# CBSE 11 Engineering Medical Maths Sequences and Series

## Preview text

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

7/4/2018

#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

7/4/2018

(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

7/4/2018

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

7/4/2018

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

7/4/2018

(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

7/4/2018

(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

7/4/2018

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

7/4/2018

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 