝑑𝑑 = 545

(viii) Given that, 𝑎𝑎𝑛𝑛 = 4,𝑑𝑑 = 2, 𝑆𝑆𝑛𝑛 = −14

We know that,



4 = 𝑎𝑎 + (𝑛𝑛 − 1)2

⇒ 4 = 𝑎𝑎 + 2𝑛𝑛 − 2

⇒ 𝑎𝑎 + 2𝑛𝑛 = 6

⇒ 𝑎𝑎 = 6 − 2𝑛𝑛 …………………. (i)

We also know that,








−14 =𝑛𝑛2

[𝑎𝑎 + 4]

⇒ −28 = 𝑛𝑛(𝑎𝑎 + 4)

⇒ −28 = 𝑛𝑛(6 − 2𝑛𝑛 + 4) {From equation (i)}

⇒ −28 = 𝑛𝑛(−2𝑛𝑛 + 10)

⇒ −28 = −2𝑛𝑛2 + 10𝑛𝑛

⇒ 2𝑛𝑛2 − 10𝑛𝑛 − 28 = 0

⇒ 𝑛𝑛2 − 5𝑛𝑛 − 14 = 0

⇒ 𝑛𝑛2 − 7𝑛𝑛 + 2𝑛𝑛 − 14 = 0 [Factorisation by splitting the middle term]

⇒ 𝑛𝑛(𝑛𝑛 − 7) + 2(𝑛𝑛 − 7) = 0

⇒ (𝑛𝑛 − 7)(𝑛𝑛 + 2) = 0

Either 𝑛𝑛 − 7 = 0 or 𝑛𝑛 + 2 = 0

⇒ 𝑛𝑛 = 7 or 𝑛𝑛 = −2

But, 𝑛𝑛 can only take positive integral values.

So, 𝑛𝑛 = 7

From equation (i), we obtain

𝑎𝑎 = 6 − 2𝑛𝑛

⇒ 𝑎𝑎 = 6 − 2(7)

= 6 − 14

= −8

Hence, 𝑛𝑛 = 7 𝑎𝑎𝑛𝑛𝑑𝑑 𝑎𝑎 = −8.

(ix) Given that, 𝑎𝑎 = 3,𝑛𝑛 = 8, 𝑆𝑆 = 192

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)d]


192 =82

[2 × 3 + (8 − 1)𝑑𝑑]

⇒ 192 = 4[6 + 7𝑑𝑑]

⇒ 48 = 6 + 7𝑑𝑑





⇒ 42 = 7𝑑𝑑

So, 𝑑𝑑 = 6

(x) Given that, 𝑙𝑙 = 28, 𝑆𝑆 = 144 and there are total 9 terms.

We know that,




144 =92

(𝑎𝑎 + 28)

⇒ (16) × (2) = 𝑎𝑎 + 28

⇒ 32 = 𝑎𝑎 + 28

Hence, 𝑎𝑎 = 4

4. How many terms of the AP : 9, 17, 25, . .. must be taken to give a sum of 636?

Solution:

Let the number of terms in the given AP = 𝑛𝑛

It is given that for this AP, 𝑎𝑎 = 9

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 17 − 9 = 8

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑)


636 =𝑛𝑛2

[2 × 9 + (𝑛𝑛 − 1)8]

⇒ 636 =𝑛𝑛2

[18 + (𝑛𝑛 − 1)8]

⇒ 636 = 𝑛𝑛[9 + 4𝑛𝑛 − 4]

⇒ 636 = 𝑛𝑛(4𝑛𝑛 + 5)

⇒ 4𝑛𝑛2 + 5𝑛𝑛 − 636 = 0

⇒ 4𝑛𝑛2 + 53𝑛𝑛 − 48𝑛𝑛 − 636 = 0 [Factorisation by splitting the middle term]

⇒ 𝑛𝑛(4𝑛𝑛 + 53) − 12(4𝑛𝑛 + 53) = 0





⇒ (4𝑛𝑛 + 53)(𝑛𝑛 − 12) = 0

Either 4𝑛𝑛 + 53 = 0 or 𝑛𝑛 − 12 = 0

⇒ 𝑛𝑛 = −534

or 𝑛𝑛 = 12

𝑛𝑛 cannot be −534

. As the number of terms can neither be negative nor fractional, Hence, 𝑛𝑛 = 12.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:


𝑎𝑎 = 5

𝑎𝑎𝑛𝑛 = 𝑙𝑙 = 45

𝑆𝑆𝑛𝑛 = 400

We know that,




400 =𝑛𝑛2

(5 + 45)

400 =𝑛𝑛2

(50)

So, 𝑛𝑛 = 16

We also know that,

𝑎𝑎𝑛𝑛 = 𝑙𝑙 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑

45 = 5 + (16 − 1)𝑑𝑑

40 = 15𝑑𝑑

Hence, 𝑑𝑑 = 4015

= 83

6. 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:






𝑎𝑎 = 17

𝑎𝑎𝑛𝑛 = 𝑙𝑙 = 350

𝑑𝑑 = 9

Let the number of terms in the given AP = 𝑛𝑛

We know that,



350 = 17 + (𝑛𝑛 − 1)9

⇒ 333 = (𝑛𝑛 − 1)9

⇒ (𝑛𝑛 − 1) = 37

So, 𝑛𝑛 = 38

We also know that,




⇒ 𝑆𝑆38 =382

(17 + 350) = 19(367) = 6973

Hence, the given AP contains 38 terms and their sum is 6973.

7. Find the sum of first 22 terms of an AP in which 𝑑𝑑 = 7 and 22nd term is 149.

Solution:

According to the question, for this AP,

𝑑𝑑 = 7

𝑎𝑎22 = 149

𝑆𝑆22 = ?

We know that,


⇒ 𝑎𝑎22 = 𝑎𝑎 + (22 − 1)𝑑𝑑

⇒ 149 = 𝑎𝑎 + 21 × 7





⇒ 149 = 𝑎𝑎 + 147

⇒ 𝑎𝑎 = 2

We also know that,


(𝑎𝑎 + 𝑎𝑎𝑛𝑛)


𝑆𝑆22 =222

(2 + 149)

⇒ 𝑆𝑆22 = 11(151) = 1661

Therefore, the sum of first 22 terms of the given AP is 1661.

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

Solution:

As per the question, for the given AP:

𝑎𝑎2 = 14

𝑎𝑎3 = 18

𝑑𝑑 = 𝑎𝑎3 − 𝑎𝑎2 = 18 − 14 = 4

𝑎𝑎2 = 𝑎𝑎 + 𝑑𝑑

⇒ 14 = 𝑎𝑎 + 4

⇒ 𝑎𝑎 = 10

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]

⇒ 𝑆𝑆51 =512

[2 × 10 + (51 − 1)4]

=512

[20 + (50)(4)]

=51(220)

2= 51(110)

= 5610

Therefore, the sum of first 51 terms of the given AP is 5610.





9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first 𝑛𝑛 terms.

Solution:

As per the question, for the given AP:

𝑆𝑆7 = 49, 𝑆𝑆17 = 289

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]

⇒ 𝑆𝑆7 =72

[2𝑎𝑎 + (7 − 1)𝑑𝑑]

⇒ 49 =72

(2𝑎𝑎 + 6𝑑𝑑)

⇒ 7 = (𝑎𝑎 + 3𝑑𝑑)

⇒ 𝑎𝑎 + 3𝑑𝑑 = 7 … (i)

Similarly, 𝑆𝑆17 = 172

[2𝑎𝑎 + (17 − 1)𝑑𝑑]

⇒ 289 =172

[2𝑎𝑎 + 16𝑑𝑑]

⇒ 17 = (𝑎𝑎 + 8𝑑𝑑)

⇒ 𝑎𝑎 + 8𝑑𝑑 = 17 … (ii)

By subtracting equation (i) from equation (ii),

⇒ 5𝑑𝑑 = 10

⇒ 𝑑𝑑 = 2

From equation (i),

𝑎𝑎 + 3(2) = 7

⇒ 𝑎𝑎 + 6 = 7

⇒ 𝑎𝑎 = 1


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


=𝑛𝑛2

[2(1) + (𝑛𝑛 − 1)(2)]





=𝑛𝑛2

(2 + 2𝑛𝑛 − 2)

=𝑛𝑛2

(2𝑛𝑛)

= 𝑛𝑛2

10. Show that 𝑎𝑎1,𝑎𝑎2, … , 𝑎𝑎𝑛𝑛, …form an AP where 𝑎𝑎𝑛𝑛 is defined as below:

(i) 𝑎𝑎𝑛𝑛 = 3 + 4𝑛𝑛

(ii) 𝑎𝑎𝑛𝑛 = 9 − 5𝑛𝑛

Also find the sum of the first 15 terms in each case.

Solution:

(i) 𝑎𝑎𝑛𝑛 = 3 + 4𝑛𝑛 …… (1)

𝑎𝑎𝑘𝑘+1 = 3 + 4(𝑘𝑘 + 1)

= 3 + 4𝑘𝑘 + 4

Here, 𝑎𝑎𝑘𝑘+1 − 𝑎𝑎𝑘𝑘 = (3 + 4𝑘𝑘 + 4) − (3 + 4𝑘𝑘) = 4 which is independent of 𝑘𝑘.

So, this is an AP with common difference, 𝑑𝑑 = 4.

Also, by substituting 𝑛𝑛 = 1 in equation (1), we get first term of AP i.e., 𝑎𝑎 = 7.

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


𝑆𝑆15 =152

[2(7) + (15 − 1)4]

=152

[(14) + 56]

=152

(70)

= 15 × 35

= 525

Therefore, the sum of the first 15 terms of this AP is 525.





(ii) 𝑎𝑎𝑛𝑛 = 9 − 5𝑛𝑛 ………… (1)

Here 𝑎𝑎𝑘𝑘+1 − 𝑎𝑎𝑘𝑘 = (9 − 5𝑘𝑘 − 5) − (9 − 5𝑘𝑘)

= −5 which is independent of 𝑘𝑘.

So, this is an AP with common difference, 𝑑𝑑 = −5 and

We obtain first term, 𝑎𝑎 = 4, by substituting 𝑛𝑛 = 1 in equation (1)

As, 𝑆𝑆𝑛𝑛 = 𝑛𝑛2

[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


𝑆𝑆15 =152

[2(4) + (15 − 1)(−5)]

=152

[8 + 14(−5)]

=152

(8 − 70)

=152

(−62) = 15(−31)

= −465

Therefore, the sum of the first 15 terms of this AP is −465.

11. If the sum of the first 𝑛𝑛 terms of an AP is 4𝑛𝑛 − 𝑛𝑛2, what is the first term (that is 𝑆𝑆1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the 𝑛𝑛th terms.

Solution:

According to the question, for a given AP: 𝑆𝑆𝑛𝑛 = 4𝑛𝑛 − 𝑛𝑛2

First term, 𝑎𝑎 = 𝑆𝑆1 = 4(1) − (1)2 = 4 − 1 = 3

Sum of first two terms = 𝑆𝑆2

= 4(2) − (2)2 = 8 − 4 = 4

We know that,

𝑎𝑎𝑛𝑛 = 𝑆𝑆𝑛𝑛 − 𝑆𝑆𝑛𝑛−1


Second term, 𝑎𝑎2 = 𝑆𝑆2 − 𝑆𝑆1 = 4 − 3 = 1

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 𝑎𝑎2 − 𝑎𝑎 = 1 − 3 = −2






= 3 + (𝑛𝑛 − 1)(−2)

= 3 − 2𝑛𝑛 + 2

= 5 − 2𝑛𝑛

Thus, 𝑎𝑎3 = 5 − 2(3) = 5 − 6 = −1

𝑎𝑎10 = 5 − 2(10) = 5 − 20 = −15

Therefore, the first term is 3 and the sum of first two terms is 4. The second term is 1. 3rd, 10th and 𝑛𝑛th terms are −1,−15 and 5 − 2𝑛𝑛 respectively.

12. Find the sum of the first 40 positive integers divisible by 6.

Solution:

The positive integers that are divisible by 6 are 6, 12, 18, 24 …

Now, an AP is formed with these terms, whose first term is 6 and common difference is 6.

𝑎𝑎 = 6,𝑑𝑑 = 6, 𝑆𝑆40 =?

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]

⇒ 𝑆𝑆40 =402

[2(6) + (40 − 1)6]

= 20[12 + (39)(6)]

= 20(12 + 234)

= 20 × 246

= 4920

Hence, the sum of the first 40 positive integers divisible by 6 is 4920.

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

Solution:

Multiples of 8 are 8, 16, 24, 32 …

Now, these terms are in an AP, whose first term is 8 and common difference is 8.

Therefore, 𝑎𝑎 = 8, 𝑑𝑑 = 8, 𝑆𝑆15 =?





We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


𝑆𝑆15 =152

[2(8) + (15 − 1)8]

=152

[16 + 14(8)]

= 152

(16 + 112)

= 15(128)2

= 15 × 64

= 960

Hence, the sum of the first 15 multiples of 8 is 960.

14. Find the sum of the odd numbers between 0 and 50.

Solution:

We can observe clearly that the odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49

So, it can be noticed that these odd numbers are in an AP.

𝑎𝑎 = 1,𝑑𝑑 = 2, 𝑙𝑙 = 49

We know that,


⇒ 49 = 1 + (𝑛𝑛 − 1)2

⇒ 48 = 2(𝑛𝑛 − 1)

⇒ 𝑛𝑛 − 1 = 24

So, 𝑛𝑛 = 25

We also know that,



𝑆𝑆25 =252

(1 + 49)

=25(50)

2= (25)(25)





= 625

Hence, the sum of the odd numbers between 0 and 50 is 625.

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 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:

As per the question, penalties are in AP having first term as 200 and common difference as 50.

𝑎𝑎 = 200,𝑑𝑑 = 50

The penalty that has to be paid if contractor has delayed the work by 30 days =𝑆𝑆30

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


𝑆𝑆30 =302

[2(200) + (30 − 1)50]

= 15[400 + 1450]

= 15(1850)

= 27750

Hence, the contractor has to pay ₹ 27750 as penalty, if he has delayed the work by 30 days.

16. A sum of ₹ 700 is to be used to 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 of the prizes.

Solution:

Let the cost of 1st prize be 𝐶𝐶.

Cost of 2nd prize = 𝐶𝐶 − 20

and cost 3ed prize = (𝐶𝐶 − 20) − 20





Cost of the prizes are forming an AP with common difference, −20 and first term as 𝐶𝐶.

𝑎𝑎 = 𝐶𝐶, 𝑑𝑑 = −20

Given that, 𝑆𝑆7 = 700

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


72

[2𝐶𝐶 + (7 − 1)𝑑𝑑] = 700

⇒[2𝐶𝐶 + (6)(−20)]

2= 100

⇒ 𝐶𝐶 + 3(−20) = 100

⇒ 𝐶𝐶 − 60 = 100

⇒ 𝐶𝐶 = 160

Hence, the values of the prizes (in ₹) are 160, 140, 120, 100, 80, 60 and 40.

17. In a school, students thought of planting 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 the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

To solve this question, let’s first calculate the number of trees planted by students of one section of all the classes from I to XII.

Then the number of trees planted by students of all the three sections of all the classes from I to XII gives us the total number of trees planted by the students.

As per the question, number of trees planted by the students of one section of all classes forms as AP as follows:

1, 2, 3, 4, 5, … ,12


Common difference, 𝑑𝑑 = 2 − 1 = 1





We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


𝑆𝑆12 =122

[2(1) + (12 − 1)(1)]

= 6(2 + 11)

= 6(13)

= 78

Thus, the number of trees planted by one section of the classes = 78

Then the number of trees planted by three sections of the classes = 3 × 78 = 234

Hence, the total number of trees planted by the school students is 234.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . .. as shown in Fig. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22

7)

[Hint: Length of successive semicircles is 𝑙𝑙1, 𝑙𝑙2, 𝑙𝑙3, 𝑙𝑙4, … with centres at A, B, A, B, . . .,respectively.]

Solution:

Semi-perimeter of circle = 𝜋𝜋𝜋𝜋

𝑙𝑙1 = 𝜋𝜋(0.5) =𝜋𝜋2

cm

𝑙𝑙2 = 𝜋𝜋(1) = 𝜋𝜋 cm

𝑙𝑙3 = 𝜋𝜋(1.5) =3𝜋𝜋2

cm





So, 𝑙𝑙1, 𝑙𝑙2, 𝑙𝑙3, … . i. e. the lengths of the semi-circles form an AP,

𝜋𝜋2

,𝜋𝜋,3𝜋𝜋2

, 2𝜋𝜋, …

𝑎𝑎 =𝜋𝜋2

𝑑𝑑 = 𝜋𝜋 −𝜋𝜋2

=𝜋𝜋2

𝑆𝑆13 =?

We know that the sum of 𝑛𝑛 terms of an AP is given by


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]

⇒ S13 =132�2 �

𝜋𝜋2� + (13 − 1) �

𝜋𝜋2��

=132�𝜋𝜋 +

12𝜋𝜋2�

= �132� (7𝜋𝜋)

=91𝜋𝜋

2

=91 × 22

2 × 7= 13 × 11

= 143

Hence, the total length of such spiral of thirteen consecutive semi-circles is 143 cm.

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row,18 in the row next to it and so on (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

According to the question, here the number of logs in each row are in an AP

20, 19, 18 …

For the given AP,





𝑎𝑎 = 20

𝑑𝑑 = 𝑎𝑎2 − 𝑎𝑎1 = 19 − 20 = −1

Let a total of 200 logs be placed in 𝑛𝑛 rows.

𝑆𝑆𝑛𝑛 = 200

We know that,


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]


200 =𝑛𝑛2

[2(20) + (𝑛𝑛 − 1)(−1)]

⇒ 400 = 𝑛𝑛(40 − 𝑛𝑛 + 1)

⇒ 400 = 𝑛𝑛(41 − 𝑛𝑛)

⇒ 400 = 41𝑛𝑛 − 𝑛𝑛2

⇒ 𝑛𝑛2 − 41𝑛𝑛 + 400 = 0

⇒ 𝑛𝑛2 − 16𝑛𝑛 − 25𝑛𝑛 + 400 = 0 [Factorisation by splitting the middle term]

⇒ 𝑛𝑛(𝑛𝑛 − 16) − 25(𝑛𝑛 − 16) = 0

⇒ (𝑛𝑛 − 16)(𝑛𝑛 − 25) = 0

Either (𝑛𝑛 − 16) = 0 or (𝑛𝑛 − 25) = 0

⇒ 𝑛𝑛 = 16 or 𝑛𝑛 = 25


𝑎𝑎16 = 20 + (16 − 1)(−1)

𝑎𝑎16 = 20 − 15

𝑎𝑎16 = 5

Similarly,

𝑎𝑎25 = 20 + (25 − 1)(−1)

𝑎𝑎25 = 20 − 24

= −4

Clearly, the number of logs in 16th row i.e., the top row is 5.

But, the number of logs in 25th row is negative, which is not possible.





Hence, 200 logs can be placed in 16 rows and the number of logs in the 16th row (top) is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Figure).

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 she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Solution:

The total distance covered by a competitor to pick the first potato = 2 × 5 = 10

The total distance covered by a competitor to pick the second potato = 2 × (5 +3) = 16

So, the distances covered by a competitor to pick the potatoes form an AP as follows: 10, 16, 22, 28 , …

Here, a = 10, d = 16 − 10 = 6 and n = 10.

We know that the sum of n terms of an AP is given by Sn = n2

[2a + (n − 1)d]

⇒ S10 =102

[2(10) + (10 − 1)(6)]

= 5[20 + 54]

= 5(74)

= 370 m

Hence, the total distance covered by the competitor is 370 m.

♦ ♦ ♦

EXERCISE 5.4 (Optional)*





1. Which term of the AP : 121, 117, 113, . . ., is its first negative term?

[Hint: Find 𝑛𝑛 for 𝑎𝑎𝑛𝑛 < 0]

Solution:

Given AP is 121, 117, 113 …

𝑎𝑎 = 121

𝑑𝑑 = 117 − 121 = −4


= 121 + (𝑛𝑛 − 1)(−4)

= 121 − 4𝑛𝑛 + 4

= 125 − 4𝑛𝑛

We have to find the first negative term of this AP

Therefore, 𝑎𝑎𝑛𝑛 < 0

⇒ 125 − 4𝑛𝑛 < 0

⇒ 125 < 4𝑛𝑛

⇒ 𝑛𝑛 >125

4

⇒ 𝑛𝑛 > 31.25

Therefore, 32nd term will be the first negative term of this AP

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Solution:

We know that,


𝑎𝑎3 = 𝑎𝑎 + (3 − 1)𝑑𝑑

⇒ 𝑎𝑎3 = 𝑎𝑎 + 2𝑑𝑑

Similarly, 𝑎𝑎7 = 𝑎𝑎 + 6𝑑𝑑

Given that, 𝑎𝑎3 + 𝑎𝑎7 = 6

⇒ (𝑎𝑎 + 2𝑑𝑑) + (𝑎𝑎 + 6𝑑𝑑) = 6

⇒ 2𝑎𝑎 + 8𝑑𝑑 = 6

⇒ 𝑎𝑎 + 4𝑑𝑑 = 3





⇒ 𝑎𝑎 = 3 − 4𝑑𝑑 … (i)

Also, it is given that (𝑎𝑎3) × (𝑎𝑎7) = 8

⇒ (𝑎𝑎 + 2𝑑𝑑) × (𝑎𝑎 + 6𝑑𝑑) = 8

From equation (i),

(3 − 4𝑑𝑑 + 2𝑑𝑑) × (3 − 4𝑑𝑑 + 6𝑑𝑑) = 8

⇒ (3 − 2𝑑𝑑) × (3 + 2𝑑𝑑) = 8

⇒ 9 − 4𝑑𝑑2 = 8

⇒ 4𝑑𝑑2 = 9 − 8 = 1

⇒ 𝑑𝑑2 =14

⇒ 𝑑𝑑 = ±12

⇒ 𝑑𝑑 = 12 or −1

2

From equation (i),

(When 𝑑𝑑 is 12)

𝑎𝑎 = 3 − 4𝑑𝑑

⇒ 𝑎𝑎 = 3 − 4 �12�

= 3 − 2 = 1

(When 𝑑𝑑 is −12)

𝑎𝑎 = 3 − 4 �−12�

⇒ 𝑎𝑎 = 3 + 2 = 5


[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]

(When 𝑎𝑎 is 1 and 𝑑𝑑 is 12)

𝑆𝑆16 =162�2(1) + (16 − 1) �

12��

= 8 �2 +152�

= 4(19) = 76





(When 𝑎𝑎 is 5 and 𝑑𝑑 is −12)

𝑆𝑆16 =162�2(5) + (16 − 1) �−

12��

= 8 �10 + (15) �−12��

= 8 �52�

= 20

3. A ladder has rungs 25 cm apart. (See Fig.). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2 1

2m apart, what is the length of the wood required for the rungs?

[Hint: Number of rungs = 25025

+ 1]

Solution:

It is given that the rungs are 25 cm apart and top and bottom rungs are 2 12

m apart.

∴ Total number of rungs =212×100

25+ 1 = 250

25+ 1 = 11

Now, as the lengths of the rungs decrease uniformly, they will be in an AP

The length of the wood required for the rungs equals the sum of all the terms of this AP


Last term, 𝑙𝑙 = 25







∴ 𝑆𝑆11 =112

(45 + 25) =112

(70) = 385 cm

Therefore, the length of the wood required for the rungs is 385 cm.

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of 𝑥𝑥 such that the sum of the numbers of the houses preceding the house numbered 𝑥𝑥 is equal to the sum of the numbers of the houses following it. Find this value of 𝑥𝑥.

[Hint: 𝑆𝑆𝑥𝑥−1 = 𝑆𝑆49 − 𝑆𝑆𝑥𝑥]

Solution:

The number of houses was 1, 2, 3, … , 49

It can be observed that the number of houses is in an AP having 𝑎𝑎 as 1 and 𝑑𝑑 also as 1.

We know that,

Sum of 𝑛𝑛 terms in an AP = 𝑛𝑛2

[21 + (𝑛𝑛 − 1)𝑑𝑑]

Sum of number of houses preceding 𝑥𝑥th house = 𝑆𝑆𝑥𝑥−1

=(𝑥𝑥 − 1)

2[2𝑎𝑎 + (𝑥𝑥 − 1 − 1)𝑑𝑑]

=𝑥𝑥 − 1

2[2(1) + (𝑥𝑥 − 2)(1)]

=𝑥𝑥 − 1

2[2 + 𝑥𝑥 − 2]

=(𝑥𝑥)(𝑥𝑥 − 1)

2

Sum of number of houses following 𝑥𝑥th house 𝑆𝑆49 − 𝑆𝑆𝑥𝑥

=492

[2(1) + (49 − 1)(1)] −𝑥𝑥2

[2(1) + (𝑥𝑥 − 1)(1)]

=492

(2 + 49 − 1) −𝑥𝑥2

(2 + 𝑥𝑥 − 1)

= �492� (50) −

𝑥𝑥2

(𝑥𝑥 + 1)





= 25(49) −𝑥𝑥(𝑥𝑥 + 1)

2

It is given that these sums are equal to each other.

Hence, 𝑥𝑥(𝑥𝑥−1)2

= 25(49) − 𝑥𝑥 (𝑥𝑥+1)2

𝑥𝑥2

2−𝑥𝑥2

= 1225 −𝑥𝑥2

2−𝑥𝑥2

𝑥𝑥2 = 1225

𝑥𝑥 = ±35

However, the house numbers are positive integers.

The value of 𝑥𝑥 will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of 14

m and a tread of 12

m. (see Fig.). Calculate the total volume of concrete required to build the terrace.

[Hint: Volume of concrete required to build the first step = 14

× 12

× 50 m3]

Solution:

From the figure, we see that

1st step has 14

m height,

2nd step has 12

m height,

3ed step has 34

m height.





Therefore, the height of each step is increasing by 14

m each time whereas their

width 12

m and length 50 m remains the same.

Volume of concrete in 1st step = 14

× 12

× 50 = 254

Volume of concrete in 2ed step = 12

× 12

× 50 = 252

Volume of concrete in 3ed step = 12

× 34

× 50 = 754

It can be observed that the volumes of concrete in these steps are in an AP

254

,252

,754

, …

𝑎𝑎 =254

𝑑𝑑 =252−

254

=254

and 𝑆𝑆𝑛𝑛 = 𝑛𝑛2

[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]

𝑆𝑆15 =152�2 �

254� +

(15 − 1)254

�

=152�252

+(14)25

4�

=152�252

+175

2�

=152

(100) = 750

Volume of concrete required to build the terrace is 750 m3.

♦ ♦ ♦



CBSE NCERT Solutions for Class 10 Mathematics Chapter 5 · 2020-05-25 · CBSE NCERT Solutions for...

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