Arithmetic Progression

2, 4, 6, 8, . . .

Here, the first term, a₁ = 2

Second term, a₂ = 4

a₃ = 6

Now, common difference = a₂ – a₁ = 4 – 2 = 2

Also, a₃ – a₂ = 6 – 4 = 2

Since, the common difference is same.

Hence the terms are in Arithmetic progression with common difference, d = 2.

Here, the first term, a₁ = 2

Second term,

Third Term, a₃ = 3

Now, common difference =

Also,

Since, the common difference is same.

Hence the terms are in Arithmetic progression with common difference,

– 10, – 6, – 2,2, . . .

Here, the first term, a₁ = – 10

Second term, a₂ = – 6

a₃ = – 2

Now, common difference = a₂ – a₁ = – 6 – ( – 10) = – 6 + 10 = 4

Also, a₃ – a₂ = – 2 – ( – 6) = – 2 + 6 = 4

Since, the common difference is same.

Hence the terms are in Arithmetic progression with common difference, d = 4.

0.3, 0.33, 0.333,…..

Here, the first term, a₁ = 0.3

Second term, a₂ = 0.33

a₃ = 0.333

Now, common difference = a₂ – a₁ = 0.33 – 0.3 = 0.03

Also, a₃ – a₂ = 0.333 – 0.33 = 0.003

Since, the common difference is not same.

Hence the terms are not in Arithmetic progression

0, – 4, – 8, – 12, . . .

Here, the first term, a₁ = 0

Second term, a₂ = – 4

a₃ = – 8

Now, common difference = a₂ – a₁ = – 4 – 0 = – 4

Also, a₃ – a₂ = – 8 – ( – 4) = – 8 + 4 = – 4

Since, the common difference is same.

Hence the terms are in Arithmetic progression with common difference, d = – 4.

Here, the first term,

Second term,

Now, common difference

Also,

Since, the common difference is same.

Hence the terms are in Arithmetic progression with common difference, .

3, 3 + √2, 3 + 2√2, 3 + 3√2, ….

Here, the first term, a₁ = 3

Second term, a₂ = 3 + √2

a₃ = 3 + 2√2

Now, common difference = a₂ – a₁ = 3 + √2 – 3 = √2

Also, a₃ – a₂ = 3 + 2√2 –(3 + √2) = 3 + 2√2 – 3 – √2 = √2

Since, the common difference is same.

Hence the terms are in Arithmetic progression with common difference, d = √2 .

127, 132, 137, . . .

Here, the first term, a₁ = 127

Second term, a₂ = 132

a₃ = 137

Now, common difference = a₂ – a₁ = 132 – 127 = 5

Also, a₃ – a₂ = 137 – 132 = 5

Since, the common difference is same.

Hence the terms are in Arithmetic progression with common difference, d = 5.

a = 10, d = 5

Let a₁ = a = 10

Since, the common difference d = 5

Using formula a_{n + 1} = a_n + d

Thus, a₂ = a₁ + d = 10 + 5 = 15

a_{3 =} a₂ + d = 15 + 5 = 20

a₄ = a₃ + d = 20 + 5 = 25

Hence, An A.P with common difference 5 is 10, 15, 20, 25,….

a = – 3, d = 0

Let a₁ = a = – 3

Since, the common difference d = 0

Using formula a_{n + 1} = a_n + d

Thus, a₂ = a₁ + d = – 3 + 0 = – 3

a_{3 =} a₂ + d = – 3 + 0 = – 3

a₄ = a₃ + d = – 3 + 0 = – 3

Hence, An A.P with common difference 0 is – 3, – 3, – 3, – 3,….

a = – 7,

Let a₁ = a = – 7

Since, the common difference

Using formula a_{n + 1} = a_n + d

Thus,

Hence, An A.P with common difference is

a = – 1.25, d = 3

Let a₁ = a = – 1.25

Since, the common difference d = 3

Using formula a_{n + 1} = a_n + d

Thus, a₂ = a₁ + d = – 1.25 + 3 = 1.75

a_{3 =} a₂ + d = 1.75 + 3 = 4.75

a₄ = a₃ + d = 4.75 + 3 = 7.75

Hence, An A.P with common difference 3 is – 1.25, 1.75, 4.75, 7.75

a = 6, d = – 3

Let a₁ = a = 6

Since, the common difference d = – 3

Using formula a_{n + 1} = a_n + d

Thus, a₂ = a₁ + d = 6 + ( – 3) = 6 – 3 = 3

a_{3 =} a₂ + d = 3 + ( – 3) = 3 – 3 = 0

a₄ = a₃ + d = 0 + ( – 3) = – 3

Hence, An A.P with common difference – 3 is 6, 3, 0, – 3…

a = – 19, d = – 4

Let a₁ = a = – 19

Since, the common difference d = – 4

Using formula a_{n + 1} = a_n + d

Thus, a₂ = a₁ + d = – 19 + ( – 4) = – 19 – 4 = – 23

a_{3 =} a₂ + d = – 23 + ( – 4) = – 23 – 4 = – 27

a₄ = a₃ + d = – 27 + ( – 4) = – 27 – 4 = – 31

Hence, An A.P with common difference – 4 is – 19, – 23, – 27, – 31,….

5, 1, – 3, – 7, . . .

First term a₁ = 5

Second term a₂ = 1

Third term a₃ = – 3

We know that d = a_{n + 1} – a_n

Thus, d = a₂ – a₁ = 1 – 5 = – 4

Hence, the common difference d = – 4 and first term is 5

0.6, 0.9, 1.2, 1.5, . . .

First term a₁ = 0.6

Second term a₂ = 0.9

Third term a₃ = 1.2

We know that d = a_{n + 1} – a_n

Thus, d = a₂ – a₁ = 0.9 – 0.6 = 0.3

Hence, the common difference d = 0.3 and first term is 0.6

127, 135, 143, 151, . . .

First term a₁ = 127

Second term a₂ = 135

Third term a₃ = 143

We know that d = a_{n + 1} – a_n

Thus, d = a₂ – a₁ = 135 – 127 = 8

Hence, the common difference d = 8 and first term is 127

First term

Second term

Third term

We know that d = a_{n + 1} – a_n

Thus,

Hence, the common difference and first term is

1, 8, 15, 22, . . .

First term a = 1

Second term t₁ = 8

Third term t₂ = 15

Fourth term t₃ = 22

We know that d = t_{n + 1} – t_n

Thus, t₂ – t₁ = 15 – 8 = 7

t₃ – t₂ = 22 – 15 = 7

Thus, d = 7

3,6,9,12, . . .

First term a = 3

Second term t₁ = 6

Third term t₂ = 9

Fourth term t₃ = 12

We know that d = t_{n + 1} – t_n

Thus, t₂ – t₁ = 9 – 6 = 3

t₃ – t₂ = 12 – 9 = 3

Thus, d = 3

– 3, – 8, – 13, – 18, . . .

First term a = – 3

Second term t₁ = – 8

Third term t₂ = – 13

Fourth term t₃ = – 18

We know that d = t_{n + 1} – t_n

Thus, t₂ – t₁ = – 13 – ( – 8) = – 13 + 8 = – 5

t₃ – t₂ = – 18 – ( – 13) = – 18 + 13 = – 5

Thus, d = – 5

70, 60, 50, 40, . . .

First term a = 70

Second term t₁ = 60

Third term t₂ = 50

Fourth term t₃ = 40

We know that d = t_{n + 1} – t_n

Thus, t₂ – t₁ = 50 – 60 = – 10

t₃ – t₂ = 40 – 50 = – 10

Thus, d = – 10

Given A.P. is – 12, – 5, 2, 9, 16, 23, 30, . . .

Where first term a = – 12

Second term t₁ = – 5

Third term t₂ = 2

Common Difference d = t₂ – t₁ = 2 – ( – 5) = 2 + 5 = 7

We know that, n^th term of an A.P. is

t_n = a + (n – 1)d

We need to find the 20^th term,

Here n = 20

Thus, t₂₀ = – 12 + (20 – 1)× 7

t₂₀ = – 12 + (19)× 7 = – 12 + 133 = 121

Thus, t₂₀ = 121

Given A.P. is 12, 16, 20, 24, . . .

Where first term a = 12

Second term t₁ = 16

Third term t₂ = 20

Common Difference d = t₂ – t₁ = 20 – 16 = 4

We know that, n^th term of an A.P. is

t_n = a + (n – 1)d

We need to find the 24^th term,

Here n = 24

Thus, t₂₄ = 12 + (24 – 1)× 4

t₂₄ = 12 + (23)× 4 = 12 + 92 = 104

Thus, t₂₄ = 104

Given A.P. is 7, 13, 19, 25, . . .

Where first term a = 7

Second term t₁ = 13

Third term t₂ = 19

Common Difference d = t₂ – t₁ = 19 – 13 = 6

We know that, n^th term of an A.P. is

t_n = a + (n – 1)d

We need to find the 19^th term,

Here n = 19

Thus, t₁₉ = 7 + (19 – 1)× 6

t₁₉ = 7 + (18)× 6 = 7 + 108 = 115

Thus, t₁₉ = 115

Given A.P. is 9, 4, – 1, – 6, – 11, . . .

Where first term a = 9

Second term t₁ = 4

Third term t₂ = – 1

Common Difference d = t₂ – t₁ = – 1 – 4 = – 5

We know that, n^th term of an A.P. is

t_n = a + (n – 1)d

We need to find the 27^th term,

Here n = 27

Thus, t₂₇ = 9 + (27 – 1)× ( – 5)

t₂₇ = 9 + (26)× ( – 5) = 9 – 130 = – 121

Thus, t₂₇ = – 121

List of three digit number divisible by 5 are

100, 105,110,115,……….. 995

Let us find how many such number are there?

From the above sequence, we know that

t_n = 995, a = 100

t₁ = 105, t₂ = 110

Thus, d = t₂ – t₁ = 110 – 105 = 5

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

we can find value of “n”

Thus, on substituting all the value in formula we get,

995 = 100 + (n – 1)× 5

⇒ 995 – 100 = (n – 1)× 5

⇒ 895 = (n – 1) × 5

⇒ n = 179 + 1 = 180

Given: t₁₁ = 16 and t₂₁ = 29

To find: t₄₁

Using n^th term of an A.P. formula

t_n = a + (n – 1)d

we will find value of “a” and “d”

Let, t₁₁ = a + (11 – 1) d

⇒ 16 = a + 10 d …..(1)

t₂₁ = a + (21 – 1) d

⇒ 29 = a + 20 d …..(2)

Subtracting eq. (1) from eq. (2), we get,

⇒ 29 – 16 = (a – a) + (20 d – 10 d)

⇒ 13 = 10 d

Substitute value of “d” in eq. (1) to get value of “a”

⇒ 16 = a + 13

⇒ a = 16 – 13 = 3

Now, we will find the value of t₄₁ using n^th term of an A.P. formula

⇒ t₄₁ = 3 + 4 × 13 = 3 + 52 = 55

Thus, t₄₁ = 55

By, given A.P. 11, 8, 5, 2, . . .

we know that

a = 11, t₁ = 8, t₂ = 5

Thus, d = t₂ – t₁ = 5 – 8 = – 3

Given: t_n = – 151

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

we can find value of “n”

Thus, on substituting all the value in formula we get,

– 151 = 11 + (n – 1)× ( – 3)

⇒ – 151 – 11 = (n – 1)× ( – 3)

⇒ – 162 = (n – 1) × ( – 3)

⇒ n = 54 + 1 = 55

List of number divisible by 4 in between 10 to 250 are

12, 16,20,24,……….. 248

Let us find how many such number are there?

From the above sequence, we know that

t_n = 248, a = 12

t₁ = 16, t₂ = 20

Thus, d = t₂ – t₁ = 20 – 16 = 4

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

we can find value of “n”

Thus, on substituting all the value in formula we get,

248 = 12 + (n – 1)× 4

⇒ 248 – 12 = (n – 1)× 4

⇒ 236 = (n – 1) × 4

⇒ n = 59 + 1 = 60

Given: t₁₇ = 7 + t₁₀ ……(1)

In t₁₇, n = 17

In t₁₀, n = 10

By using n^th term of an A.P. formula,

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th term

Thus, on using formula in eq. (1) we get,

⇒ a + (17 – 1)d = 7 + (a + (10 – 1)d)

⇒ a + 16 d = 7 + (a + 9 d)

⇒ a + 16 d – a – 9 d = 7

⇒ 7 d = 7

Thus, common difference “d” = 1

Given: First term a = 6

Common Difference d = 3

To find: S₂₇ where n = 27

By using sum of n^th term of an A.P. is

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, Substituting given value in formula we can find the value of S₂₇

Thus, S₂₇ = 1215

List of first 123 even natural number is

2,4,6,…….

Where first term a = 2

Second term t₁ = 4

Third term t₂ = 6

Thus, common difference d = t₂ – t₁ = 6 – 4 = 2

n = 123

By using sum of n^th term of an A.P. is

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, Substituting given value in formula we can find the value of S_n

Thus, S_n = 15252

List of even natural number between 1 to 350 is

2,4,6,…….348

Where first term a = 2

Second term t₁ = 4

Third term t₂ = 6

Thus, common difference d = t₂ – t₁ = 6 – 4 = 2

t_n = 348 (As we have to find the sum of even numbers between 1 and 350 therefore excluding 350 )

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

we can find value of “n” by substituting all the value in formula we get,

⇒ 348 = 2 + (n – 1) × 2

⇒ 348 – 2 = 2(n – 1)

⇒ 346 = 2(n – 1)

⇒ n = 173 + 1 = 174

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, Substituting given value in formula we can find the value of S_n

Thus, S₁₇₄ = 30,450

Given: t₁₉ = 52 and t₃₈ = 128

To find: value of “a” and “d”

Using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

we will find value of “a” and “d”

Let, t₁₉ = a + (19 – 1) d

⇒ 52 = a + 18 d …..(1)

t₃₈ = a + (38 – 1) d

⇒ 128 = a + 37 d …..(2)

Subtracting eq. (1) from eq. (2), we get,

⇒ 128 – 52 = (a – a) + (37 d – 18 d)

⇒ 76 = 19 d

Substitute value of “d” in eq. (1) to get value of “a”

⇒ 52 = a + 18 ×4

⇒ 52 = a + 72

⇒ a = 52 – 72 = – 20

Now, to find value of S₅₆ we will using formula of sum of n terms

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, Substituting given value in formula we can find the value of S_n

⇒S₅₆ = 28 × [ – 40 + 55×4]

⇒S₅₆ = 28 × [ – 40 + 220]

⇒S₅₆ = 28 × 180 = 5040

Thus, S₅₆ = 5040

List of natural number divisible by 4 between 1 to 140 is

4,8,12,…….136

Where first term a = 4

Second term t₁ = 8

Third term t₂ = 12

Thus, common difference d = t₂ – t₁ = 12 – 8 = 4

t_n = 136

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

we can find value of “n” by substituting all the value in formula we get,

⇒ 136 = 4 + (n – 1) × 4

⇒ 136 – 4 = 4(n – 1)

⇒ 132 = 4(n – 1)

⇒ n = 33 + 1 = 34

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, Substituting given value in formula we can find the value of S₃₄

⇒S₃₄ = 17 × [8 + 33×4]

⇒S₃₄ = 17 × [8 + 132]

⇒S₃₄ = 17 × 140 = 2380

Thus, S₃₄ = 2380

Given: S₅₅ = 3300 where n = 55

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, on substituting the given value in formula we get,

⇒ 3300 = 55 × [ a + 27d]

⇒ a + 27d = 60 ……(1)

We need to find value of 28^th term i.e t₂₈

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

we can find value of t₂₈ by substituting all the value in formula we get,

⇒ t₂₈ = a + (28 – 1) d

⇒ t₂₈ = a + 27 d

From eq. (1) we get,

⇒ t₂₈ = a + 27 d = 60

⇒ t₂₈ = 60

Let the first term be a – d

the second term be a

the third term be a + d

Given: sum of consecutive three term is 27

⇒ (a – d) + a + (a + d) = 27

⇒ 3 a = 27

Also, Given product of three consecutive term is 504

⇒ (a – d)× a × (a + d) = 504

⇒ (9 – d) × 9 × (9 + d) = 504 (since, a = 9)

⇒ 9² – d² = 56 (since, (a – b)(a + b) = a² – b²)

⇒ 81 – d² = 56

⇒ d² = 81 – 56 = 25

⇒ d = √25 = ± 5

Case 1:

Thus, if a = 9 and d = 5

Then the three terms are,

First term a – d = 9 – 5 = 4

Second term a = 9

Third term a + d = 9 + 5 = 14

Thus, the A.P. is 4, 9, 14

Case 2:

Thus, if a = 9 and d = – 5

Then the three terms are,

First term a – d = 9 – ( – 5) = 9 + 5 = 14

Second term a = 9

Third term a + d = 9 + ( – 5) = 9 – 5 = 4

Thus, the A.P. is 14, 9, 4

Let the first term be a – d

the second term be a

the third term be a + d

the fourth term be a + 2 d

Given: sum of consecutive four term is 12

⇒ (a – d) + a + (a + d) + (a + 2d) = 12

⇒ 4 a + 2d = 12

⇒ 2(2 a + d) = 12

⇒ 2a + d = 6 …..(1)

Also, sum of third and fourth term is 14

⇒ (a + d) + (a + 2d) = 14

⇒ 2a + 3d = 14 ……(2)

Subtracting eq. (1) from eq. (2) we get,

⇒(2a + 3d) – (2a + d) = 14 – 6

⇒2a + 3d – 2a – d = 8

⇒ 2d = 8

⇒ d = 4

Substituting value of “d” in eq. (1) we get,

⇒ 2a + 4 = 6

⇒ 2a = 6 – 4 = 2

⇒ a = 1

Thus, a = 1 and d = 4

Hence, first term a – d = 1 – 4 = – 3

the second term a = 1

the third term a + d = 1 + 4 = 5

the fourth term a + 2 d = 1 + 2×4 = 1 + 8 = 9

Thus, the A.P. is – 3, 1, 5, 9

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Given: t₉ = 0

⇒ t₉ = a + (9 – 1)d

⇒ 0 = a + 8d

⇒ a = – 8d

To Show: t₂₉ = 2× t₁₉

Now,

⇒ t₂₉ = a + (29 – 1)d

⇒ t₂₉ = a + 28d

⇒ t₂₉ = – 8d + 28d = 20 d (since, a = – 8d )

⇒ t₂₉ = 20 d

⇒ t₂₉ = 2 × 10 d ….(1)

Also,

⇒ t₁₉ = a + (19 – 1)d

⇒ t₁₉ = a + 18d

⇒ t₁₉ = – 8d + 18d = 10 d (since, a = – 8d )

⇒ t₁₉ = 10 d …..(2)

From eq. (1) and eq. (2) we get,

t₂₉ = 2× t₁₉

By given information we can form an A.P.

10, 11, 12, 13, ……

Hence, the first term a = 10

Second term t₁ = 11

Third term t₂ = 12

Thus, common difference d = t₂ – t₁ = 12 – 11 = 1

Here, number of terms from 1^st Jan 2016 to 31^st Dec 2016 is,

n = 366

We need to find S₃₆₆

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, on substituting the given value in formula we get,

⇒S₃₆₆ = 183 [ 20 + 365]

⇒S₃₆₆ = 183 × 385

⇒S₃₆₆ = Rs 70,455

Given: A man borrows = Rs. 8000

Repay with total interest = Rs 1360

In 12 months, thus n = 12

Thus, S₁₂ = 8000 + 1360 = 9360

Each installment being less than preceding one

Thus, d = – 40

We need to find “a”

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, on substituting the given value in formula we get,

⇒ 9360 = 6 [ 2a – 11 × 40]

⇒ 1560 = 2a – 440

⇒ 1560 + 440 = 2a

⇒ 2a = 2000

Thus, first installment a = Rs. 1000

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

S_n = sum of n terms

Thus, on substituting the given value in formula we get,

Let a = first term, t_n = last term

⇒ 9360 = 6 [ 1000 + t_n]

⇒ t_n = 1560 – 1000 = 560

Thus, last installment t_n = 560

By given information we can form an A.P.

5000, 7000, 9000, ……

Hence, the first term a = 5000

Second term t₁ = 7000

Third term t₂ = 9000

Thus, common difference d = t₂ – t₁ = 9000 – 7000 = 2000

Here, number of terms n = 12

We need to find S₁₂

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, on substituting the given value in formula we get,

⇒S₁₂ = 6 × [ 10,000 + 11 × 2000]

⇒S₁₂ = 6 × [10,000 + 22,000]

⇒S₁₂ = 6 × 32,000

⇒S₁₂ = Rs. 192000

Given: first term a = 20

Second term t₁ = 22

Third term t₂ = 24

Common difference d = t₂ – t₁ = 24 – 22 = 2

We need to find t₁₅ thus n = 15

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

On substituting all value in n^th term of an A.P.

⇒ t₁₅ = 20 + (15 – 1) × 2

⇒ t₁₅ = 20 + 14 × 2

⇒ t₁₅ = 20 + 28 = 48

We have been given that, there are 27 rows in an auditorium

Thus, we need to find total seats in auditorium i.e. S₂₇

Now, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Thus, on substituting the given value in formula we get,

⇒S₂₇ = 27 × 46

⇒S₂₇ = 1242

Let Monday be the first term i.e. a = t₁

Let Tuesday be the second term i.e t₂

Let Wednesday be the third term i.e t₃

Let Thursday be the fourth term i.e t₄

Let Friday be the fifth term i.e t₅

Let Saturday be the sixth term i.e t₆

Given: t₁ + t₆ = 5 + (t₂ + t₆)

⇒ a = 5 + (t₂ + t₆) – t₆

⇒ a = 5 + t₂ …..(1)

We know that,

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Thus, t₂ = a + (2 – 1) d

⇒ t₂ = a + d

Now substitute value of t₂ in (1) we get,

⇒ a = 5 + (a + d)

⇒ d = a – 5 – a = – 5

Given: t₃ = – 30°

Thus, t₃ = a + (3 – 1) × ( – 5)

⇒ – 30 = a + 2 × ( – 5)

⇒ – 30 = a – 10

⇒ a = – 30 + 10 = – 20°

Thus, Monday, a = t₁ = – 20°

Using formula t_{n + 1} = t_n + d

We can find the value of the other terms

Tuesday, t₂ = t₁ + d = – 20 – 5 = – 25°

Wednesday, t₃ = t₂ + d = – 25 – 5 = – 30°

Thursday, t₄ = t₃ + d = – 30 – 5 = – 35°

Friday, t₅ = t₄ + d = – 35 – 5 = 40°

Saturday, t₆ = t₅ + d = – 40 – 5 = – 45°

Thus, we obtain an A.P.

– 20°, – 25°, – 30°, – 35°, – 40°, – 45°

First term a = 1

Second term t₁ = 2

Third term t₃ = 3

Common difference d = t₃ – t₂ = 3 – 2 = 1

We need to find total number of trees when n = 25

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find S₂₅

Thus, on substituting the given value in formula we get,

⇒S₂₅ = 25 × 13 = 325

First term a = – 10

Second term t₁ = – 6

Third term t₂ = – 2

Fourth term t₃ = 2

Common difference d = t₁ – a = – 6 – ( – 10) = – 6 + 10 = 4

Common difference d = t₂ – t₁ = – 2 – ( – 6) = – 2 + 6 = 4

Common difference d = t₃ – t₂ = 2 – ( – 2) = 2 + 2 = 4

Since, the common difference is same

∴ The given sequence is A.P. with common difference d = 4

Hence, correct answer is (B)

Given first term t₁ = – 2

Common difference d = – 2

By using formula t_{n + 1} = t_n + d

t₂ = t₁ + d = – 2 + ( – 2) = – 2 – 2 = – 4

t₃ = t₂ + d = – 4 + ( – 2) = – 4 – 2 = – 6

t₄ = t₃ + d = – 6 + ( – 2) = – 6 – 2 = – 8

Hence, the A.P. is – 2, – 4, – 6, – 8

∴ correct answer is (C)

List of first 30 natural number is

1, 2, 3,……..,30

First term a = 1

Second term t₁ = 2

Third term t₂ = 3

Common difference d = t₃ – t₂ = 3 – 2 = 1

number of terms n = 30

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find S₃₀

⇒ S₃₀ = 15 [ 2 + 29]

⇒ S₃₀ = 15 × 31

⇒ S₃₀ = 465

Hence, Correct answer is (B)

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

⇒ t₇ = a + (7 – 1) × ( – 4)

⇒ 4 = a + 6 × ( – 4)

⇒ 4 = a – 24

⇒ a = 24 + 4 = 28

Thus, the correct answer is (D)

Given: a = 3.5, d = 0, n = 101

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Substituting all given value in the formulae we get,

⇒ t_n = 3.5 + (101 – 1) × 0

⇒ t_n = 3.5

Thus, correct answer is (B)

Given: first term a = – 3

Second term t₁ = 4

Common difference d = t₁ – a = 4 – ( – 3) = 4 + 3 = 7

We need to find t₂₁ where n = 21

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Substituting all given value in the formulae we get,

⇒ t₂₁ = – 3 + (21 – 1) × 7

⇒ t₂₁ = – 3 + 20 × 7

⇒ t₂₁ = – 3 + 140 = 137

Hence, correct answer is (C)

Given d = 5

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Thus, t₁₈ – t₁₃ = [a + (18 – 1) × 5] – [ a + (13 – 1) × 5]

⇒ t₁₈ – t₁₃ = [ 17 × 5] – [ 12 × 5]

⇒ t₁₈ – t₁₃ = 85 – 60 = 25

Thus, correct answer is (C)

First five multiples of 3 are

3, 6, 9, 12, 15

First term a = 3

Second term t₁ = 6

Third term t₂ = 9

Common difference d = t₂ – t₁ = 9 – 6 = 3

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find S₅

Thus, correct answer is (A)

First term a = 15

Second term t₁ = 10

Third term t₂ = 5

Common difference d = t₂ – t₁ = 5 – 10 = – 5

No. of terms n = 10

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find S₁₀

⇒S₁₀ = 5 [ 30 + 9 × ( – 5)]

⇒S₁₀ = 5 [ 30 – 45]

⇒S₁₀ = 5 × ( – 15) = – 75

Hence, correct answer is (A)

Given, first term = 1

Last term = 20

Sum of n terms, S_n = 399

We need to find no. of terms n

Using Sum of n terms of an A.P. formula

where n = no. of terms

S_n = sum of n terms

Now, on substituting given value in formula we get,

∴ correct answer is (B)

First term from end a = 49

t_n = – 11

t_{n – 1} = – 8

Common difference d = t_n – t_{n – 1} = – 11 – ( – 8) = – 11 + 8 = – 3

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

no. of terms n = 4

⇒ t₄ = 49 + (4 – 1) × ( – 3)

⇒ t₄ = 49 + 3 × ( – 3)

⇒ t₄ = 49 – 9 = 40

Given: t₁₀ = 46

t₅ + t₇ = 52

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Hence, by given condition we get,

⇒ t₁₀ = 46

⇒ a + (10 – 1)d = 46

⇒ a + 9d = 46 ……(1)

⇒ t₅ + t₇ = 52

⇒ [a + (5 – 1)d] + [a + (7 – 1)d] = 52

⇒ [a + 4d] + [a + 6d] = 52

⇒ 2a + 10d = 52 ……(2)

Multiply eq. (2) by 2 we get,

⇒ 2a + 18d = 92 ……(3)

Subtract eq. (2) by eq. (3)

⇒ [2a + 18d] – [ 2a + 10d] = 92 – 52

⇒ 8d = 40

Substitute “d” in (1)

⇒ a + 9 × 5 = 46

⇒ a + 45 = 46

⇒ a = t₁ = 46 – 45 = 1

we know that, t_{n + 1} = t_n + d

⇒ t₂ = t₁ + d = 1 + 5 = 6

⇒ t₃ = t₂ + d = 6 + 5 = 11

Hence, an A.P. is 1, 6, 11, . . .

t₄ = – 15 and t₉ = – 30

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Hence, by given condition we get,

t₄ = – 15

⇒ a + (4 – 1)d = – 15

⇒ a + 3d = – 15 …..(1)

t₉ = – 30

⇒ a + (9 – 1)d = – 30

⇒ a + 8d = – 30 …..(2)

Subtracting eq. (1) from eq. (2)

⇒ [a + 8d] – [a + 3d] = – 30 – ( – 15)

⇒ 5d = – 30 + 15 = – 15

Substituting, “d” in eq. (1)

⇒ a + 3 × ( – 3) = – 15

⇒ a + – 9 = – 15

⇒ a = – 15 + 9 = – 6

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find S₁₀

⇒S₁₀ = 5 [ – 12 + 9 × ( – 3)]

⇒S₁₀ = 5 [ – 12 – 27]

⇒S₁₀ = 5 × ( – 39) = – 195

Given A.P. is 9, 7, 5….

Whose first tern a = 9

Second term t₁ = 7

Third term t₃ = 5

Common difference d = t₃ – t₂ = 5 – 7 = – 2

Another A.P. is 24, 21, 18, . . .

Whose first tern a = 24

Second term t₁ = 21

Third term t₃ = 18

Common difference d = t₃ – t₂ = 18 – 21 = – 3

We have been given, n^th term of both the A.P. is same

thus, by using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Hence, by given condition we get,

⇒ 9 + (n – 1) × ( – 2) = 24 + (n – 1) × ( – 3)

⇒ 9 – 2n + 2 = 24 – 3n + 3

⇒ 11 – 2n = 27 – 3n

⇒ 3n – 2n = 27 – 11

⇒ n = 16

Thus, value of n^th term where a = 9, d = – 2, n = 16 is

⇒ t_n = 9 + (16 – 1) × ( – 2)

⇒ t_n = 9 – 15 × 2

⇒ t_n = 9 – 30 = – 21

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Hence, by given condition we get,

t₃ + t₈ = 7

⇒ [a + (3 – 1)d] + [a + (8 – 1)d] = 7

⇒ [a + 2d] + [a + 7d] = 7

⇒ 2a + 9d = 7 …..(1)

t₇ + t₁₄ = – 3

⇒ [a + (7 – 1)d] + [a + (14 – 1)d] = – 3

⇒ [a + 6d] + [ a + 13d] = – 3

⇒ 2a + 19d = – 3 …..(2)

Subtracting eq. (1) from eq. (2)

⇒ [2a + 19d] – [2a + 9d] = – 3 – 7

⇒ 10d = – 10

Substituting, “d” in eq. (1)

⇒ 2a + 9 × ( – 1) = 7

⇒ 2a – 9 = 7

⇒ 2a = 7 + 9 = 16

Now, we can find value of t₁₀

Thus, t₁₀ = 8 + (10 – 1)× ( – 1)

⇒ t₁₀ = 8 – 9 = – 1

Given, first term a = – 5

Last term t_n = 45

Sum of n terms S_n = 120

To find no of terms “n”

Using Sum of n terms of an A.P. formula

where n = no. of terms

S_n = sum of n terms

Now, on substituting given value in formula we get,

⇒ 120 = 20 n

To find the common difference ‘d’

We use n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Thus, on substituting all values we get,

⇒ t₆ = – 5 + (6 – 1)d

⇒ 45 = – 5 + 5d

⇒ 5d = 45 + 5 = 50

Thus, common difference is 10

List of n natural number is

1, 2, 3, ……n

First term a = 1

Second term t₁ = 2

Third term t₃ = 3

Thus, common difference d = t₃ – t₂ = 3 – 2 = 1

Given S_n = 36

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find no. of terms n

⇒ n(1 + n) = 36 × 2 = 72

⇒ n² + n – 72 = 0

⇒ n² + 9n – 8n – 72 = 0

⇒ n(n + 9) – 8(n + 9) = 0

⇒ (n – 8)(n + 9) = 0

⇒ n – 8 = 0 or n + 9 = 0

⇒ n = 8 or n = – 9

Since, number of terms n can’t be negative

∴ n = 8

Let 3 parts of 207 be a – d, a, a + d such that,

⇒ (a – d) + a + (a + d) = 207

⇒ 3a = 207

Since, product of two smaller terms is 4623

⇒ (a – d) × a = 4623

⇒ (69 – d) × 69 = 4623

⇒ d = 69 – 67 = 2

Thus, a – d = 69 – 2 = 67

a = 69

a + d = 69 + 2 = 71

Thus, the A.P so formed is 67, 69, 71

Let first term = a

Common difference = d

Since, A.P. consist of 37 terms, therefor the middle most term is

Thus, three middle most term are t₁₈ = 18^th term, t₁₉ = 19^th term,

t₂₀ = 20^th term

We use n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Thus, on substituting all values we get,

Given, t₁₈ + t₁₉ + t₂₀ = 225

⇒ [a + (18 – 1)d] + [a + (19 – 1)d] + [a + (20 – 1)d] = 225

⇒ [a + 17d] + [a + 18d] + [a + 19d] = 225

⇒ 3a + 54d = 225

Dividing by 3

⇒ a + 18d = 75 …….(1)

Given, sum of last three term is 429

⇒ t₃₅ + t₃₆ + t₃₇ = 429

⇒ [a + (35 – 1)d] + [a + (36 – 1)d] + [a + (37 – 1)d] = 429

⇒ [a + 34d] + [a + 35d] + [a + 36d] = 429

⇒ 3a + 105d = 429

Dividing by 3

⇒ a + 35d = 143 …….(2)

Subtracting eq. (1) from eq. (2) we get,

⇒ [a + 35d] – [a + 18d] = 143 – 75

⇒ 17d = 68

Substituting value of ‘d’ in eq. (1) we get,

⇒ a + 18 × 4 = 75

⇒ a + 72 = 75

⇒ a = 75 – 72 = 3

⇒ a = t₁ = 3

We know that, t_{n + 1} = t_n + d

t₂ = t₁ + d = 3 + 4 = 7

t₃ = t₂ + d = 7 + 4 = 11

t₄ = t₃ + d = 11 + 4 = 15

t₃₇ = 3 + (37 – 1) × 4

t₃₇ = 3 + 36 × 4

t₃₇ = 3 + 144 = 147

Thus, the A.P. is 3, 7, 11, . . . . ., 147

Given first term = a

Second term = b

Last term = c

Common difference d = second term – first term = b – a

We will first find the number of terms

We use n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Thus, on substituting all values we get,

⇒ c = a + (n – 1)(b – a)

⇒ c = a + (b – a)n + a – b

⇒ c = 2a – b + (b – a)n

⇒ (b – a)n = c + b – 2a

Using Sum of n terms of an A.P. formula

where n = no. of terms

S_n = sum of n terms

On substituting all the values we get,

Hence, proved

We know that, sum of n^th term of an A.P. we will find it’s

sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

Now, Sum of p terms is

And, Sum of q terms is

Given: S_p = S_q

Multiply by 2 on both sides, we get,

⇒p[ 2a + (p – 1)d] = q[ 2a + (q – 1)d]

⇒2ap + p(p – 1)d = 2aq + q(q – 1)d

⇒ 2ap – 2aq + p(p – 1)d – q(q – 1)d = 0

⇒ 2a(p – q) + d[p² – p– q² + q] = 0

⇒ 2a(p – q) + d[(p²– q² ) – p + q] = 0

⇒ 2a(p – q) + d[(p– q )(p + q) – (p – q)] = 0

(since, (a – b)(a + b) = a² – b²)

⇒ 2a(p – q) + d(p – q) [p + q – 1 ] = 0

⇒ (p – q)[2a + d (p + q – 1) ] = 0

Since, p ≠ q

∴ p – q ≠ 0

⇒ 2a + d (p + q – 1) = 0

Multiply both side by

⇒ S_{p + q} = 0

Hence proved

We use n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Thus m^th term = t_m = a + (m – 1)d

Given: m × t_m = n × t_n

⇒ m × [ a + (m – 1)d] = n × [a + (n – 1)d]

⇒ am + m(m – 1)d = an + n(n – 1)d

⇒ am – an + m(m – 1)d – n(n – 1)d = 0

⇒ a(m – n) + d[m(m – 1) – n(n – 1)] = 0

⇒ a(m – n) + d[ m² – m – n² + n] = 0

⇒ a(m – n) + d[ (m² – n²) – m + n] = 0

⇒ a(m – n) + d[ (m – n)(m + n) –(m – n)] = 0

(since, (a – b)(a + b) = a² – b²)

⇒ a(m – n) + d(m – n)[(m + n) –1] = 0

⇒ (m – n) [a + d(m + n –1)] = 0

Since, m ≠ n

∴ m – n ≠ 0

⇒ a + d(m + n –1) = 0

⇒ t_{m + n} = 0

Hence proved

Given: Principal Amount P = 1000

Rate of interest R = 10%

Also,

Simple interest after 1 year

Simple interest after 2 year

Simple interest after 3 year

According to this the simple interest for 4, 5, 6 years will be 400,

500, 600 respectively.

Let first term a = 100

Second term t₁ = 200

Third term t₃ = 300

Common difference d = t₃ – t₂ = 300 – 200 = 100

Amount of simple interest after 20 years

We use n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

⇒ t₂₀ = 100 + (20 – 1) × 100

⇒ t₂₀ = 100 + 19 × 100

⇒ t₂₀ = 100 + 1900 = 2000