CBSE 11 Engineering Medical Maths Sequences and Series


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NCERT Solutions for Class 11 Subjectwise ● Class 11 Mathematics ● Class 11 Physics ● Class 11 Biology ● Class 11 Chemistry ● Class 11 English ● Class 11 Accountancy ● Class 11 Business Study ● Class 11 Economics

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#422402 Topic: Arithmetic Progression In an A.P.,the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is − 112.
Solution First term = 2. Let d be the common difference of the A.P. Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, . . . Sum of first five terms = 10 + 10d Sum of next five terms = 10 + 35d According to the given condition ,
1 10 + 10d = 4 (10 + 35d) ⇒ 40 + 40d = 10 + 35d ⇒ 30 = − 5d ⇒ d = − 6 ∴ a20 = a + (20 − 1)d = 2 + (19)( − 6) = 2 − 114 = − 112 Thus, the 20th term of the A.P. is − 112.

#423193 Topic: Arithmetic Progression
The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that (q − r)a + (r − p)b + (p − q)c = 0

Solution

Let t and d be the first term and the common difference of the A.P. respectively, The nth term of an A.P. is given by an = t + (n − 1)d Therefore, ap = t + (p − 1)d = a . . . (1)aq = t + (q − 1)d = b . . . (2)ar = t + (r − 1)d = c . . . (3) Subtracting equation (2) from (1), we obtain

a−b
(p − 1 − q + 1)d = a − b ⇒ (p − q)d = a − b ∴ d = p − q Subtracting equation (4) from (3), we obtain

. . . (4)

b−c
(q − 1 − r + 1)d = b − c ⇒ (q − r)d = b − c ⇒ d = q − r

. . . (5)

Equating both the values of d obtained in (4) and (5), we obtain a−b b−c p − q = q − r ⇒ (a − b)(q − r) = (b − c)(p − q) ⇒ aq − bq − ar + br = bp − bq − cp + cq ⇒ ( − aq + ar) + (bp − br) + ( − cp + cq) = 0 Thus, the given result is proved.

⇒ − a(q − r) − b(r − p) − c(p − q) = 0 ⇒ a(q −

#466081 Topic: Arithmetic Progression In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is Rs.15 for the first km and Rs. 8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1 of the air remaining in the cylinder at a time.
4 (iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre. (iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8 % per annum.
Solution

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(i) Fare for first km = Rs. 15 Fare for second km = Rs. 15 + 8 = Rs 23

Fare for third km = Rs. 23 + 8 = 31

Here, each subsequent term is obtained by adding a fixed number (8) to the previous term.

Hence, it is in A.P.

(ii) Let us assume, initial quantity of air = 1 .....1) Therefore, quantity removed in first step = 1
4 Remaining quantity after first step 1 − 1 = 3 ....2)
44 Quantity removed in second step
31 3 = ×=
4 4 16 Remaining quantity after second step = 3 − 3 = 9 ....3)
4 16 16 Here, each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

(iii) Cost of digging of 1st meter = 150 Cost of digging of 2nd meter = 150 + 50 = 200 Cost of digging of 3rd meter = 200 + 50 = 250 Here, each subsequent term is obtained by adding a fixed number (50) to the previous term.

Hence, it is an AP.

(iv) Amount in the beginning = Rs. 10000

Interest at the end of 1st year @ 8% = 10000 × 8 Thus, amount at the end of 1st year = 10000 + 800 = 10800 Interest at the end of 2nd year @ 8% = 10800 × 8 Thus, amount at the end of 2nd year = 10800 + 864 = 11664 Since, each subsequent term is not obtained by adding a fixed number to the previous term; hence, it is not an AP.

#466082 Topic: Arithmetic Progression Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10 (ii) a = − 2, d = 0 (iii) a = 4, d = − 3 (iv) a = − 1, d = 1
2 (v) a = − 1.25, d = − 0.25
Solution

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An arithmetic progression is given by a, (a + d), (a + 2d), (a + 3d), . where a = the first term, d = the common difference (i) 10, 10 + 10, 10 + 2(10) and 10 + 3(10) = 10, 20, 30 and 40

(ii) − 2, − 2 + (0), − 2 + 2(0) and − 2 + 3(0) = − 2, − 2, − 2 and − 2 (this is not an A.P)

(iii) 4, 4 + ( − 3), 4 + 2( − 3) and 4 + 3( − 3) = 4, 4 − 3, 4 − 6 and 4 − 9 = 4, 1, − 2 and − 5

(iv)− 1,



1

+

(

1 2

)

,



1

+

(2

1 2

)

and



1

+

(3

1 2

)

=

− 1,



1 , 0 and
2

1 2

(v)− 1.25, − 1.25 + ( − 0.25), − 1.25 + 2( − 0.25) and − 1.25 + 3( − 0.25) = − 1.25, − 1.50, − 1.75 and − 2.0

#466083 Topic: Arithmetic Progression For the following APs, write the first term and the common difference: (i) 3, 1, − 1, − 3, . . . . (ii) − 5, − 1, 3, 7, . . . . .
1 5 9 13 (iii) 3 , 3 , 3 , 3 , . . . . (iv) 0.6, 1.7, 2.8, 3.9, . . .
Solution (i) 3, 1, − 1, − 3… Here, first term, a = 3 Common difference, d = Second term − First term d=1−3= −2

(ii) − 5, − 1, 3, 7… Here, first term, a = − 5 Common difference, d = Second term − First term d = ( − 1) − ( − 5) = − 1 + 5 = 4

(iii)

1 ,

5 ,

9 ,

13

333 3

Here, first term, a = 1 3

Common difference, d = Second term − First term

51 4 d= − =
33 3

(iv) 0.6, 1.7, 2.8, 3.9… Here, first term, a = 0.6 Common difference, d = Second term − First term d = 1.7 − 0.6 = 1.1

#466084 Topic: Arithmetic Progression

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Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (i) 2, 4, 8, 16, . . . .
57 (ii) 2, 2 , 3, 2 , . . . (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, . . .
1111 (viii) − 2 , − 2 , − 2 , − 2 , . . . (ix) 1, 3, 9, 27 (x) a, 2a, 3a, 4a, . . . (xi) a, a2, a3, a4, . . .
(xii) √2, √8, √18, √32, . . . (xiii) √3, √6, √9, √12, . . .
(xiv) 12, 32, 52, 72, . . (xv) 12, 52, 72, 73, . .

Solution a, b, c are said to be in AP if the common difference between any two consecutive number of the series is same ie b − a = c − b ⇒ 2b = a + c

(i) It is not in AP, as the difference between consecutive terms is different.

(ii) It is in AP with common difference d = 5 − 2 = 1 ,

2

2

tn = a + (n − 1)d

a=2
1 t5 = 2 + (5 − 1) 2 Next three terms are 4, 9 , 5
2

(iii) It is in AP with common difference d = − 3.2 + 1.2 = − 2 ,and a = − 1.2 Next three terms are a + (5 − 1)d = − 9.2, a + (6 − 1)d = − 11.2, a + (7 − 1)d = − 13.2

(iv) It is in AP with common difference d = − 6 + 10 = 4, and a = − 10 Next three terms are a + (5 − 1)d = 6, a + (6 − 1)d = 10, a + (7 − 1)d = 14

(v) It is in AP with common difference d = 3 + √2 − 3 = √2, and
a=3 Next three terms are
a + (5 − 1)d = 3 + 4√2, a + (6 − 1)d = 3 + 5√2, a + (7 − 1)d = 3 + 6√2
(vi) It is not in AP since 0.22 − 0.2 ≠ 0.222 − 0.22 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=466086%2C+466084%2C+46… 4/36

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(vii) It is in AP with common difference d = − 4 − 0 = − 4 and a = 0, Next three terms are a + (5 − 1)d = − 16, a + (6 − 1)d = − 20, a + (7 − 1)d = − 24
(viii) It is in AP, with common difference 0, therefore next three terms will also be same as previous ones, i.e., − 1 2
(ix) It is not in AP since 3 − 1 ≠ 9 − 3
(x) It is in AP with common difference d = 2a − a = a and first term is a, Next three terms are a + (5 − 1)d = 5a, a + (6 − 1)d = 6a, a + (7 − 1)d = 7a
(xi) It is not in AP, as the difference is not constant.
(xii) It is in AP with common difference d = √2 and a = √2,
Next three terms are
a + (5 − 1)d = 5√2 = √50, a + (6 − 1)d = √72, a + (7 − 1)d = √98
(xiii) It is not in AP as difference is not constant.
(xiv) It is not in AP as difference is not constant.
(xv) It is in AP with common difference d = 52 − 1 = 24 and a = 1, Next three terms are a + (5 − 1)d = 97, a + (6 − 1)d = 121, a + (7 − 1)d = 145

#466085 Topic: Arithmetic Progression

Sr. no. a

d

n

an

(i)

7

3

8

----

(ii)

-18

---- 10 0

(iii)

----

-3 18 -5

(iv)

-18.9 2.5 ---- 3.6

(v)

3.5

0

105 ----

Fill in the blanks in the following table, given that a is the first term, d the common difference and an is the nth term of the AP:

Solution

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(i) a = 7, d = 3, n = 8, an = ? We know that, For an A.P. an = a + (n − 1)d

= 7 + (8 − 1)3 = 7 + (7)3 = 7 + 21 = 28 Hence, an = 28

(ii) Given that
a = − 18, n = 10, an = 0, d = ? We know that,
an = a + (n − 1)d 0 = − 18 + (10 − 1)d
18 = 9d
18
d= 9 =2 Hence, common difference, d = 2

(iii) Given that d = − 3, n = 18, an = − 5 We know that, an = a + (n − 1)d − 5 = a + (18 − 1)( − 3) − 5 = a + (17)( − 3) − 5 = a − 51 a = 51 − 5 = 46 Hence, a = 46

(iv) a = − 18.9, d = 2.5, an = 3.6, n = ? We know that, an = a + (n − 1)d 3.6 = − 18.9 + (n − 1)2.5
3.6 + 18.9 = (n − 1)2.5
22.5 = (n − 1)2.5
22.5
(n − 1) = 2.5 = 9 n−1=9
n = 10 Hence, n = 10

(v) a = 3.5, d = 0, n = 105, an = ? We know that, an = a + (n − 1)d an = 3.5 + (105 − 1)0 an = 3.5 + 104 × 0 an = 3.5

#466086 Topic: Arithmetic Progression

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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
1 (ii) 11th term of the AP: − 3, − 2 , 2,...., is
1 (A) 28 (B) 22 (C) 38 (D) − 48 2

Solution (i) Given that A.P. 10, 7, 4, … First term, a = 10 Common difference, d = a2 = a1 = 7 − 10 = − 3 We know that an = a + (n − 1)d a30 = 10 + (30 − 1)( − 3) = 10 + (29)( − 3) = 10 − 87 = − 77 Hence, the correct answer is option C.

(ii) A.P. is − 3, − 1 , 2, . . . . . . 2

First term a = − 3

−1

5

Common difference, d = a2 − a1 = − ( − 3) =

2

2

We know that, an = a + (n − 1)d

5

5

a11 = − 3 + (11 − 1) 2 = − 3 + (10) 2 = − 3 + 25 = 22

Hence, the answer is option B.

#466087 Topic: Arithmetic Progression

In the following APs, find the missing terms in the boxes :
Solution (i) For this A.P., a=2 a3 = 26 We know that, an = a + (n − 1)d a3 = 2 + (3 − 1)d 26 = 2 + 2d 24 = 2d d = 12 a2 = 2 + (2 − 1)12 = 14
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Therefore, 14 is the missing term.

(ii) For this A.P., a2 = 13 and a4 = 3 We know that, an = a + (n − 1)d a2 = a + (2 − 1)d 13 = a + d ... (i) a4 = a + (4 − 1)d 3 = a + 3d... (ii) On subtracting (i) from (ii), we get, − 10 = 2d d= −5 From equation (i), we get, 13 = a + ( − 5) a = 18 a3 = 18 + (3 − 1)( − 5) = 18 + 2( − 5) = 18 − 10 = 8 Therefore, the missing terms are 18 and 8 respectively.
(iii) For this A.P., a1 = 5 and
1 a4 = 9 2 We know that, an = a + (n − 1)d a4 = 5 + (4 − 1)d
1 9 = 5 + 3d
2 3
d= 2
a2 = a + d 3
a2 = 5 + 2
13 a2 =
2 3
a3 = a2 + 2 a3 = 8 Therefore, the missing terms are 6 1 and 8 respectively.
2
(iv) For this A.P., a = − 4 and a6 = 6 We know that, an = a + (n − 1)d a6 = a + (6 − 1)d 6 = − 4 + 5d 10 = 5d d=2 a2 = a + d = − 4 + 2 = − 2 a3 = a + 2d = − 4 + 2(2) = 0 a4 = a + 3d = − 4 + 3(2) = 2
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CBSE 11 Engineering Medical Maths Sequences and Series