Arithmetic Progressions

For a sequence to be AP, the difference of consecutive terms should remain constant and that is called the common difference of the AP
(i) Fare for 1^st km = Rs. 15

Fare for 2^nd km = Fare of first km + Additional fare for 1 km
= Rs. 15 + 8
= Rs 23

Fare for 3^rd km = Fare of first km + Fare of additional second km + Fare of additional third km
= Rs. 23 + 8
= Rs 31

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

Hence, it is an AP

(ii) Let us take initial quantity of air = 1

Hence, quantity of air removed in first step = 1/4

Remaining quantity after 1^st step

Quantity removed after 2^nd step = Quantity removed in first step x initial quantity

Remaining quantity after 2^nd step would be:

Here a fixed number is not added to each subsequent term.

Hence, it is not an AP

(iii) Cost of digging of 1st meter = Rs 150

Cost of digging of 2nd meter = 150 + 50

= Rs 200

Cost of digging of 3rd meter = 200 + 50

= Rs 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 at rate of 8%

= 10000 x 8%

= 800

Hence, amount at the end of 1^styear

= 10000 + 800

= 10800

Interest at the end of 2^nd year at rate of 8%

= 10800 x 8%

= 864

Thus, amount at the end of 2^ndyear

= 10800 + 864 = 11664

Since, each subsequent term is not obtained by adding a unique number to the previous term; hence, it is not an AP

(i) Here, first term a₁ = 10 and common difference, d = 10

Hence,

2^nd term a₂ = a₁ + d

= 10 + 10

= 20

3^rd term a₃ = a₁ + 2d

= 10 + 2 x 10

= 30

4^th term a₄ = a₁ + 3d

= 10 + 30

= 40

Therefore,

first four terms of the AP are:

10, 20, 30, 40, ……

(ii) Here,

First term a = -2 and Common difference = 0

Therefore, first four terms of the given AP are:

a₁ = - 2, a₂ = - 2, a₃ = - 2 and a₄ = - 2

(iii) Here, first term a₁ = 4 and common difference d = - 3

We know that a_n = a + (n – 1)d, where n = number of terms

Thus, second term a₂ = a + (2 – 1)d

a₂ = 4 + (2-1)×(-3)

= 4 - 3 = 1

3^rd term a₃ = a + (3 – 1)d

= 4 + (3-1) × (-3)

= 4 - 6

= -2

4^th term a₄ = a + (4-1)d

= 4 + (4 - 1) ×( -3)

= 4 - 9 = -5

Therefore,

First four terms of given AP are:

4, 1, - 2, - 5

(iv) We have,

1^st term = - 1 and d = 1/2

Hence,

2^nd term a₂ = a₁ + d

= -1 + 1/2

= - 1/2

3^rdterm a₃ = a₁ + 2d

= -1 + 2 * 1/2

= 0

4^th term a₄ = a₁ + 3d

= -1 + 3 * 1/2

= 1/2

Therefore,

The four terms of A.P. are -1, - 1/2, 0, 1/2

(v) We have

1^st term = - 1.25 and d = - 0.25

2^nd term a₂ = a + d

= -1.25 - 0.25

= -1.5

3^rd term a₃ = a + 2d

= -1.25 + 2 × (-0.25)

= -1.25 - 0.5

= -1.75

4^th term a₄ = a + 3d

= -1.25 + 3 × (-0.25)

= -2

Therefore, first four terms of the A.P. are: -1.25, -1.5, -1.75 and – 2

(i) Here, first term a = 3

Now,

The Common difference of the A.P. can be calculated as:

a₄ – a₃

= - 3 – ( -1)

= - 3 + 1

= - 2

a₃ – a₂

= - 1 – 1

= - 2

a₂ – a₁

= 1 – 3

= - 2

Now, here a_k+1 – a_k = - 2 for all values of k

Hence, first term = 3 and common difference = - 2

(ii) a₄ – a₃

= 7 – 3

= 4

a₃ – a₂

= 3 – (-1)

= 3 + 1

= 4

a₂ – a₁

= - 1 – (-5)

= - 1 + 5

= 4

Now, here, a_k+1 – a_k = 4 for all values of k

Therefore, first term = - 5 and common difference = 4

(iii) From the question,

a₄ – a₃

=

a₃ – a₂

=

a₂ – a₁

=

Now, a_k+1 – a_k = for all values of k

Therefore,

First term = 1/3 and common difference = 4/3

(iv) a₄ – a₃

= 3.9 – 2.8

= 1.1

a₃ – a₂

= 2.8 – 1.7

= 1.1

a₂ – a₁

= 1.7 – 0.6

= 1.1

Now, here, a_k+1 – a_k = 1.1 for all values of k

Therefore,

First term = 0.6 and common difference = 1.1

For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP

(i) if a_k+1 – a_k is same for different values of k then the series is an AP.

We have, a₁ = 2, a₂ = 4, a₃ = 8 and a₄ = 16

a₄ – a₃ = 16 – 8 = 8

a₃ – a₂ = 8 – 4 = 4

a₂ – a₁= 4 – 2 = 2

Here, a_k+1 – a_k is not same for all values of k.

Hence, the given series is not an AP.

(ii) As per the question:

a1 = 2,

a2 = 5/2,

a3 = 3

And

a4 = 7/2

a₄ – a₃ = – 3= 1/2

a₃ – a₂ = 3 – = 1/2

a₂ – a₁ = – 2 = 1/2

Now, we can observe that a_k+1 – a_k is same for all values of k.

Hence, it is an AP.

And, the common difference = 1/2

Next three terms of this series are:

a₅ = a + 4d

= 2 + 4* 1/2

= 4

a₆ = a + 5d

= 2 + 5 * 1/2

=

a₇ = a + 6d

= 2 + 6 * 1/2

= 5

Hence,

The next three terms of the AP are: 4, 9/2 and 5

(iii) a₄ – a₃ = - 7.2 + 5.2 = - 2

a₃ – a₂ = - 5.2 + 3.2 = - 2

a₂ – a₁ = - 3.2 + 1.2 = - 2

In this, a_k+1 – a_k is same for all values of k.

Hence, the given series is an AP.

Common difference = - 2

Next three terms of the series are:

a₅ = a + 4d

= -1.2 + 4 × (-2)

= -1.2 - 8 = -9.2

a₆ = a + 5d

= -1.2 + 5 × (-2)

= -1.2 - 10 = -11.2

a₇ = a + 6d

= -1.2 + 6 × (-2)

= -1.2 - 12 = -13.2

Next three terms of AP are: - 9.2, - 11.2 and – 13.2

(iv)a₄ – a₃ = 2 + 2 = 4

a₃ – a₂ = - 2 + 6 = 4

a₂ – a₁ = - 6 + 10 = 4

Here, a_k+1 – a_k is same for all values of k

Hence, the given series is an AP.

Common difference = 4

Next three terms of the AP are:

a₅ = a + 4d

= -10 + 4 × 4

= -10 + 16 = 6

a₆ = a + 5d

= -10 + 5 × 4

= -10 + 20 = 10

a₇ = a + 6d

= -10 + 6 × 4

= -10 + 24 = 14

Next three terms of AP are: 6, 10 and 14.

(v) a₄ – a₃ = 3 + 3√2 – 3 - 2√2 = √2

a₃ – a₂ = 3 + 2√2 – 3 - √2 = √2

a₂ – a₁ = 3 + √2 – 3 = √2

Here, a_k+1 – a_k is same for all values of k

Hence, the given series is an AP.

Common difference = √2

Next three terms of the AP are

a₆ = a + 5d = 3 + 5√2

a₇ = a + 6d = 3 + 6√2

Next three terms of AP are: 3 + 4√2, 3 + 5√2 and 3 + 6√2

a₄ – a₃ = 0.2222 – 0.222 = 0.0002

a₃ – a₂ = 0.222 – 0.22 = 0.002

a₂ – a₁ = 0.22 – 0.2 = 0.02

Here, a_k+1 – a_k is not same for all values of k

Hence, the given series is not an AP.

(vii) Here; a₄ – a₃ = - 12 + 8 = - 4

a₃ – a₂ = - 8 + 4 = - 4

a₂ – a₁ = - 4 – 0 = - 4

Since a_k+1 – a_k is same for all values of k.

Hence, this is an AP.

The next three terms can be calculated as follows:

a₅ = a + 4d = 0 + 4(- 4) = - 16

a₆ = a + 5d = 0 + 5(- 4) = - 20

a₇ = a + 6d = 0 + 6(- 4) = - 24

Thus, next three terms are; - 16, - 20 and – 24

(viii) Here, it is clear that d = 0

Since a_k+1 – a_k is same for all values of k.

Hence, it is an AP.

The next three terms will be same, i.e. – 1/2

(ix) a₄ – a₃ = 27 – 9 = 18

a₃ – a₂ = 9 – 3 = 6

a₂ – a₁ = 3 – 1 = 2

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP.

(x) a₄ – a₃ = 4a – 3a = a

a₃ – a₂ = 3a – 2a = a

a₂ – a₁ = 2a – a = a

Since a_k+1 – a_k is same for all values of k.

Hence, it is an AP.

Next three terms are:

a₅ = a + 4d = a + 4a = 5a

a₆ = a + 5d = a + 5a = 6a

a₇ = a + 6d = a + 6a = 7a

Next three terms are; 5a, 6a and 7a.

(xi) Here, the exponent is increasing in each subsequent term.

since, the difference is not same,

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP.

(xii) Different terms of this AP can also be written as follows:

√2, 2√2, 3√2, 4√2, ………..

a4 – a3 = 4√2 - 3√2 = √2

a3 – a2 = 3√2 - 2√2 = √2

a2 – a1 = 2√2 - √2 = √2

Since ak+1 – ak is same for all values of k.

Hence, it is an AP.

Next three terms can be calculated as follows:

a5 = a + 4d = √2 + 4√2 = 5√2

a6 = a + 5d = √2 + 5√2 = 6√2

a7 = a + 6d = √2 + 6√2 = 7√2

Next three terms are; 5√2, 6√2 and 7√2

(xiii) a₄ – a₃ = √12 - √9 = 2√3 – 3 ( √12 =√ 2 x 2 x 3 = 2√3)

a₃ – a₂ = √9 - √6 = 3 - √6

a₂ – a₁ = √6 - √3

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP

(xiv) The given terms can be written as follows:

1, 9, 25, 49, …

Here, a₄ – a₃ = 49 – 25 = 24

a₃ – a₂ = 25 – 9 = 16

a₂ – a₁ = 9 – 1 = 8

Since a_k+1 – a_k is not same for all values of k.

Hence, it is not an AP.

(xv) 1², 5², 7², 73.....

a = 1

d = 5² - 1 = 25 - 1 = 24

d = 7² - 5² = 49 - 25 = 24

d = 74 - 49 = 24

As, common difference is same. The sequence is in A.P.

(i) Given: a = 7, d = 3 and n = 8,

a_n = ?

We know:

a_n = a + (n – 1)d

Thus, a_n = 7 + (8 – 1)3 = 7 + 21 = 28

(ii) Given a = - 18, n = 10, a_n = 0, d = ?

We know that a_n = a + (n – 1)d

Thus, 0 = - 18 + (10 – 1)d

0 = - 18 + 9d

Or, 9d = 18

Or, d = 18/9 = 2

(iii) Given d = - 3, n = 18, a_n = - 5, a = ?

We know that, a_n = a + (n – 1)d

Or, - 5 = a + (18 – 1) (- 3)

Or, - 5 = a – 51

Or, a = - 5 + 51 = 46

(iv) Given a = - 18.9, d = 2.5, a_n = 3.6, n = ?

We know that, a_n = a + (n – 1)d

Or, 3.6 = – 18.9 + (n – 1)2.5

Or, 2.5(n – 1) = 3.6 + 18.9 = 22.5

n – 1 = 22.5/2.5 = 9

n = 9 + 1 = 10

(v) Given that a = 3.5, d = 0, n = 105, a_n= ?

We know:,

a_n = a + (n – 1)d

a_n = 3.5 + (105 – 1)0

a_n = 3.5 + 0

= 3.5

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

Here, a = 10, d = – 3 and n = 30

We know:

a_n = a + (n – 1)d

Or, a₃₀ = 10 + (30 – 1)(- 3)

= 10 – 87
= - 77

Hence,

Answer (C) – 77

(ii) The 11th term of the AP: is
Here, a = - 3, d = 5/2 and n = 11

We know that, a_n = a + (n – 1)d

Or, a₁₁ = - 3 + (11 – 1) 5/2

= - 3 + 10 x (5/2)
= -3 + (5)(5)
= - 3 + 25
= 22

Hence,

Answer (B) 22

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Let three terms a, b and c are in A.P

Therefore, b - a = c - b

Rearranging we get,



And also nth term of an AP is given by

a_n = a + (n - 1)d
where, a = first term
n = number of terms
d = common difference



(i) We know: In AP, middle term is average of the other two terms

Hence, middle term = (2 + 26)/2 = 28/2 = 14

Thus, above AP can be written as 2, 14, 26

(ii) We know: In AP, middle term is average of the other two terms
The middle term between 13 and 3 will be;

(13 + 3)/2 = 16/2 = 8

Now, a₄ – a₃ = 3 – 8 = - 5

a₃ – a₂ = 8 – 13 = - 5

Thus, a₂ – a₁ = - 5

Or, 13 – a₁ = - 5

Or, a₁ = 13 + 5 = 18

Thus, above AP can be written as 18, 13, 8, 3

(iii)
We have, a = 5 and a₄ =
We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are first term and common difference respectively.

Now common difference:

a₄ = a + 3 d

= 5 + 3d

d =

Hence, using d, 2^nd term and 3^rd term can be calculated as:

a₂ = a + d

= 5 +

=

a₃ = a + 2d

= 5 +

= 8

Therefore, the A.P. can be written as:

(iv) Here, a = - 4 and a₆ = 6

We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are first term and common difference respectively.

Common difference:

a₆ = a + 5d

6 = -4 + 5d

5d = 6 + 4 = 10

d = 2

The second, third, fourth and fifth terms of this AP are:

a₂ = a + d = - 4 + 2 = - 2

a₃ = a + 2d = - 4 + 4 = 0

a₄ = a + 3d = - 4 + 6 = 2

a₅ = a + 4d = - 4 + 8 = 4

Thus, the given AP can be written as: - 4, - 2, 0, 2, 4, 6

(v) Given: Second term = 38 and sixth term = -22
So,

a + d = 38.......(1)
a = 38 - d
a + 5d = -22.....(2)
Putting the value of a in equation 2, we get,

38 - d + 5d = -22
38 + 4d = -22
4d = -22 - 38
4d = -60
d = -15

Putting the value of d in equation 1, we get,
a - 15 = 38
a = 38 + 15 = 53
Therefore, the series is
53, 38, 23, 8, -7, -22

Given,

First term, a = 3,

Common difference, d = a₂ – a₁ = 8 – 3 = 5,

nth term, a_n = 78

To find: n = ?

We know that nth term of an A.P is given by:

a_n = a + (n – 1)d

78 = 3 + (n – 1)5

(n – 1) 5 = 78 – 3 = 75

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

To find: Number of terms, n
(i) Given: a = 7, d = 6, a_n = 205, n = ?

We know that nth term of an AP is given by, a_n = a + (n – 1)d

205 = 7 + (n – 1)6

(n – 1)6 = 205 – 7 = 198

n – 1 = 33

n = 34

Thus, 205 is the 34th of term of this AP.

(ii) Given: a = 18, d = 15.5 – 18 = - 2.5, a_n = - 47

We know that nth term of an AP is given by, a_n = a + (n – 1)d

Or, - 47 = 18 + (n – 1)(- 2.5)

Or, (n – 1)(- 2.5) = - 47 – 18 = - 65

Or, n – 1 = 26

Or, n = 27

Thus, - 47 is the 27th term of this AP.

To find: Whether -150 is term of the AP

Given: a = 11, d = 8 – 11 = - 3, a_n = - 150, n = ?

We know that a_n = a + (n – 1)d

Or, - 150 = 11 + (n – 1)(- 3)

Or, (n – 1)(-3) = - 150 – 11 = - 161

Or, n – 1 = 161/3


n - 1 = 53.67

It is clear that 161 is not divisible by three and we shall get a fraction as a result. But number of term cannot be a fraction.

Hence, - 150 is not a term of the given AP.

To Find : 31st term.
Given: 11th term of AP, a₁₁ = 38 and 16th term of AP, a₁₆ = 73

We know that a_n = a + (n – 1)d

Hence,
a₁₁ = a + (11 - 1)d
a₁₁ = a + 10d = 38 .......eq(i)

And,
a₁₆ = a + (16 - 1)d
a₁₆ = a + 15d = 73 ........eq(ii)

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

a + 15d – (a + 10d) = 73 – 38

Or, 5d = 35

Or, d = 7

Substituting the value of d in eq(i) we get;

a + 10 x 7 = 38

Or, a + 70 = 38

Or, a = 38 – 70 = - 32

Now 31st term can be calculated as follows:

a₃₁ = a + 30d

= - 32 + 30 x 7

= - 32 + 210 = 178

Given, a₃ = 12 and a₅₀ = 106

a₃ = a + 2d = 12

a₅₀ = a + 49d = 106

Subtracting 3rd term from 50th term, we get;

a + 49d – a – 2d = 106 – 12

47d = 94

d = 2

Substituting the value of d in 3rd term, we get;

a + 2 x 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a₂₉ = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64

Given, a₃ = 4 and a₉ = - 8

a₃ = a + 2d = 4

a₉ = a + 8d = - 8

Subtracting 3rd term from 9th term, we get;

a + 8d – a – 2d = - 8 – 4 = - 12

6d = - 12

d = - 2

Substituting the value of d in 3rd term, we get;

a + 2(-2) = 4

a – 4 = 4

a = 8

Now; 0 = a + (n – 1)d

0 = 8 + (n – 1)(- 2)

(n – 1)(- 2) = - 8

n – 1 = 4

n = 5

Thus, 5th term of this AP is zero.

To find: Common Difference, d
nth term of an AP is given by,
a_n = a + (n - 1) d
where,
a is the first term, n is the number of terms in an AP and d is the common difference
Tenth and seventeenth terms of this AP can be given as follows:

a₁₀ = a + 9d

a₁₇ = a + 16d
It is give that 17th term is 7 more than the tenth term, So

Subtracting 10th term from 17th term, we get;

a + 16d – a – 9d = 7

7d = 7

d = 1
Hence common difference of the A.P is 1

To find : n such that a_n = a₅₄ + 132
Given: a = 3, d = 15 – 3 = 12

54th term can be given as follows:

a₅₄ = a + ( 54 - 1)d

= 3 + 53 x 12

= 3 + 636
= 639

771 = a + (n – 1)d

771 = 3 + (n -1)12

(n – 1)12 = 771 – 3 = 768

n - 1 = 768/12

n – 1 = 64

n = 65

Thus, the required term is 65th term.

Let the common difference of two AP's be d, their first terms as a and a'
nth term of both the AP's will be given by
a_n = a + (n - 1) d
a'_n = a' + (n - 1) d
Now 100th term of 1st AP will be given by: a₁₀₀ = a + (100 - 1) d = a + 99d
100th term of second AP will be given by: a'₁₀₀ = a' + (100 - 1) d = a' + 99d
Given, a₁₀₀ - a'₁₀₀ = (a + 99 d) - (a' + 99 d)
⇒a₁₀₀ - a'₁₀₀ = (a - a')
So, difference does not depend on number of terms.
Thus, a₁₀₀₀ - a'₁₀₀₀ = 100 = a₁₀₀-a'₁₀₀
So the difference between their 1000th terms is 100.

Since, 100 is the smallest three digit number and it gives a reminder of 2 when divided by 7, therefore, 105 is the smallest three digit number which is divisible by 7

Since, 999 is greatest three digit number, and it gives a remainder of 5, thus 999 – 5 = 994 will be the greatest three digit number which is divisible by 7

Therefore, here we have,

First term (a) = 105,

The last term (a_n) = 994

The common difference = 7

We know that, a_n = a + (n – 1)d

Or, 994 = 105 + (n – 1)7

Or, (n – 1)7 = 994 – 105 = 889

Or, n – 1 = 127

Or, n = 128

Thus, there are 128 three digit numbers which are divisible by 7.

12 is the first number after 10 which is divisible by 4

Since, 250 gives a remainder of 2 when divided by 4, thus 250 – 2 = 248 is the greatest number less than 250 which is divisible by 4.

Here, we have first term (a) = 12, last term (n) = 248 and common difference (d) = 4

Thus, number of terms (n) =?

We know that, a_n = a + (n -1)d

Or, 248 = 12 + (n – 1)4

Or, (n – 1)4 = 248 – 12 = 236

Or, n – 1 = 59

Or, n = 60

Thus, there are 60 numbers between 10 and 250 that are divisible by 4

In first AP: a = 63, d = 2

In second AP: a = 3, d = 7

As per question:

63 + (n – 1) 2 = 3 + (n – 1) 7

⇒ 63 – 3 + (n – 1) 2 = (n – 1) 7

⇒ 60 + 2n – 2 = 7n – 7

⇒ 2n + 58 = 7n – 7

⇒ 2n + 58 + 7 = 7n

⇒ 2n + 65 = 7n

⇒ 7n – 2n = 65

⇒ 5n = 65

⇒ n = 65/5 = 13

Thus, for the 13 value of n, nth term of given two APs will be equal

To Find: A.P
Given: a₃ = 16, a₇ - a₅ = 12
For this question,
We need to find third, fifth and seventh term of an AP by formula of nth term.
We know that, nth term of an AP is given by :
a_n = a + (n - 1) d
So,
Given a₃ = 16 and a₇ – a₅ = 12
where a₃ = third term of the AP, and so on

a₃ = a + 2d = 16 ...........................eq(i)

a₅ = a + 4d

a₇ = a + 6d

As per question;
7th term exceeds the fifth term by 12, So the difference of seventh and fifth term will be 12

a + 6d – a – 4d = 12

⇒ 2d = 12

⇒ d = 6

Substituting the value of d in eq (i), we get;

a + 2 × 6 = 16

Or, a + 12 = 16

Or, a = 16 – 12 = 4

Thus, the AP can be given as follows:
a, a+d, a+2d, a+3d, a+4d..... and thus,

4, 10, 16, 22, 28, …

To find: 20th term from the last
In the given AP
First term,a = 3
common difference, d = 5
we know, nth term of an AP is
a_n = a + (n - 1)d

Now, Let the no of terms in given AP is n
last term = 253

⇒ 253 = a + (n - 1) d

⇒ 253 = 3 + (n -1) x 5

⇒ 253 = 3 + 5n – 5 = – 2

⇒ 5n = 253 + 2 = 255

⇒ n = 255/5 = 51

Total number of terms in AP is 51

Therefore, 20th term from the last term will be 32th term from starting (as, 51 - 20 + 1 = 32)

a₃₂ = a + 31d

= 3 + 31 x 5

= 3 + 155 = 158

Thus, required term is 158

Given, a₈ + a₄ = 24 and a₁₀ + a₆ = 44

we know, nth term of an AP is a + (n - 1)d
where a is first term and d is common difference.
therefore,
a₈ = a + 7d

a₄ = a + 3d

As per question:

a + 7d + a + 3d = 24

Or, 2a + 10d = 24

Or, a + 5d = 12 …………(1)

a₁₀ = a + 9d

Acc. to question :
a + 9d + a + 5d = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22..........(2)

Subtracting equation (1) from equation (2);

a + 7d – a – 5d = 22 – 12

⇒ 2d = 10

⇒ d = 5

Here, a = 5000, d = 200 and a_n = 7000

We know, a_n = a + (n – 1)d

7000 = 5000 + (n – 1)200

(n -1)200 = 7000 – 5000

(n – 1)200 = 2000

n – 1 = 10

n = 11

i.e. after 11 years, starting from 1995, his salary will reach to 7000, so we have to add 10 in 1995, because these numbers are in years.

Thus, 1995 + 10 = 2005

Hence, his salary reached at Rs. 7000 in 2005.

Ramkali's saving in First week = Rs. 5

Ramkali's saving in second week = Ramkali's saving of first week + Ramkali's saving of second week = 5 + 1.75 = Rs. 6.75

Rampkali's saving in third week = 6.75 + 1.75 = Rs. 8.50

And thus an AP will be formed as 5, 6.75. 8.50,......... with common difference of 1.75

Now, Saving in nth week = Rs. 20.75

Therefore,

a = 5, d = 1.75 and a_n = 20.75

We know, a_n = a + (n -1)d

20.75 = 5 + (n – 1)1.75

(n – 1) 1.75 = 20.75 – 5

(n – 1) 1.75 = 15.75

n – 1 = 15.75/1.75 = 9

n = 10

(i) Here, a = 2, d = 5 and n = 10

Sum of n terms can be given as follows:

S₁₀

= 5(4 + 45)

= 5

= 245

Thus, sum of the 10 terms of given AP (S_n)=245

(ii) Here, a = - 37, d = 4 and n = 12

Sum of n terms can be given as follows:

S₁₂

= 6(-74 + 44)

= 6

=-180

Thus, sum of the 12 terms of given AP (S_n)= -180

(iii) Here, a = 0.6, d = 1.1 and n = 100

Sum of n terms can be given as follows:

S₁₀₀

= 50(1.2 + 108.9)

= 50

=5505

Thus, sum of the 100 terms of given AP (S100) = 5505

(iv) Here,

a= , n = 11,

d =

Sum of n terms can be given as follows:

S₁₁

= (+ )

=

=

Thus, sum of the 100 terms of given AP (S_n)=

(i) Here, a = 7, d = 3.5 and last term = 84

Number of terms can be calculated as follows;

a_n =a + (n-1)d

Or, 84 = 7 + (n-1)3.5

Or, (n-1)3.5 = 84-7

Or, n - 1 = 77/3.5 = 22

Or, n = 23

Sum of n terms can be given as follows:

S₂₃

= (14 + 77)

=

=1046

(ii) Here, a = 34, d = - 2 and last term = 10

Number of terms can be calculated as follows:

an = a + (n – 1)d

Or, 10 = 34 + (n – 1)(- 2)

Or, 10 = 34 – (n – 1)(2)

Or, (n – 1)2 = 34 – 10 = 24

Or, n – 1 = 12

Or, n = 13

Sum of n terms can be given as follows:

S₁₃

= [+ ]

=

=

Thus, sum of the given AP (S_n)=

(iii) Here, a = - 5, d = - 3 and last term = - 230

Number of terms can be calculated as follows:

a_n = a + (n – 1)d

Or, - 230 = - 5 + (n – 1)( - 3)

Or, - 230 = - 5 – (n – 1)3

Or, (n – 1)3 = - 5 + 230 = 225

Or, n – 1 = 75

Or, n = 76

Sum of n terms can be given as follows:

S₂₃

= (-10 -225)

=

=-8930

(i) To find: n, S_n
Given:
a_n = 50
a = 5
d = 3

We know, nth term of an AP is given by

a_n = a + (n - 1)d

where a and d are first term and common difference respectively and n are the number of terms of the A.P

Number of terms can be calculated as follows:

a_n = a + (n – 1)d

Or,
50 = 5 + (n – 1)3

Or, (n – 1)3 = 50 – 5 = 45

Or, n – 1 = 15

Or, n = 16

Sum of N terms can be given as follows:

S₁₆

= 8(10 + 45)

= 8 x 55

= 440

(ii) To find: d and S₁₃
Given:
a = 7
a₁₃ = 35

We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are first term and common difference respectively.

So, Common difference can be calculated as follows:

a_n = a + (n – 1)d

Or, 35 = 7 + 12d

Or, 12d = 35 – 7 = 28

Or, d = 7/3

Sum of n terms can be given as follows:

S₁₃

= (14 + 28)

=

=273
S₁₃ = 273

(iii) To Find: a, S₁₂
Given:
d = 3
a₁₂ = 37

We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are first term and common difference respectively.

Therefore, First term can be calculated as follows:

a_n = a + (n – 1)d

Or, 37 = a + 11 x 3

Or, a = 37 – 33 = 4

Sum of n terms can be given as follows:

S₁₂

= 6(8 + 33)

= 6 x 41

= 246

(iv)
To find: d, a₁₀
Given:
a₃ = 15
S₁₀ = 125
Sum of n terms can be given as follows:

S₁₀

125 = 5(2a + 9d)

25 = 2a + 9d .........................eq(i)

We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are first term and common difference respectively.

According to the question; the 3rd term is 15, which means;

a + 2d = 15 ...........eq(ii)

Now,

Subtracting equation 2 times eq(ii) from equation eq(i), we get;

(2a + 9d) – 2(a + 2d) = 25 – 30

Or, 2a + 9d - 2a - 4d = - 5
Or, 5d = - 5
d = -1

Now,

Substituting the value of d in equation (2), we get;

a + 2(- 1) = 15

Or, a – 2 = 15

Or, a = 17

10^th term can be calculated as follows;

a₁₀ = a + 9d

= 17 – 9 = 8

Thus, d = - 1 and 10th term = 8

(v)
To find: a and a₉
Given:
d = 5
S₉ = 75
Sum of n terms can be given as follows:

75

75= (2a + 40)

= 2a x 40

Now,

We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are first term and common difference respectively.

9^th term can be calculated as follows:

a₉ = a + 8d

= -

=

(vi)
To find : n and a_n
Given:
a = 2
d = 8
S_n = 90
Sum of n terms can be given as follows:

90

90= (4 + 8N - 8)

90 = N(2 + 4N - 4)

4N² - 2N - 90 =0

2N² - N - 45 =0

(2N +9)(N -5)

Hence, n = - 9/2 and n = 5

Rejecting the negative value,

We have n = 5

We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are first term and common difference respectively.

Now, 5^th term will be:

a₅ = a + 4d

= 2 + 4 x 8

= 2 + 32 = 34

(vii) To find: n and d
Given:
a = 8
a_n = 62
S_n = 210
Sum of n terms can be given as follows:

210
Since, a_n = a + (n - 1)d, we have

420 = (8 + 62)

420 = n x 70

n = 6

Now, for calculating d:

a₆ = a + 5d

Or, 62 = 8 + 5d

Or, 5d = 62 – 8 = 54

Or, d = 54/5

(viii) To find: n and a
Given:
a_n = 4
d = 2
S_n = - 14
Sum of n terms can be given as follows:

-14

-28 = (a + 4)

.................................eq(i)

We know;

We know, nth term of an AP is

a_n = a + (n - 1)d

where a and d are the first term and common difference respectively.

4 = a + (n – 1)2

Or, 4 = a + 2n – 2

Or, a + 2n = 6

Or, 2n = 6 – a

Or, n = (6 – a)/2 .............................eq(ii)

Using (i) and (ii):

-56 = (6 – a)(a + 4)

24 + 2a – a² = - 56

a² - 2a - 80 = 0

(a + 8)(a – 10)= 0

Therefore, a = - 8 and a = 10

As a is smaller than 10 and d has positive value, hence we’ll take a = - 8

Now, we can find the number of terms as follows:

a_n = a + (n – 1)d

4 = - 8 + (n – 1)2

(n – 1)2 = 4 + 8 = 12

n – 1 = 6, n = 7

Hence, n = 7 and a = - 8

(ix) To find: d
Given: a =3
n = 8
S_n = 192
Sum of n terms can be given as follows:

192 = 4(6 + 7d)

7d = 42

d = 6

(x)
To Find: a
Given: l = 28, S = 144
Sum of n terms can be given as follows:

288 = n (a + 28)

9a + 252 = 288

9a = 288 -252

9a = 36

a = 4

Given that, a = 9, d = 8 and S_n = 636

Now

We know:

⇒ 636

⇒ 636= (18 + 8N -8)

⇒ 636 = N(9 +4N -4)

⇒ 4N² + 5N -636 =0
Now to factorize the above quadratic equation, we need to make the product ( 636 . 4 = 2544) and difference should be 5
⇒ 4N² + 53N - 48N -636 = 0

⇒ N(4N + 53) - 12(4N + 53) = 0
⇒ (N - 12)(4N + 53) = 0

Now, N = - 53/4 and N = 12

Taking the integral value and rejecting the fractional value,
we have n = 12

Sum of n terms of an AP is given by,




where 'a' is first term of AP, n is the number of terms of AP and 'a_n' is last term of AP.

Given,
First term, a = 5
and last term, a_n = 45
Also, sum = 400

Putting the values in the formula, we get



⇒ 800 = 50n




⇒ n = 16


i.e. there are 16 terms in given AP.

Now the nth term of an AP is given by the formula,

a_n=a+(n-1)d

where, a is the first term, n is the number of terms, a_n is the nth term and d is the common difference.

45 = 5 + (16 - 1) x d

⇒15d = 40

We have;
First term, a = 17,
Last term, a_n = 350 and
Common difference, d = 9

We know, nth term formula:

a_n = a + (n – 1)d

350 = 17 + (n – 1)9

(n – 1)9 = 350 – 17

n – 1 = 333/9 = 37

n = 38

and we know, sum of 'n' terms of an AP if first and last term is given is

= 19*367

= 6973

We have; n = 22, d = 7 and a₂₂ = 149

We know;

a_n = a + (n – 1)d
⇒ 149 = a + (22 - 1)7
⇒ 149 = a + 147
⇒ a = 149 - 147
⇒ a = 2

The sum can be calculated as follows:

= 11*151

= 1661

To Find: S₅₁

Given: a₂ = 14, a₃ = 18


We have; a₂ = 14, a₃ = 18 and n = 51

We know,

d = a₃ – a₂ = 18 – 14 = 4

Therefore, d = 4

a₂ – a = 4

14 – a = 4

a = 10

Now, sum can be calculated as follows:

= 51 x (20 + 200)/2

= 51 x 110

= 5610

Given: S₇ = 49, S₁₇ = 289
Sum of n terms of an A.P is given by the formula,

S₇ = 49
Therefore,

⇒ 49 = 7(a + 3 d)

⇒ 7= a + 3 d
⇒ a + 3 d = 7 [1]

Similarly,

289 = 17(a + 8 d)

17 = a + 8 d

a + 8 d = 17 [2]

Subtracting [1] from [2] we get;

a + 8 d – a – 3 d = 17 – 7

⇒ 5 d = 10

⇒ d = 2

Using the value of d in the equation [1], we can find 'a' as follows:

a + 3 d =7

⇒ a + 6 = 7

⇒ a = 1

Using the values of a and d; we can find the sum of first n terms as follows:

⇒ S = n²

(i)Let us take different values for a, i.e. 1, 2, 3 and so on

a = 3 + 4 = 7

a₂ = 3 + 4 x 2 = 11

a₃ = 3 + 4 x 3 = 15

a₄ = 3 + 4 x 4 = 19

Here; each subsequent member of the series is increasing by 4 and hence it is an AP.

The sum of "n" terms is given as:

where, n = no. of terms

a= first term

d = common difference

∴ The sum of first 15 terms is given as:

= (15/2) [2 × 7 + (15-1)×4]

= (15/2) [14 + (14)×4]

= (15/2) [70]

= 525

(ii) Let us take values for a, i.e. 1, 2, 3 and so on

a = 9 – 5 = 4

a₂ = 9 – 5 x 2 = - 1

a₃ = 9 – 5 x 3 = - 6

a₄ = 9 – 5 x 4 = - 11

Here; each subsequent member of the series is decreasing by 5 and hence it is an AP.

∴ The sum of first 15 terms is given as:

= (15/2) [2 × 4 + (15-1)×-5]

= (15/2) [8 + (14)×-5]

= (15/2) [8-70]

= 15(-31)
= -465

As the sum till first term will only contain first term,
S₁ = a₁

We can find the first term as follows:
S_n = 4n - n²

So, by putting the values in place of 1 we will get the sum till that term
S₁ = 4 × 1 - 1² = 3

Now; sum of first two terms can be calculated as follows:

S₂ = 4 × 2 - 2²

= 8 - 4 = 4

Hence; second term = 4 – 3 = 1

Therefore, common difference, d = 1 – 3 = - 2

So, 3rd term can be can be calculated by just replacing n with 3
a₃ = 5 - 2 x 3
a₃ = -1
Similarly
a₁₀ = 5 - 2x 10
a₁₀ = -15

And so on we can calculate any number of terms of this AP

The smallest positive integer which is divisible by 6 is 6 itself and its 40th multiple will be 6 x 40 = 240

So, we have a = 6, d = 6, n = 40 and 40th term = 240.

Sum of first 40 positive integers divisible by 6 can be calculated as follows:

= 20(12 + 234)

= 20*246

= 4920

As we have to calculate sum of first 15 multiples of 8, series will look like:

8, 16, 24, ...........up to 15 terms

Given: First term, a = 8,Common Difference, d = 8 and nth term, n = 15

Sum of first 15 multiples of 8 can be calculated as:

= 15 (8 + 56)

= 15 x 64

= 960

To Find: Sum of odd numbers from 0 and 50

Let us write these numbers

1, 3, 5, ,................, 49

As we can clearly see this forms an AP with first term, a = 1 and common difference, d = 2 and nth term, a_n = 49

Now,

first we need to find number of terms,

for that we have the formula of nth terms of an AP given by

a_n = a + (n - 1) d

Putting the values we get

49 = 1 + (n - 1) 2

48 = (n - 1) 2

(n - 1) = 24

n = 25

So there are 25 odd numbers between 0 and 50

And sum of these 25 numbers are given by using sum of n terms of an AP


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

Given: a = 200, d = 50 and n = 30

We can find the penalty by using the sum of 30 terms:

= 15(400 + 1450)

= 15*1850

= 27750

Hence, he has to pay an amount of Rs. 27750 as penalty

As, each price is decreased by 20 rupees, we can consider amount given to each student in increasing order to be an AP.
Let the lowest price given be 'a' and then prize is increased by Rs 20.
In that case, we have
First term = a
Common difference, d = - 20
No of terms, n = No of students getting prize = 7
Sum of 'n' terms = Total prize money = 700

Now,

First there are 12 classes and each class has 3 sections

Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1.

Thus every class will plant 3 times the number of their class
(for example class (iii) will plant = 3 x 3 = 9 plants)

Similarly,

No. of trees planted by 3 sections of class 2 = 6

No. of trees planted by 3 sections of class 3 = 9

No. of trees planted by 3 sections of class 4 = 12

Its clearly an AP with first term = Number of trees planted by class 1 = 3

We have; a = 3, d = 3 and n = 12

We can find the total number of trees as follows:

= 6 (6 + 33)

= 6 x 39

= 234

Circumference of 1^st semicircle = πr = 0.5π

Circumference of 2^nd semicircle = πr = 1π = π

Circumference of 3^rd semicircle =πr = 1.5π

It is clear that a = 0.5 π, d = 0.5π and n = 13

Hence; the length of the spiral can be calculated as follows:

= (7)

=

= 143 cm

As, the rows are going up, the no of logs are decreasing,
20, 19, 18, ...,
it's an AP
Suppose 200 logs are arranged in 'n' rows, then
We have;
First term, a = 20,
Common difference, d = - 1 and
Sum of n terms, S_n = No of logs = 200

We know;

200

400= (40 - N +1)

N² -41N +400 =0

(N -16)(N –25)

Thus, n = 16 and n = 25

If number of rows is 25 then;

a₂₅ = 20 + 24 x (- 1)

= 20 – 24 = - 4

Since; negative value for number of logs is not possible hence; number of rows = 16

a₁₆ = 20 + 15 x (- 1)

= 20 – 15 = 5

Thus, number of rows = 16 and number of logs in top rows = 5

Distance covered in picking and dropping 1^st potato = 2 x 5 = 10 m

Distance covered in picking and dropping 2^nd potato = 2 (5+3) = 16 m

Distance covered in picking and dropping 3^rd potato = 2(5 + 3 + 3) = 22 m

Therefore, a = 10, d = 6 and n = 10

Total distance can be calculated as follows:

= 5(20 + 54)

= 5*74

= 370 m

To Find: S₁₆
Given: a₃ + a₇ = 6
a₃ x a₇ = 8

nth term of an AP is given by the formula
a_n = a + (n - 1) d
where, a_n = nth term
n = number of term
d = common difference
So now its given that sum of third and seventh term is 6, thus we need to find 3rd and 7th term first,
a₃ = a + 2d

a₇ = a + 6d

As per question;

a₃ + a₇ = 6

a + 2d + a + 6d = 6

2a + 8d = 6

a + 4d = 3

a = 3 – 4d ....................eq (i)

Similarly,

(a + 2d)(a + 6d) = 8

a² + 6ad + 2ad + 12d² = 8 ..........................eq(ii)

Substituting the value of a in equation (ii), we get;

(3 – 4d)² + 8(3 – 4d)d + 12d² = 8

9 – 24d + 16d² + 24d – 32d² + 12d² = 8

9 – 4d² = 8

2d = 1

d = ± 1/2

Using the value of d in equation (1), we get;

a = 3 – 4d

Or, a = 3 – 2 = 1

Sum of first 16 terms is calculated as follows:

S₁₆ = 8[ 2 + (15/2)]

= 4 x 19

S₁₆ = 76

Total distance between top and bottom rung = 2 m 50 cm

Distance between any two rungs = 25 cm




And it is also given that bottom most rungs is of 45 cm length and top most is of 25 cm length.

As it is given that the length of rungs decrease uniformly, it will for an AP with a = 25, a₁₁ = 45 and n = 11

d can be calculated as follows:

a₁₁ = a + 10d

45 = 25 + 10d

10d = 45 – 25 = 20

d = 2

Total length of wood will be equal to the sum of 11 terms:

= 11[ 25 + (10)]

= 11 x 35

= 385 cm

The AP in the above problem is

1, 2, 3, - - - , 49

With first term, a = 1

Common difference, d = 1

nth term of AP = a + (n - 1)d

a_n = 1 + (n - 1)1

a_n = n [1]

Suppose there exist a mth term such that, (m<49)

Sum of first m - 1 terms of AP = Sum of terms following the mth term

Sum of first m - 1 terms of AP = Sum of whole AP - Sum of first m terms of AP

As we know sum of first n terms of an AP is, if last term a_n is given

(m - 1)(1 + m - 1) = 49(1 + 49) - m(1 + m) [using 1]

(m - 1)m = 2450 - m(1 + m)

m² - m = 2450 - m + m²

2m² = 2450

m² = 1225

m = 35 or m = - 35 [not possible as no of terms can't be negative.]

and a_m = m = 35 [using 1]

So, sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35.

Dimensions of 1st step = 50 m x 0.25 m x 0.5 m

Volume of 1^st step = 6.25 m³

Dimensions of 2^nd step = 50 m x 0.5 m x 0.5 m

Volume of 2^ndstep = 12.5 m³

Dimensions of 3^rdstep = 50 m x 0.75 m x 0.5 m

Volume of 3^rdstep = 18.75 m³

Now, we have a = 6.25, d = 6.25 and n = 15

Sum of 15 terms can be calculated as follows:

S₁₅

= 750

Hence, the volume of concrete will be 750m³