Arithmetic Progressions

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

4 = a + (7 - 1)(- 4) …..(Given Data)

4 = a - 24

a = 24 + 4 = 28

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a_n = 3.5 + (101 - 1)0

= 3.5

(Here we have given d = 0, therefore its an constant A.P. therefore its all terms will be equal and constant)

a_{1 =} - 10

a_{2 =} - 6

a_{3 =} - 2

a_{4 =} 2

a_{2 -} a_{1 =} 4

a_{3 -} a_{2 =} 4

a_{4 -} a_{3 =} 4

a₂ - a_{1 =} a_{3 -} a_{2 =} a_{4 -} a_{3 =} 4

Therefore, its an A.P with d = 4

First term, a = - 5

Common difference,

n = 11

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a_{11 =} - 5 + (11 - 1)(5/2)

a_{11 =} - 5 + 25

= 20

First term, a = - 2

Second Term, d = - 2

a_{1 =} a = - 2

a_{2 =} a + d = - 2 + (- 2) = - 4 (using formula a_{n =} a + (n - 1)d)

similarly

a_{3 =} - 6

a_{4 =} - 8

so A.P is

- 2, - 4, - 6, - 8

first two terms of an AP are a = - 3 and a_{2 =} 4.

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a_{2 =} a + d

4 = - 3 + d

d = 7

Common difference, d = 7

a_{21 =} a + 20d

= - 3 + (20)(7)

= 137

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a_{2 =} a + d = 13 …..(1)

a_{5 =} a + 4d = 25 ……(2)

Solving 1 and 2

From 1 we have

a = 13 - d

using this in 2, we have

13 - d + 4d = 25

13 + 3d = 25

3d = 12

d = 4

a = 13 - 4 = 9

a_{7 =} a + 6d

= 9 + 6(4)

= 9 + 24 = 33

Let nth term of the given AP be 210.

Here, first term, a = 21

and common difference, d = 42 – 21 = 21 and a_n = 210

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

210 = 21 + (n - 1)21

189 = (n - 1)21

n - 1 = 9

n = 10

So the 10th term of an AP is 210.

Given, the common difference of AP i.e., d = 5

Now,

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a₁₈ -a_{13 =} a + 17d – (a + 12d)

= 5d

= 5(5)

= 25

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a₁₈ - a₁₄ = a + 17d - (a + 13d)

32 = 4d

d = 8

so common difference is 8

Let a and A be the first terms of Two Aps

a = - 1

A = - 8

Let d and D be the common differences of two Aps

D = d ……..(i)

Let a₄ and A₄ be the 4^th terms of two Aps

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a₄ -A_{4 =} a + 3d – (A + 3D)

= a + 3d - A - 3d (using (i))

= a - A

= - 1 - (- 8)

= 7

Let a and d be first term and common difference respectively

According to the question,

⇒ 7a_{7 =} 11a₁₁

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

we have,

7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a + 68d = 0

⇒ a + 17d = 0

⇒ a_{18 =} 0

18th term of AP is 0

First term, a = - 11

Common difference, a₂ -a = - 8 - (- 11) = 3

Let the last term be a_n

a_{n =} a + (n - 1)d

49 = - 11 + (n - 1)3

49 + 11 = (n - 1)3

n - 1 = 60/3

n - 1 = 20

n = 21

4^th term from last will be 18^th term from starting

And a₁₈ = a + 17d

= - 11 + 17(3)

= 40

Alternatively we can use the direct formula

a_n = l - (n - 1)d

where, a_n = nth term from last of an AP

l = last term

d = common difference

Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i.e., 1, 2, 3, ….., 100.

Given,

First term, a = - 5

Common Difference, d = 2

Using the formula,

Where S_n is the sum of first n terms

n = no of terms

a = first term

d = common difference

S₄ = (6/2)[ 2 (- 5) + (6 - 1)2]

= 3 (- 10 + 10)

= 0

Given,

First term, a = 10

Common Difference, d = 6 - 10 = - 4(a₂ - a₁)

Using the formula,

Where S_n is the sum of first n terms

n = no of terms

a = first term

d = common difference

S₁₆ = (16/2)[ 2(10) + (16 - 1) (- 4)]

= 8(20 - 60)

= - 320

Given

First term, a = 1,

Nth term, a_n = 20,

Some of n terms, S_n = 299

Using the formula

Where S_n = Sum of first n terms

n = no of terms

l = a_n = last term

2S_n = n(a + a_n)

2(399) = n(1 + 20)

n = 798/21

n = 38

The first five multiples of 3 are 3, 6, 9, 12 and 15.

Here, first term, a = 3, common difference, d = 6 – 3 = 3 and number of terms, n = 5

Using the formula,

Where S_n is the sum of first n terms

n = no of terms

a = first term

d = common difference

S₅ = (5/2)[ 2(3) + (5 - 1)(3)]

= (5/2)[ 6 + 12]

= 5(9)

= 45

We have a₁ = - 1 , a₂ = - 1, a₃ = - 1 and a₄ = - 1

a₂ - a₁ = 0

a₃ - a₂ = 0

a₄ - a₃ = 0

Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.

We have a₁ = 0, a₂ = 2, a₃ = 0 and a₄ = 2

a₂ - a₁ = 2

a₃ - a₂ = - 2

a₄ - a₃ = 2

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

We have a₁ = 1 , a₂ = 1, a₃ = 2 and a₄ = 2

a₂ - a₁ = 0

a₃ - a₂ = 1

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

We have a₁ = 11, a₂ = 22 and a₃ = 33

a₂ - a₁ = 11

a₃ - a₂ = 11

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

We have a₁ = , a₂ = and a₃ =

a₂ - a₁ =

a₃ - a₂ =

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

We have a₁ = 2 , a₂ = 2², a₃ = 2³ and a₄ = 2⁴

a₂ - a₁ = 2² - 2 = 4 - 2 = 2

a₃ - a₂ = 2³ - 2² = 8 - 4 = 4

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.

False

Clearly, the difference of successive terms in not same, all though, a₂ - a₁ = a₃ - a₂ but a₄ - a₃ = a₃ - a₂ therefore it does not form an AP.

True

Given

First term, a = - 3

Common difference, d = a₂ - a₁ = - 7 - (- 3) = - 4

a₃₀ - a₂₀ = a + 29d - (a + 19d)

= 10d

= - 40

It is so because difference between any two terms of an AP is proportional to common difference of that AP

Suppose there are two AP's with first terms a and A

And their common differences are d and D respectively

Suppose n be any term

a_n = a + (n - 1)d

A_n = A + (n - 1)D

As common difference is equal for both AP's

We have D = d

Using this we have

A_n - a_n = a + (n - - 1)d - [ A + (n - 1)D]

= a + (n - 1)d - A - (n - 1)d

= a - A

As a - A is a constant value

Therefore, difference between any corresponding terms will be equal to a - A.

No

Let 0 be the n th term of given AP. Such that a_n = 0.

Given that first term a = 31, common difference, d = 28 – 31 = - 3

The nth term of an AP, is

a_n = a + (n - 1)d

0 = 31 + (n - 1) (- 3)

Since, n should be positive integer. So, 0 is not a term of the given AP.

No, because the total fare after each km is

15, 15 + 8, 15 + 8(2), 15 + 8(3) …….

15, 23, 31, 39….

a₁ = 15, a₂ = 23, a₃ = 31 and a₄ = 39

a₂ - a₁ = 23 - 15 = 8

a₃ - a₂ = 31 - 23 = 8

a₄ - a₃ = 39 - 31 = 8

Since, all the successive terms of the given list have same difference i.e., common difference = 8.

Hence, the total fare after each km forms an AP.

(i) The fee charged from a student every month by a school for the whole session is

400, 400, 400, 400, …

which from an AP, with common difference , d = 400 – 400 = 0

(ii) The fee charged month by a school from I to XII is

250, 250 + 50, 250 + 50(2), 250 + 50(3)………

250, 300, 350, 400……

Which from an AP, with common difference d = 300 – 250 = 50

(iii)

=

= 100t

So, the amount of money in the account of Varun at the end of every year is

1000, (1000 + 100 × 1), (1000 + 100 × 2), (1000 + 100 × 3), …

i.e., 1000, 1100, 1200, 1300, …

which form an AP, with common difference , d = 1100 – 1000 = 100

(iv) Let the number of bacteria in a certain food = x

Since, they double in every second.

x, 2x, 4x, 6x . . . . .

a₂ - a₁ = 2x - x = x

a₃ - a₁ = 4x - 2x = 2x

Since, the difference between each successive term is not same. So, the list does form an AP.

Put n = 1, 2, 3, 4

We get

a₁ = 2(1) - 3 = - 1

a₂ = 2(2) - 3 = 1

a₃ = 2(3) - 3 = 3

a₄ = 2(4) - 3 = 5

List of AP is - 1, 1, 3, 5 . . . .

a₂ - a₁ = 1 - (- 1) = 2

a₃ - a₂ = 3 - 1 = 2

a₄ - a₃ = 5 - 3 = 2

which form an AP, with common difference , d = 2

Put n = 1, 2, 3, 4

We get

a₁ = 3(1)² + 5 = 8

a₂ = 3(2)² + 5 = 17

a₃ = 3(3)² + 5 = 32

a₄ = 3(4)² + 5 = 53

List of AP is 8, 17, 32, 53. . . .

a₂ - a₁ = 17 - 8 = 9

a₃ - a₂ = 32 - 17 = 15

As

a₃ - a₂ is not equal to a₂ - a₁

the list is not an AP

Put n = 1, 2, 3, 4

We get

a₁ = 1 + 1 + (1)² = 3

a₂ = 1 + 2 + (2)² = 7

a₃ = 1 + 3 + (3)² = 13

a₄ = 1 + 4 + (4)² = 21

List of AP is 3, 7, 13, 21. . . .

a₂ - a₁ = 7 - 3 = 4

a₃ - a₂ = 13 - 7 = 6

As

a₃ - a₂ is not equal to a₂ - a₁

the list is not an AP

(A₁)

AP is 2, - 2, - 6, - 10, ….

So common difference is simply

a₂ - a₁ = - 2 - 2 = - 4 = (B₃)

(A₂)

Given

First term, a = - 18

No of terms, n = 10

Last term, a_n = 0

By using the nth term formula

a_{n =} a + (n - 1)d

0 = - 18 + (10 - 1)d

18 = 9d

d = 2 = (B₅)

(A₃)

Given

First term, a = 0

Tenth term, a₁₀ = 6

By using the nth term formula

a_{n =} a + (n - 1)d

a₁₀ = a + 9d

6 = 0 + 9d

= (B₆)

(A₄)

Let the first term be a and common difference be d

Given that

a₂ = 13

a₄ = 3

a₂ - a₄ = 10

a + d - (a + 3d) = 10

d - 3d = 10

- 2d = 10

d = - 5

= (B₁)

Here

as difference of successive terms are equal therefore its an AP with common difference

next three term will be

Here

as difference of successive terms are equal therefore its an AP with common difference

next three term will be

Here

as difference of successive terms are equal therefore its an AP with common difference

next three term will be

Here

a₁ = a + b

a₂ = (a + 1) + b

a₃= (a + 1) + (b + 1)

a₂ - a₁ = (a + 1) + b - (a + b) = 1

a₃ - a₂ = (a + 1) + (b + 1) - (a + 1) - b = 1

as difference of successive terms are equal therefore its an AP with common difference

next three term will be

(a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1(2), (a + 1) + (b + 1) + 1(3)

(a + 2) + (b + 1), (a + 2) + (b + 2), (a + 3) + (b + 2)

Here a₁ = a

a₂ = 2a + 1

a₃ = 3a + 2

a₄= 4a + 3

a₂ - a₁ = a + 1

a₃ - a₂ = a + 1

a₄ - a₃ = a + 1

as difference of successive terms are equal therefore its an AP with common difference

next three term will be

4a + 3 + a + 1, 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1)

5a + 4, 6a + 5, 7a + 6

First three terms of AP are :

a + d, a + 2d, a + 3d

First three terms of AP are :

a + d, a + 2d, a + 3d

- 5 + 1 (- 3), - 5 + 2 (- 3), - 5 + 3 (- 3)

- 8, - 11, - 13y

First three terms of AP are :

a + d, a + 2d, a + 3d

For the above terms to be in AP

The difference of successive terms should be equal

i.e.,

a₅ - a₄ = a₄ - a₃ = a₃ - a₂ = a₂ - a₁ = d

where d let be common difference

7 - a = b - 7 = 23 - b = c - 23

Implies b - 7 = 23 - b

2b = 30

b = 15 (eqn 1)

Also

7 - a = b - 7

from eqn 1

7 - a = 15 - 7

a = - 1

and

c - 23 = 23 - b

c - 23 = 23 - 15

c - 23 = 8

c = 31

so a = - 1

b = 15

c = 31

and the sequence - 1, 7, 15, 23, 31 is an AP

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

Given

5^th term, a₅ = 19

And by using the nth term formula i.e. a_n = a + (n - 1)d

a + 4d = 19

a = 19 - 4d (eqn 1)

also,

term - 8^th term = 20

a + 12d - (a + 7d) = 20

5d = 20
d = 4

Using this in eq[1], we have
a = 19 - 4(4) = 3

Hence, AP is
a, a + d, a + 2d, ...
i.e. 3, 7, 11, ...

Let the first term, common difference and number of terms of an AP are a, d and n, respectively.

Let the no of terms be x

By using the nth term formula i.e. a_n = a + (n - 1)d

a₂₆ = a + 25d = 0 (given)

a = - 25d (eqn 1)

a₁₁ = a + 10d = 3 (given)

- 25d + 10d = 3 (using eqn 1)

- 15d = 3

(eqn 2)

using eqn2 and eqn1 we get

Also,

(given)

hence

Common difference = - 1/5

No of terms = 27

Let the first term, common difference be a and d respectively.

As we know

a_n = a + (n - 1)d

and Given

a₅ + a₇ = 52

a + 4d + a + 6d = 52

2a + 10d = 52

a + 5d = 26

a = 26 - 5d [ eqn i]

also we have given

a₁₀ = 46

a + 9d = 46

26 - 5d + 9d = 46 [ from eqn i]

4d = 46 - 26

4d = 20

d = 5

using this value in eqn I , we get

a = 26 - 5d

a = 26 - 5(5)

a = 1

as a = 6 and d = 5

So, required AP is a, a + d, a + 2d, a + 3d, …. i.e., 1, 1 + 5, 1 + 2(5), 1 + 3(5), …. i.e., 1, 6, 11, 16, …

Let the a be first term and d be common difference

As we know

a_n = a + (n - 1)d

Given,

a₇ = a₁₁ - 24

a + 6d = a + 10d - 24

10d - 6d = 24

4d = 24

d = 6

given a = 12

then, a₂₀ = a + 19d

a₂₀ = 12 + 19(6)

= 12 + 114

= 126

Let the first term, common difference and number of terms of an AP are a, d and n respectively.

Given : 9^th term is zero i.e. a₉ = 0

To prove : a₂₉ = 2a₁₉

Proof :

As a₉ = 0

a + 8d = 0 [ eqn i]

Using the nth term formula i.e. a_n = a + (n - 1)d

Taking LHS

a₂₉ = a + 28d

= (2 - 1)a + (36 - 8)d

= 2a - a + 36d - 8d

= 2a + 36d - (a + 8d)

= 2(a + 18d) - 0 [ using i]

= 2a₉ [ as a₉ = a + 8d]

= RHS

let the first term, common difference of an AP are a and d respectively.

a = 7

d = a₂ - a₁ = 10 - 7 = 3

Let the nth term of this AP is 55

Then

a_n = 55

a + (n - 1)d = 55

7 + (n - 1)3 = 55

3(n - 1) = 48

n - 1 = 16

n = 17

so the 55 is a term of given AP

and 55 is the 17^th term of given AP

Let a₁ = k² + 4k + 8

a₂ = 2k² + 3k + 6

a₃ = 3k² + 4k + 4

Three terms will be in an AP if

a₂ - a₁ = a₃ - a₂

2k² + 3k + 6 - (k² + 4k + 8) = 3k² + 4k + 4 - (2k² + 3k + 6)

k² - k - 2 = k² + k - 2

2k = 0

k = 0

Let the three parts of the number 207 are

a₁ = a - d

a₂ = a

a₃ = a + d

Clearly a₁, a₂ and a₃ are in AP with common difference as d.

Now, by given condition,

Sum = 207

a₁ + a₂ + a₃ = 207

(a - d) + a + (a + d) = 207

3a = 207

a = 69

Also,

a₁a₂ = 4623

(a - d)a = 4623

(69 - d)69 = 4623

69 - d = 67

d = 69 - 67

d = 2

Hence, required three parts are 67, 69, 71.

Let A, B and C be the three angles in AP (in degrees)

And A being the least and C being the greatest

Then

C = 2A [ Given][ eqn i]

B - A = C - B

2B = A + C

2B = A + 2A [ using eqn i]

2B = 3A

B = [ eqn ii]

Using the angle sum property of triangle

A + B + C = 180^O

[ using i and ii]

9A = 360

A = 40^O

[ using ii]

C = 2(40) = 80^O [ using i]

First term of first AP, a = 9

First term of second AP, A = 24

Common difference of first AP, d = 7 - 9 = - 2

Common difference of second AP, D = 21 - 24 = - 3

Given that nth term is same i.e

A_n = a_n

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

A + (n - 1)D = a + (n - 1)d

24 + (n - 1) (- 3) = 9 + (n - 1) (- 2)

24 - 3n + 3 = 9 - 2n + 2

27 - 3n = 11 - 2n

3n - 2n = 27 - 11

n = 16

i.e their 16^th term is equal in both cases,

and a₁₆ = A₁₆ = a + 15d = 9 + 15 (- 2) = - 21

Let the first term and common difference of an AP are a and d, respectively.

Given

a₃ + a₈ = 7

As we know, nth term of an AP is

a_{n =} a + (n - 1)d

where a = first term

a_n is nth term

d is the common difference

a + 2d + a + 7d = 7

2a + 9d = 7

2a = 7 - 9d [ Eqn 1]

a₇ + a₁₄ = - 3

a + 6d + a + 13d = - 3

2a + 19d = - 3

7 - 9d + 19d = - 3 [ using eqn i]

7 + 10d = - 3

10d = - 10

d = - 1

using this value in eqn i

2a = 7 - 9 (- 1)

2a = 16

a = 8

Now,

a₁₀ = a + 9d

= 8 + 9 (- 1)

= 8 - 9 = - 1

Given AP, – 2, - 4, - 6, …, - 100.

Here,

first term, a = - 2

common difference, d = - 4 – (2) = - 2

last term, l = - 100.

We know that, the nth term 'a_n' of an AP from the end is

a_n = l - (n - 1)d

where l is the last term and d is the common difference.

12th term from the end,

a₁₂ = - 100 - (12 - 1) - 2

= - 100 + 22

= - 78

Hence, the 12th term from the end is - 78

Given AP is 53, 48, 43, ….

Whose, first term, a = 53

common difference, d = 48 – 53 = - 5

Let nth term of the AP be the first negative term.

Then, we have to find the least value for which

a_n < 0

a + (n - 1)d < 0

53 + (n - 1) (- 5) < 0

53 - 5(n - 1) < 0

5(n - 1) > 53

so n will be least natural number greater than

i.e., 12th term is the first negative term of the given AP.

Here, the first number is 11, which divided by 4 leave remainder 3 between 10 and 300.

And the next number is 15 and the next is 19

Last term before 300 is 299, which divided by 4 leave remainder 3.

So, the list is

11, 15, 19, …, 299

Clearly, This is an AP

With first term, a = 11

Common difference, d = 15 - 11 = 4

Last term, a_n = 299

Using the nth term formula

a_n = a + (n - 1)d

299 = 11 + (n - 1) (- 4)

299 - 11 = (n - 1) (- 4)

288 = (n - 1) (- 4)

n - 1 = 72

n = 73

so there are 73 numbers between 10 and 300 which leaves 3 as remainder when divided by 4.

Here,

first term, a =

common difference, d

last term, a_n = 4 =

using the nth term formula

a_n = a + (n - 1)d

13 = - 4 + (n - 1) [ multiplication by 3 on the both side]

13 + 4 = n - 1

n = 18

as n is even middle terms will be

a₉ + a₁₀ = a + 8d + a + 9d

= 2a + 17d

= 3

Let the first term, common difference and the number of terms of an AP are a, d and n respectively.

Given that, first term, a = - 5

last term, a_n = 45

Sum of the terms of the AP , S_n = 120

We know that, if last term of an AP is known, then sum of n terms of an AP is,

240 = 40n

n = 6

Also we know the nth term formula

a_n = a + (n - 1)d

45 = - 5 + (6 - 1)d

50 = 5d

d = 10

Hence, number of terms and the common difference of an AP are 6 and 10 respectively.

Here First term, a = 1

Common difference, d = - 2 - 1 = - 3

Last term, a_n = - 236

As we know

a_n = a + (n - 1)d

- 236 = 1 + (n - 1) (- 3)

- 236 - 1 = (n - 1) (- 3)

- 237 = (n - 1) (- 3)

n - 1 = 79

n = 80
Using the sum of first n terms formula, if last term is given

i.e,

= 80/2(1 - 236)

= 40 (- 235)

= - 9400

First term,

Common difference,

No of terms, n = 11

Sum of first n terms,

Given, AP – 2, - 7, - 12, …

Let the nth term of an AP is - 77.

Then, first term, a = - 2

common difference, d = - 7 – (2) = - 7 + 2 = - 5.

Let nth term of an AP is - 77

a_n = a + (n - 1)d

- 77 = - 2 + (n - 1) (- 5)

- 75 = (n - 1) (- 5)

15 = n - 1

n = 16

so 16^th term of AP is - 77

S₁₆ is the sum of AP upto the term - 77 .i.e. 16^th term

As, S_n = (a + a_n) [ as last term is given]

= (- 2 - 77)

= 8 (- 79) = - 632

So 16^th term is - 77 and sum upto 16^th term is - 632

Given that, nth term of the series is a_n = 3 - 4n

For a_1,

Put n = 1 so a₁ = 3 - 4(1) = - 1

For a_2,

Put n = 2, so a₁ = 3 - 4(2) = - 5

For a_1,

Put n = 3 so a₁ = 3 - 4(3) = - 9

For a_1,

Put n = 4 so a₁ = 3 - 4(4) = - 13

So AP is - 1, - 5, - 9, - 13, …

a₂ - a₁ = - 5 - (- 1) = - 4

a₃ - a₂ = - 9 - (- 5) = - 4

a₄ - a₃ = - 13 - (- 9) = - 4

Since, the each successive term of the series has the same difference. So, it forms an AP with common difference, d = - 4

We know that, sum of n terms of an AP is

Where a = first term

d = common difference

and n = no of terms

= 10[ - 2 - 76]

= - 780

So Sum of first 20 terms of this AP is - 780.

S_n = n(4n + 1)

S_{n - 1} = (n - 1)[ 4(n - 1) - 1]

= (n - 1)[ 4n - 5]

S_n - S_{n - 1} = n(4n + 1) - (n - 1)(4n - 5)

(a₁ + a₂ + a₃ + - - - + a_{n - 1} + a_n) - (a₁ + a₂ + a₃ + - - - + a_{n - 1}) = 4n² + n - (4n² - 5n - 4n + 5)

a_n = 11n - 5

For a_1,

Put n = 1 so a₁ = 11(1) - 5 = 6

For a_2,

Put n = 2, so a₁ = 11(2) - 5 = 17

For a_1,

Put n = 3 so a₁ = 11(3) - 5 = 28

For a_1,

Put n = 4 so a₁ = 11(4) - 5 = 39

So AP is 6, 17, 28, 39, …

S_n = 3n² + 5n

S_{n - 1} = 3(n - 1)² + 5(n - 1)

= 3(n² - 2n + 1) + 5n - 5

= 3n² - 6n + 3 + 5n - 5

= 3n² - n - 2

S_n - S_{n - 1} = 3n² + 5n - (3n² - n - 2)

(a₁ + a₂ + a₃ + - - - + a_{n - 1} + a_n) - (a₁ + a₂ + a₃ + - - - + a_{n - 1}) = 6n + 2

a_n = 6n + 2

then

a_k = 6k + 2 = 164

6k = 164 - 2 = 162

k = 27

Let a be first term and d be common difference of an AP

Then

Taking LHS

= 6(2a + (n - 1)d

= (12 - 6)(2a + (n - 1)d)

= 3(4 - 2)(2a + (n - 1)d)

= 3[ (4 - 2)(2a + (n - 1)d)]

= 3[ 4(2a + (n - 1)d) - 2(2a + (n - 1)d)]

= 3(S₈ - S₄) [ By eqn 1 & eqn 2]

= RHS

Hence proved.

Let the a be first term and d be common difference of an AP

Given,

4^th term = - 15

a + 3d = - 15 [ using a_n = a + (n - 1)d]

a = - 3d - 15 [ eqn 1]

9^th term = - 30

a + 8d = - 30

- 3d - 15 + 8d = - 30 [ using 1]

5d = - 15

d = - 3

putting this value in eqn 1 we get

a = - 3 (- 3) - 15

= 9 - 15 = - 6

Also we know,

Sum of n terms,

= 17 (- 30)

= - 510

Let a and d be the first term and common difference, respectively of an AP

We know, Sum of first n terms of an AP

36 = 3[ 2a + 5d]

12 = 2a + 5d

2a = 12 - 5d [ eqn 1]

Now,

256 = 8[ 2a + 15d]

32 = 2a + 15d

32 = 12 - 5d + 15d [ using eqn 1]

20 = 10d

d = 2

using this value in eqn 1we get,

2a = 12 - 5(2)

2a = 2

a = 1

Now,

= 5(2 + 9(2))

= 5(20) = 100

So the sum of first 20 terms is 100

No of terms = 11 as it is even

Middle term will be i.e. 6^th term

Let the first term be a and common difference be d of an AP

As

a_n = a + (n - 1)d

a₆ = a + (6 - 1)d

30 = a + 5d [ eqn 1]

Now as we know,

[ using eqn 1]

First term, a = 8

Common difference, d = 10 - 8 = 2

For finding, the sum of last ten terms, we write the given AP in reverse order.

i.e., 126, 124, 122, …., 12, 10, 8

and First term A = 126 and Common difference D = - 2

Sum of ten terms of new AP is

For finding, the sum of first seven numbers which are multiples of 2 as well as of 9.

Take LCM of 2 and 9 which is 18.

So, the series becomes 18, 36, 54, ….

Clearly this series is an AP.

Here, first term, a = 18

common difference, d = 36 – 18 = 18

Now as Sum of first n terms of an AP is

= 7[ 18 + 54]

= 504

Hence the required sum is 504

Let n number of terms are needed to make the sum - 55.

Here, first term , a = - 15

common difference, d = - 13 + 15 = 2

By using the sum of n terms formula,

- 110 = n (- 30 + 2n - 2)

- 110 = n(2n - 32)

2n² - 32n + 110 = 0

n² - 16n + 55 = 0

n² - 11n - 5n + 55 = 0

n(n - 11) - 5(n - 11) = 0

(n - 5)(n - 11) = 0

So n is either 5 or 11

Hence, either 5 and 11 terms are needed to make the sum - 55.

Given that, first term of the first AP, a = 8 and that of second AP, A = - 30

and common difference of the first AP, d = 20 and that of second AP, D = 8

Given that

Sum of first n terms of first AP = Sum of first 2n terms of second AP

Let her pocket money be Rs x.

Now, she takes Rs 1 on day 1, Rs 2 on day 2, Rs 3 on day 3 and so on till the end of the month, from this money.

This makes a list

1, 2, 3, - - - , 31 [ as Jan has 31 days]

Clearly this is an AP with first term, a = 1 and d = 1 and no of terms = 31

So amount of money she puts in piggy bank in a month

Sum of n terms of this AP,

= 31(16)

= 496

So, Kanika puts Rs 496 till the end of the month from her pocket money

Also, she spent 204 of her pocket money and found that at the end of the month, she still has Rs 100 with her.

Now, according to the condition,

x - 496 - 204 = 100

x - 700 = 100

x = 800

So she gets 800 Rs as her pocket money.

Given that,

Yasmeen, during the first month, saves = 32 Rs

During the second month, saves = 36 Rs

During the third month, saves = 40 Rs

Let Yasmeen saves Rs 2000 during the n months.

Here, we have arithmetic progression 32, 36, 40, …

First term, a = 32

common difference, d = 36 – 32 = 4

Total money save by her in n months = Sum of this AP upto n terms

4000 = n[ 2(32) + (n - 1)4]

4000 = n[ 64 + 4n - 4)

4000 = n[ 4n + 60]

4n² + 60n - 4000 = 0

n² + 15n - 1000 = 0

n² + 40n - 25n - 1000 = 0

(n - 25)(n + 40) = 0

n = 25 or n = - 40

but n = - 40 as no of terms and months cannot be negatice

so it would take 25 months to save 2000 Rs.

Let the first term, common difference and the number of terms of an AP are a, d and n, respectively.

Given S₅ + S₇ = 167

Using the formula,

Where S_n is the sum of first n terms

So we have,

5(2a + 4d) + 7(2a + 6d) = 334

10a + 20d + 14a + 42d = 334

24a + 62d = 334

12a + 31d = 167

12a = 167 - 31d [ eqn 1]

Also,

S₁₀ = 235

5[ 2a + 9d] = 235

2a + 9d = 47

12a + 54d = 282 [ multiplication by 6 both side]

167 - 31d + 54d = 282 [ using equation 1]

23d = 282 - 167

23d = 115

d = 5

using this value in equation 1

12a = 167 - 31(5)

12a = 167 - 155

12a = 12

a = 1

Now

= 10[ 2 + 95]

= 970

So the sum of first 20 terms is 970.

Since, multiples of 2 as well as of 5 = LCM of (2, 5) = 10

Multiples of 2 as well as of 5 between 1 and 500 is 10, 20, 30, …., 490.

Clearly this is an AP with common difference, d = 10

And first term, a = 10

Let the no if terms in this AP are n

Then, by nth term formula

a_n = a + (n - 1)d

490 = 10 + (n - 1)10

480 = (n - 1)10

n - 1 = 48

n = 49

now, Sum of this AP

[ as last term is given]

= 49(250)

= 12250

Since, multiples of 2 as well as of 5 = LCM of (2, 5) = 10

Multiples of 2 as well as of 5 from 1 and 500 is 10, 20, 30, …., 500.

Clearly this is an AP with common difference, d = 10

And first term, a = 10

Let the no if terms in this AP are n

Then,

a_n = a + (n - 1)d

500 = 10 + (n - 1)10

490 = (n - 1)10

n - 1 = 49

n = 50

now, Sum of this AP

[ as last term is given]

= 25[ 10 + 500]

= 25(510)

= 12750

Since, multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5) i.e., 10.

Multiples of 2 or 5 from 1 to 500 = List of multiple of 2 from 1 to 500 + List of multiple of 5 from 1 to 500 - List of multiple of 10 from 1 to 500

= (2, 4, 6, …, 500) + (5, 10, 15, …, 500) - (10, 20, 30, …., 500)

All of these list form an AP.

And

Required sum = sum(2, 4, 6, …., 500) + sum(5, 10, 15, …, 500) - sum(10, 20, 30, …., 500)

Consider series

2, 4, 6, …., 500

First term, a = 2

Common difference, d = 2

Let n be no of terms

a_n = a + (n - 1)d

500 = 2 + (n - 1)2

498 = (n - 1)2

n - 1 = 249

n = 250

let the sum of this AP be S1 using the formula,

S1 = 125(502)

S1 = 62750 [ eqn 1]

Now, Consider series

5, 10, 15, …., 500

First term, a = 5

Common difference, d = 5

Let n be no of terms

By nth term formula

a_n = a + (n - 1)d

500 = 5 + (n - 1)

495 = (n - 1)5

n - 1 = 99

n = 100

Let the sum of this AP be S_2 using the formula,

S2 = 50(505)

S2 = 25250 [ eqn 2]

Consider series

10, 20, 30, …., 500

First term, a = 10

Common difference, d = 10

Let n be no of terms

a_n = a + (n - 1)d

500 = 10 + (n - 1)10

490 = (n - 1)10

n - 1 = 49

n = 50

Let the sum of this AP be S1 using the formula,

S3 = 25(510)

S3 = 12750 [ eqn 3]

So required Sum = S1 + S2 - S3

= 62750 + 25250 - 12750

= 75250

Let the a be first term and d be common difference of AP

And we know that, that the nth term is

a_n = a + (n - 1)d

Given,

2a₈ = a₂

2(a + 7d) = a + d

2a + 14d = a + d

a = - 13d [ eqn1]

Also,

3(a + 10d) = a + 3d + 3

3a + 30d = a + 3d + 3

2a + 27d = 3

2 (- 13d) + 27d = 3 [ using eqn1]

d = 3

using this value in eqn1

a = - 13(3)

= - 39

Now,

a₁₅ = a + 14d

= - 39 + 14(3)

= - 39 + 42

= 3

So 15^th term is 3.

Let the a be first term and d be common difference of an AP

As n = 37 (odd)

Middle term will be

Three middle most terms will be

18^th, 19^th and 20^th terms

Given,

a₁₈ + a₁₉ + a₂₀ = 225

a + 17d + a + 18d + a + 19d = 225 [ using a_n = a + (n - 1)d]

3a + 54d = 225

3a = 225 - 54d

a = 75 - 18d [ eqn 1]

Also last three terms will be 35^th , 36^th and 37^th terms

Given,
a₃₅ + a₃₆ + a₃₇ = 429

a + 34d + a + 35d + a + 36d = 429

3a + 105d = 429

a + 35d = 143

75 - 18d + 35d = 143 [ using eqn1]

17d = 68

d = 4

using this value in eqn 1

a = 75 - 18(4)

a = 3

so AP is a, a + d, a + 2d….

i.e. 3, 7, 11….

(i) The number (integers) between 100 and 200 which is divisible by 9 are 108, 117, 126, …198

Let n be the number of terms between 100 and 200 which is divisible by 9.

Then,

a_n = a + (n - 1)d

198 = 108 + (n - 1)9

90 = (n - 1)9

n - 1 = 10

n = 11

now, Sum of this AP

[ as last term is given]

= 11(153)

= 1683

(ii) The sum of the integers between 100 and 200 which is not divisible by 9 = (sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 which is divisible by 9.

Let the required sum be S

S = S1 - S2

Where S1 is the sum of AP 101, 102, 103, - - - , 199

And S2 is the sum of AP 108, 117, 126, - - - - , 198

For S1

First term, a = 101

Common difference, d = 199

Let n be no of terms

Then,

a_n = a + (n - 1)d

199 = 101 + (n - 1)1

98 = (n - 1)

n = 99

now, Sum of this AP

[ as last term is given]

= 99(150)

= 14850

For S1

First term, a = 108

Common difference, d = 9

Last term, a_n = 198

Let n be no of terms

Then,

a_n = a + (n - 1)d

198 = 108 + (n - 1)9

10 = (n - 1)

n = 11

now, Sum of this AP

= 11(153)

= 1683

Therefore

S = S1 - S2

= 14850 - 1683

= 13167

Let a and d the first term and common difference of an AP.

a₁₁:a₁₈ = 2:3

a + 10d : a + 17d = 2:3

3a + 30d = 2a + 34d

a = 4d [ eqn 1]

a₅ = a + 4d = 4d + 4d = 8d [ by eqn 1] [ Eqn 2]

a₂₁ = a + 20d = 4d + 20d = 24d [ by eqn 1] [ Eqn 3]

now using the formula

= 30d

= 21(14)d

= 294

S₅:S₂₁ = 30 : 294 = 5: 49

Given that, the AP is a, b, …, c.

Here, first term = a

common difference = b - a

and last term, a_n = c

as nth term of an AP

a_n = a + (n - 1)d

[ eqn 1]

Sum of an AP

So

Hence Proved.

Given equation is - 4 + (- 1) + 2 + - - - + x = 437

Its terms can be listed as

- 4, - 1, 2, - - - , x

And this is an AP with First term, a = - 4

Common difference, d = - 1 - (- 4) = 3

Let the no of terms be n

Then Sum of first n terms, S_n = 437

n[ 2 (- 4) + (n - 1)3] = 874

n (- 8 + 3n - 3) = 874

n(3n - 11) = 874

3n² - 11n - 874 = 0

Solving this quadratic equation with

A = 3

B = - 11

C = - 874

Then D = b² - 4ac = (- 11)² - 4(3) (- 874)

= 121 + 10488 = 10609

(not possible as n is a natural no)

so x is the 19^th term of AP

x = a₁₉ = a + 18d

= - 4 + 18(3)

= 50

Given that,

Jaspal singh takes total loan = Rs118000

He repays his total loan by paying every month.

His first installment = 1000

Second installment = 1000 + 100 = 1100

Third installment = 1100 + 100 = 1200 and so on

Thus, we have 1000, 1100, 1200, … which form an AP, with

first term, a = 1000

common difference, d = 1100 – 1000 = 100

nth term of an AP

So amount paid in 30 installments = sum of first 30 terms of this AP

= 15(2000 + 2900)

= 15(4900) = 73500

So he pays Rs 73500 in 30 installments

Loan left = total loan - paid lone

= 118000 - 73500 = 44500 Rs

Given that, the students of a school decided to beautify the school on the annual day by fixing colorful flags on the straight passage of the school.

Given that, the number of flags = 27 and distance between each flag = 2m.

Also, the flags are stored at the position of the middle most flag i.e., 14th flag and Ruchi was given the responsibility of placing the flags. Ruchi kept her books, where the flags were stored i.e., 14th flag and she could carry only one flag at a time.

Let she placed 13 flags into her left position from middle most flag i.e., 14th flag.

For placing second flag and return his initial position distance travelled = 2 + 2 = 4m.

Similarly, for placing third flag and return his initial position, distance travelled = 4 + 4 = 8 m.

For placing fourth flag and return his initial position, distance travelled = 6 + 6 = 12 m.

For placing fourteenth flag and return his initial position, distance travelled

= 26 + 26 = 52 m

So this becomes a series,

4, 8, 12, 16, - - - , 52

Proceed same manner into her right position from middle most flag i.e., 14 flag.

(Also, when Ruchi placed the last flag in her rightmost, she return his middle most position and collect her books. This distance also included in placed the last flag.)

We get the same series

4, 8, 12, 16, - - - , 52

Clearly this series is an AP with First term, a = 4 , common difference, d = 4 and

no of terms. n = 13

Total distance covered by Ruchi for placing these flags = 2S_n

= 13(56)

= 13(56)

= 728 m

Hence, the required is 728 m in which she did cover in completing this job and returning back to collect her books.

Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position

= 13(2)

= 26 m

Hence, the maximum distance she travelled carrying a flag is 26 m.