# NCERT Solutions For Class 10 Maths Chapter 5

## NCERT Solutions Class 10 Maths Arithmetic Progressions

NCERT Solutions for class 10 Maths chapter 5 Arithmetic Progression solutions is an important study material. Questions from arithmetic progression is not only asked in class 10th board examination. But it is also asked in most of the competitive exams after class 12th. To score good marks in class 10th board examination it is recommended to solve NCERT questions provided at the end of each chapter. Solving these questions will help you to understand the chapter in a better way.

NCERT solution for class 10 maths chapter 5 is provided here students can have a look at these solutions when they are facing any kind of difficulty while solving the questions.

### NCERT Solutions Class 10 Maths Chapter 5 Exercises

General Form of an A.P:

Let us consider a series with n number of terms:

a1 + a2 + a3 + a4 + a5 + a6 + a7 ……………………………..  = an

The above series of numbers is said to form an Arithmetic Progression (A.P) if,

a2 – a1 = a3 – a2 = a4 – a3 = …………………………. an – an-1 = d

Here, d = common difference of the A.P and it can be positive negative or zero.

Thus the general form of an A.P can be written as:

a, a+ d, a+ 2d, a+ 3d, a+ 4d, a+ 5d …………………………. a + (n – 1)d

Where a = First Term

And,    d = Common Difference

Exercise 5.1

Q-1: From the given A.P write the first term (a) and common difference (d).

(i) 2, 4, 6, 8, 10…..

Sol.

Here, a = 2

Since d = a2 – a1

Therefore, d = 4 – 2 = 2

Q-2: Check whether the following list of numbers form an A.P? If they form an A.P then write next three terms.

(i) 5, 25, 45, 65………

Sol.

Here, a = 5

Since for an A.P:     a2 – a1 = a3 – a2 = a4 – a3

Therefore,      25 – 5 = 45 – 25 = 65 – 45 = 20

Hence the given number forms an A.P with d = 20

Therefore next three terms of this A.P are: 65 + 20 = 85

85 + 20 =105

105 + 20 = 125

(ii) 12,14,18$\frac{-1}{2},\frac{-1}{4},\frac{-1}{8}……$

Sol.

Here, a = 12$\frac{-1}{2}$

Since for an A.P:      a2 – a1 = a3 – a2 = a4 – a3

Therefore, 14121814$\frac{-1}{4}-\frac{1}{2}\neq \frac{-1}{8}-\frac{1}{4}$

Hence the given numbers does not form an A.P

The nth term of an A.P is given by:

an = a + (n – 1)×d

Here,  a = first term of an A.P and d = common difference

Example-1: Let us consider an A.P 2, 7, 12, 17……. Find its 9th term.

Sol.

Here,                   a = 2

d = 7 – 2 = 17 – 12 = 5

n = 9

Therefore,   a9 = 2+ (9 – 1)5

a9 = 42

Therefore, 9th term of this A.P = 4

Example-2 Fourth term of an A.P is 12 and Sixth term of an A.P is 18. Determine the A.P

Sol.

Given, a4 = 12 and a6 = 18

Let a be the first term and d be the common difference of an A.P

Then               12 = a + (4 – 1)d

Therefore,     12 = a + 3d . .  . . (1)

And                 18 = a + (6 – 1)d

Therefore      18 = a + 5d . . . . . . . (2)

Now, equation (2) – equation (1)

18 – 12 = 5d – 3d    Therefore d = 3

Putting d = 3 in equation (2) we get

18 = a + 5×(3)

Therefore a = 3

Hence, the required A.P is 3, 6, 9, 12, 15, 18, 21………

Example -3 Determine which term of the following A.P : 14, 9, 4 . . . . . . is -96 ?

Sol.

From the given A.P,

a = 14,

d = -5,

an = -96,

n =??

Since,            an = a + (n – 1)d

Therefore   -96 = 14 + (n – 1)×-5

(or)                5n = -96 -14 -5

Therefore       n = -23 (neglecting –ve sign)

Therefore, 23rd term of this A.P is -96.

Example-4 Check whether 328 is the term in the following A.P: 76, 97, 118, 139………

Sol.

From the given A.P:

a = 76,

d = 97 – 76 = 21,

an = 328

Since,                                       an = a + (n – 1)d

Therefore                            328 = 76 + (n – 1)×21

(or)                                       21n = 328 + 21 – 76

(or)                                           n = 13

Since n is a +ve integer

Therefore 328 is the 13th term of this given A.P

Example -5 Find how many two digit numbers are divisible by 5.?

Sol.

Two digit numbers that are divisible by 5 are: 10, 15, 20, 25…………………95

Now, from the given A.P

a = 10

d = 5

an = 95

n = ??

Since,         an = a + (n – 1)d

Therefore 95 = 10 + (n – 1)×5

(or)         5n = 95 – 10 + 5

n = 18

Therefore there are 18 two digit numbers that are divisible by 5.

Example-6 From the given A.P: 15, 8, 1, -6………….. -111. Find 8th term from last term.

Sol.

On reversing the given A.P:

a = -111 (in this case, last term will now become its first term)

d = 7 (If we go from first term to last term then d = -7 (8 – 15), so if we reverse this A.P d will become +7)

n = 8

a8 = ??

Since,          an = a + (n – 1)d

Therefore, a8  = -111 + (8 – 1)×7 = -111 + 49

a8 = -62

Therefore, 8th term from last term = -62

Exercise – 5.2

Q.1 In the given A.P: 26, 40, 54, 68 ………………. which term is 208?

Sol.

From the given A.P:

a = 26,

d = 40 – 26 = 14,

an  = 208,

n = ??

Since,             an = a + (n – 1)d

Therefore, 208 = 26 + (n – 1)×14

14n = 208 + 14 – 26

Hence,            n = 14

14th term of this given A.P is 208

Q2. Find how many total numbers of terms are there in each of the following A.P

(i) 17, 26, 35, 44,……………………….179

Sol.

From the given A.P :

a = 17,

d  = (26 – 17) = 9,

an = 179

n = ??

since,         an = a + (n – 1)d

Therefore, 179 = 17 + (n – 1)×9

9n = 179 – 17 + 9

Therefore n = 19

Hence, there are 19 terms in this given A.P

(ii)13,223,433,.3373$\frac{1}{3},\frac{22}{3},\frac{43}{3},…………………….\frac{337}{3}$

From the given A.P :

a = 13$\frac{1}{3}$,

d = 22313=7$\frac{22}{3} – \frac{1}{3} = 7$,

an = 3373$\frac{337}{3}$,

n = ??

Since, an = a + (n – 1)d

Therefore, 3373=13+(n1)7$\frac{337}{3}=\frac{1}{3}+(n-1)7$

21n=337+21-1

Therefore n= 17

Hence, there are 17 terms in this given A.P

Q3. Seventh term of an A.P is -1 and fourth term is 41. Determine the A.P and hence find its 17th term.

Sol.

Given, a7 = -1 and a4 = 41

Let a be the first term and d be the common difference of an A.P

-1 = a + (7 – 1)d

-1 = a + 6d . . . . . . .  . . (1)

And   41 = a + (4 – 1)d

41 = a + 3d . . . . . . . (2)

Now, equation (1) – equation (2)

-1 – 41 = 6d – 3d    Therefore,   d = -14

Putting d = -14 in equation (2) we get

41 = a + 3×(-14)

Therefore, a = 83

Therefore the required A.P is 83, 69, 55, 12, 41, 27, 13………

Now, from the given A.P:

a = 83,

d = 69 – 83 = -14

n = 17

a17 = ??

Since,                   an = a + (n – 1)d

Therefore          a17 = 83 + (17 – 1)× -14

a17 = 208 + 14 – 26

a17 = -141

Therefore 17th term of this given A.P is -141

Q.4 There are 26 terms in an A.P of which fifth term is 61 and the last term is 292. Find 12th term

Sol.

Given,

a5 = a + (5-1)d = 61

61 = a + 4d…………………………………(1)

a26 = a + (26 – 1)d = 292

292 = a + 25d……………………………(2)

Equation (2) – Equation (1)

292 – 61 = 25d – 4d

Therefore, d = 11

Putting the value of d in equation (1) we get

61 = a + 4 × 11

Therefore, a = 17

Hence a12 = 17 + (12 – 1)×11 = 138

Therefore 12th term of this given A.P is 138

Q.5 Find which term of an A.P is zero, if fourth term and tenth of an A.P are -26 and 52 respectively.

Sol.

Given,

an = 0

a4 = a + (4 – 1)d = -26

-26 = a + 3d………………………………(1)

a10 = a + (10 – 1)d = 52

52 = a + 9d……………………………(2)

Equation (2) – Equation (1) we get:

52 + 26 = 9d – 3d

Therefore, d = 13

On putting the values of d in equation (1) we get:

-26 = a + 3×13

Therefore, a = -65

Now, let nth term be 0

Therefore,  0 = a + (n – 1)d

0 = -65 + (n – 1)×13

0 = -65 + 13n -13

n = 6

Therefore 6th term of the given A.P is zero.

Q.6 Find the common difference d, if 14th term of an A.P exceeds its 7th term by 49.

Sol.

Since,             an = a + (n – 1)d

Therefore,   a14 = a + (14 – 1)d

Similarly,      a7 = a + (7 – 1)d

Now, according to the given condition:

a14 – a7  = 49

49 = {a + (14 – 1)d} – {a + (7 – 1)d}

49 = 13d – 6d

Therefore, d = 7 (Common Difference)

Q.7 Consider a series : 3, 20, 37, 54,71 …………………………..,Which term of this given A.P will be 289 more than its 32nd term.

Sol.

Given, a = 3

d = 20 – 3 = 17

Therefore,  a32 = a + (32 – 1)×d

a32 = 3 + 31×17

a32 = 530

Now, according to the given condition:

an – a32 = 289    (Since, an > a32)

(3+ (n – 1)17) – 530 = 289

819 = 3 + 17n – 17

n = 49

Thus 49th term of this given A.P will be 289 more than its 32nd term.

Q.8 Find the first three terms of an A.P, if sum of 12th and 15th term is 219 and sum of 5th and 7th term is 114.

Sol.

Since, an = a + (n-1)d

Therefore, according to the given conditions:

a12+ a15 = 219

[a + (12-1)d] + [a + (15-1)d] = 219

2a + 25d = 219 ……………………………………………. (1)

And, a5 + a7 = 114

[a +(5-1)d] + [a+(7-1)d] = 114

2a + 10d = 114 ……………………………………………………(2)

Subtracting equation (2) from equation (1) we get

25d – 10d = 105

Therefore, d = 7

Putting values of d in equation (2) we get

2a + 10×7 = 114

Therefore a = 22

Hence 1st term of this A.P = a = 22

2nd term of this A.P = 22+(2-1)×7 = 29

3rd term of this A.P = 22+(3-1)×7 = 36

Q.9 Nikhil started his work in 2002 at an annual salary of Rs 9000 and each year he received an increment of Rs 300. Find in which year his income will reach Rs 12900?

Sol.

Since, each year Nikhil’s salary is increased by Rs 300 and his starting salary was Rs 9000.

Therefore, this forms an A.P with a = 9000 and d = 300

Given, an = 12900 and n = ??

Since, an = a +(n-1)d

Therefore, 12900 = 9000 + (n-1)×300

300n = 12900 + 300 – 9000

Therefore n = 14

Hence, the salary of Nikhil will be 12900 Rs after 14 years from 2002.

Q.10 Find the first three terms of an A.P, if sum of 11th and 14th term is 126 and sum of 4th and 6th term is 66.

Sol.

Since, an = a + (n-1)d

Therefore, according to the given conditions:

a11+ a14 = 126

[a + (11-1)d] + [a + (14-1)d] = 126

2a + 23d = 126 ……………………………………………. (1)

And, a4 + a6 = 66

[a +(4-1)d] + [a+(6-1)d] = 66

2a + 8d = 66 ……………………………………………………(2)

Subtracting equation (2) from equation (1) we get

23d – 8d = 60

Therefore, d = 4

Putting values of d in equation (2) we get

2a + 8×4 = 66

Therefore a = 17

Hence 1st term of this A.P = a = 17

2nd term of this A.P = 17+(2-1)×4 = 21

3rd term of this A.P = 17+(3-1)×4 = 25

Q.11 In the first week of a particular year Deepak saved Rs 10 and then he increased his weekly savings by 2.25Rs. If Deepak was able to increase his weekly savings to Rs 46 in nth week. Find n

Sol.

Money saved by Deepak in 1st week = 10Rs

Money saved by Deepak in 2nd week = 10 + 2.25 = 12.25Rs

Money saved by Deepak in 3rd week = 12.25 + 2.25 =14.50Rs

Therefore, the above given situation forms an A.P: 10, 12.25, 14.50, 16.75…………..

Here, a = 10

d = 2.25

an = 46

Since,       an= a+(n-1)d

Therefore 46= 10+(n-1)2.25

2.25n=46-10+2.25

Therefore n= 17

Hence in 17th week if that year Deepak was able to increase his weekly savings to Rs 46.

Q.12 Find 26th term from the last term of A.P: 7, 12, 17,22………..187.

Sol.

Here, a = 187 (In this case, last term will be first term)

d= -5 (since the given A.P is reversed, therefore d = 12 -17 = 7 – 12 = -5)

And, n = 26

Since, an= a + (n – 1)d

Therefore an = 187 + (26 – 1)×-5

an= 62

Therefore 26th term from the last term is 62.

Q.13 Consider two A.Ps :- 9, 15, 21, 27,…….. and 63,66, 69, 72,…….. Find the value of n, such that the nth term of both the A.P’s is equal.

Sol.

For First A.P, a = 9 and d = 15-9 = 6

Since          an = a+(n-1)d

Therefore an= 9+(n-1)6

(Or)             an= 6n+3

For second A.P, a = 63 and d = 66-63 = 3

Since,          an = a + (n – 1)d

Therefore,  an = 63 + (n – 1)×3

(Or)              an = 3n +60

Now, according to the given condition:

3n+60=6n+3

Therefore n = 19

Therefore, 19th term of both the A.P’s is equal.

Q.14 Find how many three digit numbers are divisible by 9

Sol.

First 3- digit number divisible by 9 = 108

Second 3- digit number divisible by 9 = 117

Similarly last 3-digit number divisible by 9 = 999

Hence it forms an A.P with d = 9,

108, 117, 126……………999

Now, from the above A.P:

a= 108, d= 9, an= 999

n=??

Since, an= a+ (n-1)×d

Therefore 999= 108+ (n-1)×9

9n=999-108+9

Therefore n= 900

Hence, there are 900 three digit numbers that are divisible by 9.

Formula for Sum of first n terms:

Let us suppose that there are n terms in an A.P, therefore sum of first n terms of an A.P is given by Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Here, a = first term, d = common difference and n = number of terms whose sum is to be determined.

REMEMBER:     an = Sn – Sn–1

Example-1: In an A.P -2, 5, 12, 19,……… Find the sum of first 8 terms.

Sol.

Here, a = -2 and   d = (12-5) = 7 and n = 8

Since,        Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore Sn = 82$\frac{8}{2}$ [2×-2+(8-1)7]

= 4 × [45]

Therefore sum of first 8 terms = 180.

Example – 2 How many terms of an A.P 34, 43, 52, 61,………………… must be taken so that there sum is equal to 1002.

Sol.

Given,  a = 34, d = 9 and Sn = 1002

Since, Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore, Sn = 1002 = n2$\frac{n}{2}$ [2×34+(n-1)×9]

1002×2 = n×[59 + 9n]

9n2 + 59n – 2004 = 0

Now, from the above quadratic equation:    a = 9, b = 59 and c = -2004

Substituting the values of a, b and c in quadratic formulae we get:

n=(59)+(59)24(9×2004)2×9$n=\frac{-(59)+\sqrt{(59)^{2}-4(9\times-2004)}}{2\times 9}$  and  n= (59)(59)24(9×2004)2×9$\frac{-(59)-\sqrt{(59)^{2}-4(9\times-2004)}}{2\times 9}$

n=59+3481+7214418$n=\frac{-59+\sqrt{3481+72144}}{18}$  and  n=593481+7214418$n=\frac{-59-\sqrt{3481+72144}}{18}$

n=59+7562518$n=\frac{-59+\sqrt{75625}}{18}$  and  n=597562518$n= \frac{-59-\sqrt{75625}}{18}$

n = 59+27518$\frac{-59+275}{18}$  and  n= 5902518$\frac{-590-25}{18}$

n=12     (neglecting the negative terms)

Therefore, the sum of first 12 terms in the given A.P will be 2004.

Example-3 Find the sum of first 100 odd positive integers.

Sol.

Odd positive integers can be written as: 3, 5, 7, 9, 11……

Here a = 3, d = 2 and n = 100

Since          Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore, Sn = 1002$\frac{100}{2}$ [2×3+(100-1)×2]

= 50[6+198]

= 10200

Therefore the sum of first 100 odd positive integers = 10200

Example – 4 The nth term of any particular series in given by (3n – 5). Find sum of first 16 terms.

Sol.

Given,             an = 3n – 5

Therefore,      a1 = 3×1 – 5 = -2

a2 = 3×2 – 5 = 1

a3 = 3×3 – 5 = 4

Therefore, the series becomes -2, 1, 4 . . . . .

Since   [1 – (-2)] = (4 – 1) = 3 = d

Hence, the above series forms an A.P with a = -2 and d = 3

Since, the sum of first n terms is given by: Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Sn = 162$\frac{16}{2}$ [2×-2+(16-1)×3]

= 8(-4 + 45)

= 328

Therefore the sum of first 16 terms is 328

Exercise 5.3

Q1. Find the sum of the following given AP’s:

(i) 3, 7, 11, 15 . . . . . . .  up to 12 terms.

Sol.

Here a = 3, d = (7 – 3) = 4 and n = 12

Since, the sum of first 12 terms is given by:  Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore,                                            Sn = 122$\frac{12}{2}$ [2×3+(12-1)×4]

= 6×(6 + 44) = 300

Therefore, the sum of first 12 terms is 300.

(ii) –17, –30, –43. . . . . . . . up to 16 terms.

Sol.

Here a = -17, d = [-30 – (-17)] = -13 and n = 16

Since, the sum of first n terms is given by:  Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore,                                                          Sn = 162$\frac{16}{2}$ [2×(-17) + (16-1)× -13]

= 8×(-34 – 195) = -1832

Therefore, the sum of first 16 terms is -1832

(iii) 0.8, 1.3, 1.8. . . . . . . up to 50 terms.

Sol.

Here a = 0.8, d = (1.3 – 0.8) = 0.5 and n = 50

Since, the sum of first n terms is given by: Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore,        Sn = 502$\frac{50}{2}$ [2×0.8+(50 – 1)×0.5]

= 25×(1.6 + 24.5) = 652.5

Therefore, the sum of first 50 terms is 652.5

Q.2 The first term of an A.P is 11 and last term of an A.P is 91. If sum of this A.P is 867, find the total number of terms in that A.P and common difference.

Sol.

Given, a = 11, l = 91 (last term) and Sn = 867

Since,              Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

(or)                  Sn = n2$\frac{n}{2}$ [a+a+(n-1)d]

(or)                  Sn = n2$\frac{n}{2}$ [a + l]       (since an = a +(n – 1)d = l)

Therefore,  867 = n2$\frac{n}{2}$ [11+91]

867×2 = 102n

Therefore,    n = 17

Since              l = a + (17 – 1)d        (since there are total 17 terms)

Therefore, 91 = 11 + 16d

d = 5

Therefore, total numbers of terms are 17 and common difference is 5

Q.3 First term of an A.P is 17 and last term of an A.P is 407. How many terms are there in an A.P also find its sum if common difference is 13.

Sol.

Given, a = 17, d = 13, l = 407 (last term) = an

Since,            an = a + (n – 1)d

Therefore, 407 = 17 + (n – 1)×13

13n = 407 + 13 – 17

Therefore, n = 31

Since               Sn = n2$\frac{n}{2}$ [2a + (n – 1)d]

(or)                  Sn = 312$\frac{31}{2}$ [a+a+(n – 1)d]

(or)                  Sn = 312$\frac{31}{2}$ [a + l]       (since an = a +(n – 1)d = l)

Therefore,     Sn = 312$\frac{31}{2}$ [17+407]

Sn = 6572

Therefore, total number of terms in the given A.P are 31 and there sum is 6572.

Q4. 2nd and 4th terms of an A.P are 30 and 44 respectively. Find the sum of first 45 terms.

Sol.

Given, a2 = 30 and a4 = 44

Since,  an = a + (n – 1)d

Therefore, a2 = a + (2-1)d

(or)             30 = a + d . . . . . . . . . . . (1)

And,     a4 = a + (4 – 1)d

44 = a + 3d . . . . . . . . . (2)

Subtracting equation (1) from equation (2) we get

14 = 2d

Therefore d = 7, on putting value of d in equation (2) we get

44 = a + 3 × 7

Therefore a = 23

Since, the sum of first n terms is given by: Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore,    for n = 45

Sn = 452$\frac{45}{2}$[2×23+(45 – 1)×7]

Sn = 452$\frac{45}{2}$[354]

Therefore, Sn = 7965

Hence, the sum of first 45 terms is 7965

Q.5 If the sum of first n terms of any particular A.P is given by 2n2 – 6n. Find first five terms.

Sol.

Given,             Sn = 2n2 – 6n

Therefore,     S1 = -4     (2×12 – 6×1)

S2 = -4     (2×22 – 6×2)

Similarly,       S3 = 0, S4 = 8, S5 =20

Since             an = Sn – Sn –1

Therefore,   a1 = S1 = -4

a2 = S2 – S1 = -4 – (-4) = 0

a3 = S3 – S2 = 0 – (-4) = 4

a4 = S4 – S3 = 8 – 0 = 8

a5 = S5 – S6 = 20-8 = 12

Therefore, first five terms of an A.P are:  -4, 0, 4, 8, 12

Q.6 A total sum of Rs 7500 is to be used to give 6 cash prizes to students of Saint Marry School for their overall excellent academic performance. If cost of each prize is Rs 200 less than its preceding prize, find the value of each of the prizes.

Sol.

Let the cost of 1st prize be = a

Given, d = -200 (since each prize is Rs 200 less than its preceding prize)

and    Sn = 7500

Since, the sum of first n terms is given by:  Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore,    for n = 6

7500 = 62$\frac{6}{2}$[2×a + (6 – 1)×-200]

7500 = 3(2a – 1000)

2500 = 2a – 1000

a = 1750

Therefore, cost of prizes will be:

1st prize = a = 1750Rs,

2nd prize = (a – d) = 1550Rs,

Similarly, 3rd prize = 1350Rs, 4th prize = 1150Rs, 5th prize = 950Rs and 6th prize = 750Rs

Q.7 In a construction contract there is a penalty for delay in completion of a contract beyond a certain date. For the delay of one day it is Rs 500, for 2nd day it is Rs 600, and Rs 700 for the 3rd day and so on, If the penalty for each succeeding day is increased by Rs 100 and the job was delayed for 28 days. Find how much money the contractor has to pay as penalty??

Sol.

Given, a = 500 (since, penalty on 1st day is 500 Rs)

d = 100 (since, penalty increases by Rs 100 for each succeeding day.)

And     n = 28

Since, the sum of first n terms is given by:  Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore,    for n = 6

Sn = 282$\frac{28}{2}$[2×500 + (28 – 1)×100]

= 14(3700)

= 51800 Rs

Therefore, the contractor paid Rs 51800 as penalty.

Q.8 Fauzia took part in a sports event; there was a race in which 8 balloons were placed in a straight line. A basket was placed at a distance of 5m from the first balloon and all 8 balloons are 2m apart. Fauzia started from basket, picks up the nearest balloon and runs back to the basket to drop it and she continues in the same way until all balloons are in the basket. Find the total distance Fauzia covered.

Sol.

To pick up 1st balloon and drop it into basket Fauzia covered distance = 2×5m = 10m

Similarly for 2nd balloon, distance covered by Fauzia = 2(5+2) = 14m

Similarly for 3rd balloon, distance covered by Fauzia = 2(5+2+2) = 18m

Since there are 8 balloons therefore, n = 8

Here, a = 10 and d = (18-14) = 4

Since, the sum of first n terms is given by: Sn = n2$\frac{n}{2}$ [2a+(n-1)d]

Therefore,    for n = 8

Sn = 82$\frac{8}{2}$[2×10 + (8 – 1)×4]

= 4(48)

= 192m

Therefore, total distance covered by Fauzia = 192m

Q.9 The product of three numbers in an A.P is -10 and their sum is 6. Find the numbers.

Sol.

Let the numbers be (a – d), a, (a + d)

Now, according to the given conditions:

(a – d) + a + (a + d) = 6

3a = 6, therefore a = 2

And,    (a – d)× a ×(a + d) = -10

(2 – d)(2)(2 + d) = -10

4 – d2 = -5

d2 = 9

Therefore d = 3 and d = -3

Now, if d = 3

Then the numbers are: [ -1, 2, 5 ]

And, if d = -3

Then the numbers are: [ 5, 2, -1]

Extra Questions

LAQ

QUESTIONS-:

1. In an A.P., given pn=4,q=2,An=14,$p_{n}=4,\:q=2,\:A_{n}=-14,$, find ‘n’ and p.

Ans.- Given-: pn=4,q=2,An=14,$p_{n}=4,\:q=2,\:A_{n}=-14,$

As we know that, pn$p_{n}$ = p + (n-1)q

$∴$ p + (n-1)2 = 4

$\Rightarrow$ p + 2n – 2 = 4

$\Rightarrow$ p + 2n = 4 + 2

$\Rightarrow$ p +2n = 6 ……(i)

Also, we know that  An$A_{n}$ = n2[2p+(n1)q]$\frac{n}{2}[2p+\left ( n-1 \right )q]$

n2[2p+(n1)2]=14$∴ \:\:\frac{n}{2}[2p+\left ( n-1 \right )2]=-14$

n[2p+(n1)2]=14×2$\Rightarrow n\left [ 2p+\left ( n-1 \right )2 \right ]=-14\times 2$

n[2p+2n2]=28$\Rightarrow n\left [ 2p+2n-2 \right ]=-28$

n[p+(p+2n)2]=28$\Rightarrow n\left [ p+\left ( p+2n \right )-2 \right ]=-28$

n[p+62]=28[Using(i)]$\Rightarrow n\left [ p+6-2 \right ]=-28\:\:\:\:…\left [ Using\:\left ( i \right ) \right ]$

n[p+4]=28$\Rightarrow n\left [ p+4 \right ]=-28$

n=28p+4(ii)$\Rightarrow n=\frac{-28}{p+4}\:\:\:\:…\left ( ii \right )$

Now, putting the value of n=28p+4$n=\frac{-28}{p+4}$ in equation (i),

We get

p+2(28p+4)=6$p+2\left ( \frac{-28}{p+4} \right )=6$

p56p+4=6$\Rightarrow p-\frac{56}{p+4}=6$

p(p+4)56p+4=6$\Rightarrow \frac{p\left ( p+4 \right )-56}{p+4}=6$

p2+4p56=6(p+4)$\Rightarrow p^{2}+4p-56=6\left ( p+4 \right )$

p2+4p56=6p+24$\Rightarrow p^{2}+4p-56=6p+24$

p2+4p566p24=0$\Rightarrow p^{2}+4p-56-6p-24=0$

p22p80=0$\Rightarrow p^{2}-2p-80=0$

p210p+8p80=0$\Rightarrow p^{2}-10p+8p-80=0$

p(p10)+8(p10)=0$\Rightarrow p\left ( p-10 \right )+8\left ( p-10 \right )=0$

(p10)(p+8)=0$\Rightarrow \left ( p-10 \right )\left ( p+8 \right )=0$

p=10,orp=8$∴ p=10,\:or\:p=-8$

Now, putting the value of p = 10 in (i), we have

10 + 2n = 6

2n=4$\Rightarrow 2n=-4$

n=2$\Rightarrow n=-2$, which is not possible

Now putting the value of p = -8 in (i), we get

-8 + 2n = 6

2n=14$\Rightarrow 2n = 14$ n=7$\Rightarrow n = 7$

Hence, p = -8 and n = 7

1. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows -: Rs 100 for the first day, Rs 150 for the second day, Rs 400 for the third day, etc.; the penalty for each succeeding day being Rs 50 more than for the preceding day. How much does a delay of 30 days cost the contractor?

Ans.- Since the penalty for each succeeding day is Rs 50 more than for the preceding day. Therefore, the amount of penalty for different days form an A.P. with first term p (= 100) and the common difference q (= 50). We have to find how much a delay of 30 days costs the contractor. In other words, we have to find the sum of 30 terms of the A.P.

Requiredsum=302(2×100+(301)×5)$∴ Required\:sum=\frac{30}{2} (2\times 100+ ( 30-1 ) \times 5)$ [ becauseAn=n2[2p+(n1)q]$[ \ because A_{n}=\frac{n}{2} [ 2p+ ( n-1 ) q ]$ Requiredsum=15(200+29×50)$\Rightarrow Required \:sum=15 ( 200+29\times 50)$

Requiredsum=15(200+1450)$\Rightarrow Required \:sum=15 ( 200+1450 )$ Requiredsum=15×1650=24750$\Rightarrow Required \:sum=15\times 1650=24750$

Thus, a delay of 30 days will cost the contractor of Rs. 24750.

1. A sum of Rs. 560 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find out the value of each of the prizes.

Ans.- Total amount of seven prizes = Rs.560

Let us assume the value of first prize be Rs.p

By the given condition, we have

P, p-20, p-40,…

Now, m2m1=p20p=20$m_{2}-m_{1}=p-20-p=-20$

Similarly,m3m2=p40p+20=20$Similarly,\:m_{3}-m_{2}=p-40-p+20=-20$ Sincem2m1=m3m2=20,whichisconstant.$Since\:m_{2}-m_{1}=m_{3}-m_{2}=-20,\:which\:is\:constant.$

Therefore, it is an A.P. with common difference (q) = -20

Let K = p + (p-20) + (p-40) + … upto 7 terms

Here, m = p, q =-20 and n = 7

Now,  Kn=n2[2m+(n1)q]$K_{n}=\frac{n}{2}\left [ 2m+\left ( n-1 \right )q \right ]$

K7=72[2p+(71)(20)]$∴ K_{7}=\frac{7}{2}\left [ 2p+\left ( 7-1 \right )\left ( -20 \right ) \right ]$

K7=72[2p+(6)(20)]$\Rightarrow K_{7}=\frac{7}{2}\left [ 2p+\left ( 6 \right )\left ( -20 \right ) \right ]$

560=72[2p120]$\Rightarrow 560=\frac{7}{2}\left [ 2p-120 \right ]$

[Totalamountof7prizes=Rs.560=K7]$\left [ ∴ Total\: amount\: of\: 7\: prizes=Rs.560=K_{7} \right ]$

2p120=560×27$\Rightarrow 2p-120=560\times \frac{2}{7}$

2(p60)=160$\Rightarrow 2\left ( p-60 \right )=160$

p60=1602=80$\Rightarrow p-60 =\frac{160}{2}=80$

p=80+60=140$\Rightarrow p=80+60=140$

Hence, the amount of each prize (in rs.) respectively is-:

140, 140-20, 140-40, 140-60, 140-80, 140-100, 140-120

i.e, Rs.140, Rs.120, Rs.100, Rs.80, Rs.60, Rs.40, Rs.20 respectively.

1. A sum of Rs. 1000 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, then find out the value of each of the prizes.

Ans.- Here, the total amount of ten prizes is = Rs.1000

Now, Let us assume the value of first prize be RS. P

So, by the given condition, we have prizes of denomination (in Rs.) p, p – 20, p – 40, …

Now, m2m1=p20p=20$m_{2}-m_{1}=p-20-p=-20$

m3m2=p40p+20=20$m_{3}-m_{2}=p-40-p+20=-20$

Since, m2m1=m3m2=20$m_{2}-m_{1}=m_{3}-m_{2}=-20$, which is constant.

Therefore, it is an A.P. with common difference (q) = -20

Hence, A10=p+(p20)+(p40)+.upto10terms$A_{10}=p+\left ( p-20 \right )+\left ( p-40 \right )+….upto\:10\:terms$

Here, m = p, q = -20 and n = 10

Now, An=n2[2m+(n1)q][ becauseA10=1000]$A_{n}=\frac{n}{2}\left [ 2m+\left ( n-1 \right )q \right ]\left [ \ because A_{10}=1000 \right ]$

1000=5(2p180)$\Rightarrow 1000=5\left ( 2p-180 \right )$ 1000=10p900$\Rightarrow 1000=10p-900$ 10p=1900$\Rightarrow 10p=1900$ p=190$\Rightarrow p=190$

Hence, amount of each prize (in Rs.) respectively is 190, (190-20), (190-40), (190-60), (190-80), (190-100), (190-120), (190-140), (190-160) and  (190-180)

i.e., Rs.190, Rs.170, Rs.150, Rs.130, Rs.110, Rs.90, Rs.70, Rs.50, Rs.30 and Rs.10.

1. A spiral is made-up of successive semicircles, with centres alternately at P and Q, starting with centre at P, of radii 0.5 cm, 0.1 cm, 1.5 cm, 2.0 cm, …. What is the total length of such a spiral made-up of 13 consecutive semicircles ? (Takeπ=227)$\left ( Take\:\pi =\frac{22}{7} \right )$

Ans.- Since radii of semicircles are 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm …

Now, the length of semicircle (l1)$\left ( l_{1} \right )$ = Perimeter  of first semicircle = πr=227×0.5cm$\pi r=\frac{22}{7}\times 0.5cm$

=227×510=117cm$=\frac{22}{7}\times \frac{5}{10}=\frac{11}{7}cm$

So now, the length of the second semicircle(l2)$semicircle\left ( l_{2} \right )$ = Perimeter of second semicircle

= πr=227×1.0cm=227cm$\pi r=\frac{22}{7}\times 1.0cm=\frac{22}{7}cm$

Similarly, we can get the length of other semicircle as,

l3=337cmandl4=447cm$l_{3}=\frac{33}{7}cm\:and\:l_{4}=\frac{44}{7}cm$ and so on upto 13 semicircles.

Let us assume ‘A’ be the total length of all semicircles.

i.e., p= l3=337cmandl4=447cm$l_{3}=\frac{33}{7}cm\:and\:l_{4}=\frac{44}{7}cm$ … upto 13 terms

Here, p=117,q=227117=117andn=13$p=\frac{11}{7},q=\frac{22}{7}-\frac{11}{7}=\frac{11}{7}\:and\:n=13$

Since, An=n2[2p+(n1)q]$A_{n}=\frac{n}{2}\left [ 2p+\left ( n-1 \right )q \right ]$

A13=132[2×117+(131)×117]$∴ A_{13}=\frac{13}{2}\left [ 2\times \frac{11}{7}+\left ( 13-1 \right )\times \frac{11}{7}\right ]$ =132[227+1327]$= \frac{13}{2}\left [ \frac{22}{7} +\frac{132}{7}\right ]$ =132×1547=143cm$= \frac{13}{2}\times \frac{154}{7}=143cm$

1. 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. Find out in how many rows are 200 logs placed and how many logs are in the top row ?

Ans.- Suppose 200 logs are stacked in ‘s’ rows.

There are 20 logs in the first row and the number of logs in a row is one less than the number of logs in the preceding row. So, number of logs in various rows form an A.P. with first term p (=20) and the common difference q (= -1). As there are 200 logs in all rows.

$∴$ Sum of ‘s’ terms of an A.P. with p = 20 and q = -1 is 200

s2{2p+(s1)q}=200$\Rightarrow \frac{s}{2}\left \{ 2p+\left ( s-1 \right )q \right \}=200$ s2{2×20+(s1)×1}=200$\Rightarrow \frac{s}{2}\left \{ 2\times 20+\left ( s-1 \right )\times -1 \right \}=200$ s2(40s+1)=200$\Rightarrow \frac{s}{2}\left ( 40-s+1 \right )=200$ n(41s)=400$\Rightarrow n\left ( 41-s \right )=400$ s241s+400=0(s25)(s16)=0$\Rightarrow s^{2}-41s+400=0\:\:\Rightarrow \left ( s-25 \right )\left ( s-16 \right )=0$ s=16ors=25$\Rightarrow s=16\:or\:s=25$

Now, If s =25, then number of logs in 25th row is equal to 25th terms of an A.P. with first term 20 and the common difference (-1).

$∴$ number of logs in 25th row = p + 24q = 20 – 24 = -4

Clearly, this is not meaningful.

$∴$ s = 16

Thus, logs are placed in 16 rows.

Number of logs in top row = Number of logs in 16th row

= 16th term of an A.P. with p = 20 and q = -1

Hence, there are 5 logs in the top row.

1. The sum of 4thand 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Calculate A.P.

Ans.Let us assume ‘p’  be the first term and ‘q’  be the common difference of the required A.P.

A4+A8=24$∴ A_{4}+A_{8}=24$

42{2p+(41)q}+82{2p+(81)q}=24$\Rightarrow \frac{4}{2}\left \{ 2p+\left ( 4-1 \right )q \right \}+\frac{8}{2}\left \{ 2p+\left ( 8-1 \right )q \right \}=24$

2{2p+3q}+4{2p+7q}=24$\Rightarrow 2\left \{ 2p+3q \right \}+4\left \{ 2p+7q \right \}=24$

4p+6q+8p+28q=2412p+34q=24$\Rightarrow 4p+6q+8p+28q=24\:\Rightarrow 12p+34q=24$

Or, 6p + 17q = 12 ….(i)

And, A6+A10=44$A_{6}+A_{10}=44$

62{2p+(61)q}+102{2p+(101)q}=44$\Rightarrow \frac{6}{2}\left \{ 2p+\left ( 6-1 \right )q \right \}+\frac{10}{2}\left \{ 2p+\left ( 10-1 \right )q \right \}=44$

3{2p+5q}+5{2p+9q}=44$\Rightarrow 3\left \{ 2p+5q \right \}+5\left \{ 2p+9q \right \}=44$

6p+15q+10p+45q=4416p+60q=44$\Rightarrow 6p+15q+10p+45q=44\:\Rightarrow 16p+60q=44$

Or, 4p+15q = 11 ….(ii)

Now, multiplying (i) by 2 and (ii) by 3, we get

12p + 34q = 24 …..(iii)

12p + 45q = 33 ……(iv)

Now, Subracting (iii) from (iv), we have 11q = 9 or q = 911$\frac{9}{11}$

From (ii), we have 4p+15×911=11$4p+15\times \frac{9}{11}=11$

4p=1113511=12113511=1411$\Rightarrow 4p=11-\frac{135}{11}=\frac{121-135}{11}=\frac{-14}{11}$ p=1411×14=722$\Rightarrow p=\frac{-14}{11}\times \frac{1}{4}=\frac{-7}{22}$

Hence, the required A.P. is -:

722,722+911,722+1811$\frac{-7}{22},\frac{-7}{22}+\frac{9}{11},\frac{-7}{22}+\frac{18}{11}$, ….

Or, 722,12,2922$\frac{-7}{22},\frac{1}{2},\frac{29}{22}$, ….

1. The 17thterm of an A.P. is 5 more than twice its 8th If the 11th term of the A.P. is 43, then calculate its ‘nth’ term.

Ans.- Given, p11=43$p_{11}=43$

P + 10q = 43 ….(i)

[Usingpn=p+(n1)q]$\left [ Using\:p_{n}=p+\left ( n-1 \right )q \right ]$

And, p17=5+2p8$p_{17}=5+2p_{8}$

P + 16q = 5 + 2[p + 7q]

[Usingpn=p+(n1)q]$\left [ Using\:p_{n}=p+\left ( n-1 \right )q \right ]$

P + 16q = 5 + 2p + 14q

$\Rightarrow$ p – 2q = -5     ….(ii)

Subtracting (ii) from (i), we get

p+10q – (p – 2q) = 43 – (-5)

p+10q-p+2q = 43+5    12q=48q=4$\Rightarrow 12q=48\Rightarrow q=4$ …(iii)

From (iii) and (i), we have

P + 40 = 43   p=3$\Rightarrow p=3$

We know that, pn=p+(n1)q$p_{n}=p+\left ( n-1 \right )q$

pn=3+(n1)4=3+4n4=4n1$p_{n}=3+\left ( n-1 \right )4=3+4n-4=4n-1$

Hence, the ‘nth’ term of given A.P. is 4n – 1.

1. Determine the common difference of an A.P. whose first term is 5 and the sum of it first four term is half the sum of the next four terms.

Ans.- Given, First term = p = 5

And, A4=12(thesumofp5,p6,p7,p8)$A_{4}=\frac{1}{2}\left ( the\:sum\:of\:p_{5},p_{6},p_{7},p_{8} \right )$

= 12(A8A4)2A4=A8A43A4=A8$\frac{1}{2}\left ( A_{8} -A_{4}\right )\:\Rightarrow \:2A_{4}=A_{8}-A_{4}\:\Rightarrow 3A_{4}=A_{8}$

A4=42[2p+(41)q]$A_{4}=\frac{4}{2}\left [ 2p+\left ( 4-1 \right )q \right ]$ [UsingAn=n2(2p+(n1)q)]$\left [ Using A_{n} =\frac{n}{2}\left ( 2p+\left ( n-1 \right )q \right )\right ]$

= 2 (10 + 3q) = 20 + 6q

A8=82(10+7q)=4(10+7q)=40+28q$A_{8}=\frac{8}{2}\left ( 10+7q \right )=4\left ( 10+7q \right )=40+28q$

So, 3A4=3(20+6q)=60+18q$3A_{4}=3\left ( 20+6q \right )=60+18q$

3A4=A8$3A_{4}=A_{8}$   (According to question)

60 + 18q = 40 + 28q

10q=20q=2$\Rightarrow 10q=20\:\:\:\Rightarrow q=2$

$∴$ The common difference (q) = 2.

1. The sum of first six terms of an A.P. is 42. The ratio of its 10thterm to its 30th term is 1 : 3. Determine the first and the 13th term of the A.P.

Ans.- Let us assume ‘p’ be the first tern and ‘q’ be the common difference.

Here, A6=42$A_{6} =42$

62{2p+(61)q}=42$\frac{6}{2}\left \{ 2p+\left ( 6-1 \right )q \right \}=42$

2p+5q = 14 …..(i)

Also, p10p30=13$\frac{p_{10}}{p_{30}}=\frac{1}{3}$

p+9qp+29q=13$\Rightarrow \frac{p+9q}{p+29q}=\frac{1}{3}$ 3p+27q=p+29q$\Rightarrow 3p+27q = p+29q$ 3pp=29q27q$\Rightarrow 3p-p = 29q-27q$

2p=2q$\Rightarrow 2p=2q$ …..(ii)

Subtracting (ii) from (i), we get

5q + 2q = 14

7q = 14

q = 2

Now, from (ii), we have -:

2p2×2=0p=2$2p-2\times 2=0\:\:\:\Rightarrow p=2$

So now,  p13=p+12q=2+12×22+24=26$p_{13}=p+12q=2+12\times 2\:\:\:\:\Rightarrow 2+24=26$

Therefore, the first term of the given A.P. is 2 and its 13th term is 26.

1. If the sum of first ‘s’ terms of an A.P. is ‘n’ and the sum of first n terms is ‘s’, then prove that the sum of its first (s+n) terms is –(s+n).

Ans.- Let us assume ‘p’ be the first term and ‘q’ be the common difference

As=n$∴ A_{s}=n$ s2[2p+(s1)q]=n$\Rightarrow \frac{s}{2}\left [ 2p+\left ( s-1 \right ) q\right ]=n$

2p+(s1)q=2ns$\Rightarrow 2p+\left ( s-1 \right )q=\frac{2n}{s}$ ….(i)

And, An=s$A_{n}=s$

n2[2p+(n1)q]=s$\Rightarrow \frac{n}{2}\left [ 2p+\left ( n-1 \right )q \right ]=s$

2p+(n1)q=2sn$\Rightarrow 2p+\left ( n-1 \right )q=\frac{2s}{n}$ ….(ii)

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

(s – 1 – n + 1)q = 2ns2sn$\frac{2n}{s}-\frac{2s}{n}$

(s – n)q = 2(n2s2)sn$\frac{2\left ( n^{2} -s^{2}\right )}{sn}$ = 2(n+s)(ns)sn$\frac{2\left ( n+s \right )\left ( n-s \right )}{sn}$

q=2(s+n)sn$\Rightarrow q=\frac{-2\left ( s+n \right )}{sn}$ …..(iii)

Also, from (i), we get -: 2p=2ns(s1)q$2p=\frac{2n}{s}-\left ( s-1 \right )q$ …..(iv)

And now, the sum of first (s + n) terms

= s+n2[2p+(s+n1)q]$\frac{s+n}{2}\left [ 2p+\left ( s+n-1 \right )q \right ]$

= s+n2[2ns(s1)q+(s+n1)q]$\frac{s+n}{2}\left [ \frac{2n}{s}-\left ( s-1 \right )q+(s+n-1)q \right ]$ [Using (iv)]

= s+n2[2ns+nq]=n(s+n)2[2s+q]$\frac{s+n}{2}\left [ \frac{2n}{s} +nq\right ]=\frac{n\left ( s+n \right )}{2}\left [ \frac{2}{s}+q \right ]$

= n(s+n2)[2s2(s+n)sn]$n\left ( \frac{s+n}{2} \right )\left [ \frac{2}{s}-\frac{2\left ( s+n \right )}{sn} \right ]$ [Using(iii)]

= n(s+n2)[2n2s2nsn]$n\left ( \frac{s+n}{2} \right )\left [ \frac{2n-2s-2n}{sn} \right ]$

= n(s+n2)[2ssn]=(s+n)$n\left ( \frac{s+n}{2} \right )\left [ \frac{-2s}{sn} \right ]=-(s+n)$

1. In an A.P., the sum of first terms is given by -:

An=3n22+5n2$A_{n}=\frac{3n^{2}}{2}+\frac{5n}{2}$. Determine the 25th term of the A.P.

Ans.- Here, An=3n22+5n2=3n2+5n2$A_{n}=\frac{3n^{2}}{2}+\frac{5n}{2}=\frac{3n^{2}+5n}{2}$

An1=3(n1)2+5(n1)2$\Rightarrow A_{n-1}=\frac{3\left ( n-1 \right )^{2}+5\left ( n-1 \right )}{2}$ 3n2+36n+5n52$\Rightarrow \frac{3n^{2}+3-6n+5n-5}{2}$ 3n2n22$\Rightarrow \frac{3n^{2}-n-2}{2}$

Now,

pn=AnAn1$p_{n}=A_{n}-A_{n-1}$

=3n2+5n23n2n22$= \frac{3n^{2}+5n}{2}-\frac{3n^{2}-n-2}{2}$

=5n+n+22=6n+22=2(3n+1)2=3n+1$=\frac{5n+n+2}{2}=\frac{6n+2}{2}=\frac{2\left ( 3n+1 \right )}{2}=3n+1$

pn=3n+1$p_{n}=3n+1$

p1=3×1+1=4$∴ p_{1}=3\times 1+1=4$

p2=3×2+1=7$∴ p_{2}=3\times 2+1=7$

p3=3×3+1=10$∴ p_{3}=3\times 3+1=10$

Hence, the common difference (q) = 7 – 4 = 3

p2=3×2+1=7$∴ p_{2}=3\times 2+1=7$

1. A manufacturer of T.V. sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increase uniformly by a fixed number of every year, calculate-:

(i) The production in the 1st year.

(ii) The production in the 10thyear.

(iii) The total production in first 7years.

Ans.- Let us assume the number of sets produced in 1st year be ‘x’ and ‘y’ be the increase in population every year.

We are given, x + 2y = 600 …..(i)

And, x + 6y = 700 …..(ii)

Now subtracting equation (i) from (ii), we get

4y = 100 or y = 25

Now, Substituting y = 25 in equation (i), we get

x = 550

(1)        Production in the first year = x = 550

(2)        Production in 10th year = x+9y = 550+(9)(25) = 550 + 225 = 775

(3)        Total production in first 7years = x + (x+y) +(x+2y) + …..+ (x+6y)

We know that,

Sum = n2[2x+(n1)y]$\frac{n}{2}\left [ 2x+\left ( n-1 \right )y \right ]$

Here, x = 550, y = 25, n = 7

Sum=n2[2x+(n1)y]$∴ Sum\:=\:\frac{n}{2}\left [ 2x+\left ( n-1 \right )y \right ]$

= 72[2×550+(71)(25)]$\frac{7}{2}\left [ 2\times 550+\left ( 7-1 \right )\left ( 25 \right ) \right ]$

= 72[1100+(6)(25)]$\frac{7}{2}\left [ 1100+\left ( 6 \right ) \left ( 25 \right )\right ]$

= 72[1100+(6)(25)]$\frac{7}{2}\left [ 1100+\left ( 6 \right ) \left ( 25 \right )\right ]$

= 4375

$∴$ Total production in first 7 years = 4375

SAQ 2 MARK

QUESTIONS-:

1. The amount of money in the account every year, when Rs 5000 is deposited at compound interest at 5% per annum.

Ans.-  Here, Principal = Rs.5000

Rate of interest = 5% per annum

Therefore, Annual (A) = 5000(1+5100)$5000\left ( 1+\frac{5}{100} \right )$

Now, the sequence becomes  5000(1+5100)$5000\left ( 1+\frac{5}{100} \right )$,

5000(1+5100)2,5000(1+5100)3,$5000\left ( 1+\frac{5}{100} \right )^{2},5000\left ( 1+\frac{5}{100} \right )^{3},…$

Here, a2a1a3a2$a_{2}-a_{1}\neq a_{3}-a_{2}$

Therefore, the given amount of money in the account every year does not form an A.P.

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

(i) The amount of air present in a cylinder when a vaccum pump removes 16$\frac{1}{6}$ of air remaining in the cylinder at a time.

(ii) The cost of digging a well after every metre of digging, when it costs Rs.250 for the first metre and rises by Rs. 100 for each subsequent metre.

Ans.- Let us assume, x1$x_{1}$= p units

x2=p16p=56punits$∴ x_{2}= p-\frac{1}{6}p=\:\frac{5}{6}p\:units$ x3=56p16×56p=56p536p=2536punits$∴ x_{3}= \frac{5}{6}p-\frac{1}{6}\times \frac{5}{6}p=\:\frac{5}{6}p-\frac{5}{36}p=\frac{25}{36}p\:units$

Similarly, x4=2536p16×2536p=2536p25216p=125216punits$∴ x_{4}= \frac{25}{36}p-\frac{1}{6}\times \frac{25}{36}p=\:\frac{25}{36}p-\frac{25}{216}p=\frac{125}{216}p\:units$

So, the list of numbers is p, 56p,2536p,125216p,$\frac{5}{6}p, \frac{25}{36}p, \frac{125}{216}p, …$

Now here, x2x1x3x2$x_{2}-x_{1}\neq \:x_{3}-x_{2}$

Hence, it is not an A.P.

1. Write first four terms of the A.P. whose first term is 20 and common difference is 20.

Ans.- In this case we have, p= 20, q= 20

Firstterm(p1)=p=20$∴ First\: term (p_{1})=p=20$

Secondterm(p2)=p1+q=20+20=40$Second \:term (p_{2})=p_{1}+q=20+20=40$

Similarly,Thirdterm(p3)=p2+q=40+20=60$Similarly, \:Third \:term (p_{3})=p_{2}+q=40+20=60$

Now,Fourthterm(p4)=p3+q=60+20=80$Now, \:Fourth \:term (p_{4})=p_{3}+q=60+20=80$

Hence, the required four terms are 20, 40, 60, 80.

4.Write the first four terms of the A.P. whose first term is -1 and the common difference is 12$\frac{1}{2}$.

Ans.-  Here we have, p=-1, q=12$\frac{1}{2}$

So, Firstterm(p1)=p=1$First \:term\:(p_{1})=p=-1$

Secondterm(p2)=p1+q=1+12=12$Second \:term\:(p_{2})=p_{1}+q=-1+\frac{1}{2}=-\frac{1}{2}$

Similarly,Thirdterm(p3)=p2+q=12+12=0$Similarly,\:Third \:term\:(p_{3})=p_{2}+q=-\frac{1}{2}+\frac{1}{2}=0$

Similarly,Fourthterm(p4)=p3+q=0+12=12$Similarly,\:Fourth \:term\:(p_{4})=p_{3}+q=0+\frac{1}{2}=\frac{1}{2}$

Hence, the required four terms are 1,12,0,12$-1, -\frac{1}{2}, 0, \frac{1}{2}$.

1. Write four terms of the A.P. whose first term is -1.25 and the common difference is -0.25.

Ans.- Here we have, p=-1.25, q=-0.25

So, Firstterm(p1)=p=1.25$First \:term\:(p_{1})=p=-1.25$

Secondterm(p2)=p1+q=1.25=(0.25)=1.250.25=1.50$Second \:term\:(p_{2})=p_{1}+q= -1.25= (-0.25)= -1.25- 0.25= -1.50$

Similarly,Thirdterm(p3)=p2+q=1.50+(0.25)=1.500.25=1.75$Similarly,\:Third \:term\:(p_{3})=p_{2}+q= -1.50+(-0.25)= -1.50-0.25= -1.75$

Similarly,Fourthterm(p4)=p3+q=1.75+(0.25)=1.750.25=2.0$Similarly,\:Fourth \:term\:(p_{4})=p_{3}+q= -1.75 +(-0.25)=-1.75-0.25= -2.0$

Hence, the required four terms are -1.25, -1.50, -1.75, -2.0

1. Write first four terms of the A.P. when the first term ‘p’ and the common difference ‘q’ are given as follows-:

(i) p = -2, q = 0   (ii) p = 4, q = -3

Ans.- (i)Here,  First term = p = -2

Second term = p+q = -2+0 = -2

Third term = p+2q = -2+2(0) =-2

Similarly, Fourth term = p+3q = -2+3(0) =-2

Hence, the required four terms are -2, -2, -2, -2.

(ii) We have, First term = p = 4

Second term =p+q =4+(-3)=1

Third term = p+2q = 4+2(-3)= -2

Similarly, Fourth term = p+3q= 4+3(-3)= -5

Hence, the required four terms are 4, 1, -2, -5.

1. For the following A.P., write the first term and the common difference of the following 3, 1, -1, -3, …

Ans.- Here A.P. is 3, 1, -1, -3, …

So, the first term p1$p_{1}$= 3

Now, p2p1=13=2,$p_{2}-p_{1}=1-3=-2,$

p3p2=11=2,$p_{3}-p_{2}=-1-1=-2,$ p4p3=3(1)=3+1=2$p_{4}-p_{3}=-3-(-1)=-3+1=-2$

i.e., pn+1pnissameeverytime.$p_{n+1}-p_{n} is\: same\: everytime.$

So therefore, for the given A.P. common difference (q) = -2

1. For the following A.P., write the first term (p1$_{1}$) and the common difference (q) 13,53,93,133,$\frac{1}{3},\:\frac{5}{3},\:\frac{9}{3},\:\frac{13}{3},…$

Ans.- Here the given A.P. is  13,53,93,133,$\frac{1}{3},\:\frac{5}{3},\:\frac{9}{3},\:\frac{13}{3},…$

So, First term ((p1)$(p_{1})$)= 13$\frac{1}{3}$

Now, p2p1=5313=513=43$p_{2}-p_{1}=\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}$

p3p2=9353=953=43$p_{3}-p_{2}=\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}$ p4p3=13393=1393=43$p_{4}-p_{3}=\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}$

i.e. , pn+1pnissameeverytime.$p_{n+1}-p_{n}\: is\: same\: everytime.$

So therefore, for the given A.P. common difference(q)=43$(q) = \frac{4}{3}$

1. For the following A.P.s, write the first term (p) and the common difference (q)-:

(i) -5, -1, 3, 7, …  (ii) 0.6, 1.7, 2.8, 3.9, ….

Ans.- Given here, First term (p) = -5

Now the common difference (q) = -1-(-5) = 3-(-1) = 7-3 = 4

$∴$ Thus, for the given A.P., first term (p) is -5 and the common difference (q) is 4.

(ii) Given here, First term (p) = 0.6

Similarly, the common difference (q) = 1.7 – 0.6 = 2.8 – 1.7 = 3.9 – 2.8 = 1.1

$∴$ Thus, for the given A.P., first term (p) is 0.6 and the common difference (q) is 1.1.

1. Find whether the following list of numbers form an A.P If they form an A.P., find the common difference (q) and write three more terms p,p2,p3,p4,$p,p^{2},p^{3},p^{4},…$

Ans.-  Here the given list of numbers is p,p2,p3,p4,$p,p^{2},p^{3},p^{4},…$

So, first term (p1$p_{1}$) = p

Now, p2p