Class 10 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 3 - Arithmetic Progression

(1)

In the sequence 2, 4, 6, 8, …

First term = t₁ = 2, t₂ = 4, t₃ = 6, …

t₂ - t₁ = 4 - 2 = 2

t₃ - t₂ = 6 - 4 = 2

Here the first term is 2 and difference is constant which is 2.

∴ This sequence is an A.P.

(2)

In the sequence

First term = t₁ = 2,

Here the first term is 2 and difference is constant which is

∴ This sequence is an A.P.

(3)

In the sequence -10, -6, -2, 2, …

First term = t₁ = -10, t₂ = -6, t₃ = -2, …

t₂ - t₁ = -6 - (-10) = 4

t₃ - t₂ = -2 - (-6) = 4

Here the first term is 2 and common difference is 4.

∴ This sequence is an A.P.

(4)

In the sequence 0.3, 0.33, 0.333, …

First term = t₁ = 0.3, t₂ = 0.33, t₃ = 0.333, …

t₂ - t₁ = 0.33 - 0.3 = 0.03

t₃ - t₂ = 0.333 - 0.33 = 0.003

Here the difference between any two consecutive terms is not constant.

∴ This sequence is not an A.P.

(5)

In the _sequence 0, -4, -8, -12, …

First term = t1 = 0, t₂ = -4, t₃ = -8, …

t₂ - t₁ = -4 - 0 = -4

t₃ - t₂ = -8 - (-4) = -4

Here the first term is 0 and difference is constant which is -4.

∴ This sequence is an A.P.

(6)

In the sequence

First term

Here first term is and difference is constant which is 0.

∴ This sequence is an A.P.

(7)

In the sequence 3, 3 + , 3 + 3 , t₃ = 3 + .

∴ This sequence is an A.P.

(8)

In the sequence 127, 132, 137, …

First term t₁ = 127, t₂ = 132, t₃ = 137

t₂ - t₁ = 132 - 127 = 5

t₃ - t₂ = 137 - 132 = 5

Here first term is 127 and difference is constant which is 5.

∴ This sequence is an A.P.

(1)

Given: a = 10, d = 5

t₁ = 10

t₂ = t₁ + d = 10 + 5 = 15

t₃ = t₂ + d = 15 + 5 =20

t₄ = t₃ + d = 20 + 5 = 25

∴ A.P. is 10, 15, 20, 25, …

(2)

Given: a = -3, d = 0

t₁ = -3

t₂ = t₁ + d = -3 + 0 = -3

t₃ = t₂ + d = -3 + 0 = -3

t₄ = t₃ + d = -3 + 0 = -3

∴ A.P. is -3, -3, -3, -3, …

(3)

Given:

t₁ = -7

(4)

Given: a = -1.25, d = 3

t₁ = -1.25

t₂ = t₁ + d = -1.25 + 3 = 1.75

t₃ = t₂ + d = 1.75 + 3 = 4.75

t₄ = t₃ + d = 4.75 + 3 = 7.75

∴ A.P. is -1.25, 1.75, 4.75, 7.75, …

(5)

Given: a = 6, d = -3

t₁ = 6

t₂ = t₁ + d = 6 + (-3) = 3

t₃ = t₂ + d = 3 + (-3) = 0

t₄ = t₃ + d = 0 + (-3) = -3

∴ A.P. is 6, 3, 0, -3, …

(6)

Given: a = -19, d = -4

t₁ = -19

t₂ = t₁ + d = -19 + (-4) = -23

t₃ = t₂ + d = -23 + (-4) = -27

t₄ = t₃ + d = -27 + (-4) = -31

∴ A.P. is -19, -23, -27, -31, …

(1)

The given A.P. is 5, 1, -3, -7, …

Here, t₁ = 5, t₂ = 1

d = t₂ - t₁ = 1 - 5 = -4

∴ First term (a) = 5 and common difference (d) = -4

(2)

The given A.P. is 0.6, 0.9, 1.2, 1.5, …

Here, t₁ = 0.6, t₂ = 0.9

d = t₂ - t₁ = 0.9 - 0.6 = 0.3

∴ First term (a) = 0.6 and common difference (d) = 0.3

(3)

The given A.P. is 127, 135, 143, 151, …

Here, t₁ = 127, t₂ = 135

d = t₂ - t₁ = 135 - 127 = 8

∴ First term (a) = 127 and common difference (d) = 8

(4)

∴ First term (a) = and common difference (d) =

(i) 1, 8, 15, 22, …

Here a = , t1 = , t2 = , t3 = ,

t₂ - t₁ = - =

t₃ - t₂ = - = ∴ d =

(ii) 3, 6, 9, 12, ...

Here t₁ = , t₂ = , t₃ = , t₄ = ,

t₂ - t₁ = , t₃ - t₂ = ∴ d =

(iii) -3, -8, -13, -18, ...

Here t₁ = , t₂ =, t₄ = , t₄ =,

t₂ - t₁ =, t₃ - t₂ = ∴ a = , d =

(iv) 70, 60, 50, 40, …

Here t₁ =, t₂ =, t₃ = , …

∴ a = , d =

The given sequence is -12, -5, 2, 9, 16, 23, 30, …

Here, t₁ = -12, t₂ = -5, t₃ = 2, t₄ = 9

∴ t₂ - t₁ = -5 - (-12) = 7

t₃ - t₂ = 2 - (-5) = 7

t₄ - t₃ = 9 - 2 = 7

∴ t₂ - t₁ = t₃ - t₂ = … = 7 = d

∴ This sequence is an A.P.

Now, t_n = a + (n - 1)d

∴ t₂₀ = -12 + (20 - 1)7

= -12 + 19 × 7

= -12 + 133

= 121

∴ 20^th term of the given A.P. is 121.

Given A.P. is 12, 16, 20, 24, …

Here, t₁ = a = 12, d = 16 - 12 = 4

Now, t_n = a + (n - 1)d

∴ t₂₄ = 12 + (24 - 1)4

= 12 + 23 × 4

= 12 + 92

= 104

∴ The 24^th term of the given A.P. is 104.

Given A.P. is 7, 13, 19, 25, …

Here, t₁ = a = 7, d = 13 - 7 = 6

Now, t_n = a + (n - 1)d

∴ t₁₉ = 7 + (19 - 1)6

= 7 + 18 × 6

= 7 + 108

= 115

∴ The 15^th term of the given A.P. is 115.

Given A.P. is 9, 4, -1, -6, -11, …

Here, t₁ = a = 9, d = 4 - 9 = -5

Now, t_n = a + (n - 1)d

∴ t₂₇ = 9 + (27 - 1)(-5)

= 9 + 26 × (-5)

= 9 - 130

= -121

∴ The 24^th term of the given A.P. is -121.

The three digit natural numbers divisible by 5 are 100, 105, 110, …, 995

The above sequence is an A.P. as the difference between any two consecutive terms is constant.

∴ a = 100, d = 105 - 100 = 5

Let the number of terms in this A.P. be n.

∴ t_n = 995

Now, t_n = a + (n - 1)d

∴ 995 = 100 + (n - 1)5

∴ 995 - 100 = (n - 1)5

∴ 895 = (n - 1)5

∴ n - 1 = 179

∴ n = 180

∴ There are 180 three digit natural numbers which are divisible by 5.

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

Given: t₁₁ = 16, t₂₁ = 29

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

∴ t₁₁ = a + (11 - 1)d

∴ 16 = a + 10d

∴ a + 10d = 16 … (i)

Also, t₂₁ = a + (21 - 1)d

∴ 29 = a + 20d

∴ a + 20d = 29 …(ii)

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

Substituting in equation (i), we get

The 41^st term is

∴ t₄₁ = 55

∴ 41^st term of the A.P. is 55.

The given A.P. is 11, 8, 5, 2, …

Here, a = 11, d = 8 - 11 = -3

Let the n^th term of this A.P. be -151.

∴ t_n = -151

∴ a + (n - 1)d = -151

∴ 11 + (n - 1)(-3) = -151

∴ (n - 1)(-3) = -151 - 11

∴ (n - 1)(-3) = -162

∴ n - 1 = 54

∴ n = 55

∴ The 55^th term of the given A.P. is -151.

The natural numbers from 10 to 250 divisible by 4 are 12, 16, 20, …, 248

The above sequence is an A.P. as the difference is constant.

∴ a = 12, d = 16 - 12 = 4

Let the total number of terms in this A.P. be n.

∴ t_n = 248

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

∴ 248 = 12 + (n - 1)4

∴ 248 - 12 = (n - 1)4

∴ 236 = (n - 1)4

∴ n - 1 = 59

∴ n = 60

∴ There are 60 natural numbers from 10 to 250 which are divisible by 4.

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

As per the condition,

t₁₇ = t₁₀ + 7

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

∴ a + (17 - 1)d = a + (10 - 1)d + 7

∴ a + 16d = a + 9d + 7

∴ a + 16d - a - 9d = 7

∴ 7d = 7

∴ d = 1

Thus, the common difference is 1.

a = 6, d = 3, S₂₇ = ?

The even natural numbers are 2, 4, 6, 8, …

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 2, d = 2, n = 123

We know that,

Therefore,

∴ Sum of first 123 even natural numbers is 15252.

The even numbers between 1 and 350 are 2, 4, 6, 8, …, 348.

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 2, d = 2, t_n = 348

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

∴ 348 = 2 + ( n - 1)2

∴ 348 = 2 + 2n - 2

∴ 348 = 2n

∴ n = 174

Also,

∴ The sum of all even numbers between 1 and 350 is 30450.

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

Given: t₁₉ = 52, t₃₈ = 128

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

∴ t₁₉ = a + (19 - 1)d

∴ 52 = a + 18d

∴ a + 18d = 52 … (I)

Also, t₃₈ = a + (38 - 1)d

∴ 128 = a + 37d

∴ a + 37d = 128 … (II)

Adding equations (I) and (II), we get

Now,

∴ The sum of first 56 terms is 5040.

Sum of numbers from 1 to 140, which are divisible by 4 =.

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

Given: S₅₅ = 3300

Since,

And, t_n = a + (n - 1)d

∴ t₂₈ = a + (28 - 1)d

∴ t₂₈ = a + 27d

∴ t₂₈ = 60 … From (I)

∴ 28^th term of the A.P. is 60.

Let the three consecutive terms in an A.P. be a - d, a and a + d.

From the first condition, we have

a - d + a + a + d = 27

∴ 3a = 27

∴ a = 27/3

∴ a = 9 … (I)

From the second condition, we have

(a - d) a (a + d) = 504

∴ a(a²- d²) = 504

∴ 9(a²- d²) = 504 … From (I)

∴ 9(81 - d²) = 504

∴ 81 - d² = 504/9

∴ 81 - d² = 56

∴ d² = 81 - 56

∴ d² = 25

Taking square root on both the sides, we get

d = ± 5

When d = 5 and a =9, we get

a - d = 9 - 5 = 4

a = 9

a + d = 9 + 5 = 14

When d = -5 and a = 9,

a - d = 9 - (-5) = 9 + 5 = 14

a = 9

a + d = 9 - 5 = 4

∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.

Let the four consecutive terms in an A.P. be a - d, a, a + d and a + 2d.

From the first condition, we have

a - d + a + a + d + a + 2d = 12

∴ 4a + 2d =12

∴ 2(2a + d) = 12

∴ 2a + d = 12/2

∴ 2a + d = 6 … (I)

From the second condition,

a + d + a + 2d = 14

∴ 2a + 3d = 14 … (II)

Subtracting equation (I) from (II), we get

∴ d = 4

Substituting d = 4 in equation (I), we get

2a + 4 = 6

∴ 2a = 2

∴ a = 1

∴ a - d = 1 - 4 = -3

Also, a + d = 1 + 4 = 5

And, a + 2d = 1 + 8 = 9

∴ The four consecutive terms are -3, 1, 5 and 9.

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

Given: t₉ = 0

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

∴ t₉ = a + (9 - 1)d

∴ 0 = a + 8d

∴ a = -8d … (I)

Also, t₁₉ = a + (19 - 1)d

∴ t­₁₉ = a + 18d

∴ t­₁₉ = -8d + 18d … From (I)

∴ t₁₉ = 10d … (II)

And, t₂₉ = a + (29 - 1)d

i.e. t₂₉ = a + 28d

i.e. t₂₉ = -8d + 28d … From (I)

∴ t₂₉ = 20d = 2(10d)

∴ t₂₉ = 2(t₁₉) … From (II)

∴ The 29^th term is twice the 19^th term.

Sanika's daily savings of 2016 are as follows:

10, 11, 12, 13, …

This is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 10, d = 11 - 10 = 1

2016 is a leap year, so the number of days is 366.

∴ n =366

Now,

∴ Sanika' total savings on 31^st Dec 2016 will be Rs. 70455.

Each instalment being less than the preceding one by Rs. 40.

∴ The instalments are in A.P.

Amount repaid in 12 instalments

= Amount borrowed + total interest

= 8000 + 1360

∴ S₁₂ = 9360

Number of instalments (n) = 12

Since each instalment is less than the preceding one by Rs. 40.

∴ d = -40

Now,

Also, t_n = a + (n - 1)d

∴ t₁₂ = 1000 + (12 - 1)(-40)

= 1000 + 11(-40)

= 1000 - 440

= 560

∴ Amount of the first and last instalments are Rs. 1000 and Rs. 560 respectively.

Amount invested by Sachin in each year are as follows:

5000, 7000, 9000, …

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 5000, d = 7000 - 5000 = 2000, n = 12

Now,

∴ The total amount that Sachin invested in 12 years is Rs. 192000.

The number of seats arranged row-wise are as follows:

20, 22, 24, …

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 20, d = 22 - 20 = 2, n = 27

Now, t_n = a + (n - 1)d

∴ t₁₅ = 20 + (15 - 1)2

= 20 + 14 × 2

= 20 + 28

= 48

∴ t₁₅ = 48

∴ The number of seats in the 15th row is 48.

Also,

∴ The total number of seats in the auditorium is 1242.

Let the temperatures from Monday to Saturday in A.P. be

a, a + d, a + 2d, a + 3d, a + 4d, a + 5d.

According to the first condition, we have

(a) + (a + 5d) = (a + d) + (a + 5d) + 5°

∴ d = -5°

According to the second condition, we have

a + 2d = -30°

∴ a + 2(-5°) = -30°

∴ a - 10° = -30°

∴ a = -30° + 10° = -20°

∴ a + d = -20° - 5° = -25°

a + 3d = -20° + 3(-5°) = -20° - 15° = -35°

a + 4d = -20° + 4(-5°) = -20° - 20° = -40°

a + 5d = -20° + 5(-5°) = -20° - 25° = -45°

∴ The temperatures on the other five days are

-20°C, -25° C, -35° C, -40° C and -45° C.

The number of trees planted row-wise are as follows:

1, 2, 3, …

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 1, d = 2 - 1 = 1, n = 25

Now,

∴ The total number of trees in 25 rows is 325.

In the sequence -10, -6, -2, 2, …

First term = t₁ = -10, t₂ = -6, t₃ = -2, …

t₂ - t₁ = -6 - (-10) = 4

t₃ - t₂ = -2 - (-6) = 4

Here the first term is -10 and difference is constant which is 4.

∴ This sequence is an A.P. and the reason is d = 4.

Given: first term (a) = -2 and common difference (d) = -2

t₁ = -2

t₂ = t₁ + d = -2 - 2 = -4

t₃ = t₂ + d = -4 - 2 = -6

t₄ = t₃ + d = -6 - 2 = -8

∴ First four terms of the A.P. are -2, -4, -6, -8.

The first 30 natural numbers are 1, 2, 3, …, 30.

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 1, d = 2 - 1 = 1, n = 30

Now,

∴ The sum of the first 30 natural numbers is 465.

Given: t₇ = 4, d = -4

Now, t_n = a + (n - 1)d

∴ t₇ = a + (7 - 1)(-4)

∴ 4 = a + 6(-4)

∴ 4 = a - 24

∴ a = 4 + 24

∴ a = 28

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

Now, t_n = a + (n - 1)d

∴ t₁₀₁ = 3.5 + (101 - 1)(0)

= 3.5 + 100×0

= 3.5 + 0

= 3.5

∴ t₁₀₁ = 3.5

Given: a = t₁ = -3, t₂ = 4

Here, d = t₂ - t₁ = 4 - (-3) = 4 + 3 = 7

Now, t_n = a + (n - 1)d

∴ t₂₁ = -3 + (21 - 1)(7)

= -3 + 20×7

= -3 + 140

= 137

∴ 21^st term is 137.

Given: d = 5

Now, t_n = a + (n - 1)d

∴ t₁₈ - t₁₃ = a + (18 - 1)d - [a + (13 - 1)d]

= a + 17d - [a + 12d]

= a + 17d - a - 12d

= 5d

= 5×5

= 25

∴ t₁₈ - t₁₃ = 25

First five multiples of 3 are 3, 6, 9, 12, 15.

The above sequence is an A.P.as the difference between every two consecutive terms is constant.

∴ a = 3, d = 6 - 3 = 3, n = 5

Now,

∴ Sum of first five multiples of 3 is 45.

The sequence is 15, 10, 5, …

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 15, d = 10 - 15 = -5, n = 10

Now,

∴ The sum of first 10 terms is -75.

Given: First term (t₁) = 1, last term (t_n) = 20, S_n = 399

Now,

The given A.P. is -11, -8, -5, …, 49

Reversing the above A.P., we get 49, …, -5, -8, -11

This is also an A.P.

Here, a = 49, d = -11 - (-8) = -11 + 8 = -3

Now, t_n = a + (n - 1)d

∴ t₄ = 49 + (4 - 1)(-3)

= 49 + (3)(-3)

= 49 - 9

= 40

∴ Fourth term from the end in the given A.P. is 40.

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

Given: t₁₀ = 46, t₅ + t₇ = 52

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

∴ t₁₀ = a + (10 - 1)d

∴ 46 = a + 9d

i.e. a + 9d = 46 … (I)

Also, t₅ + t₇ = 52

∴ a + (5 - 1)d + a + (7 - 1)d = 52

∴ a + 4d + a + 6d = 52

∴ 2a + 10d = 52

∴ 2 (a + 5d) = 52

Dividing both the sides by 2, we get

a + 5d = 26 … (II)

Subtracting equation (II) from (I), we get

Substituting d = 5 in equation (II), we get

a + 5×5 = 26

∴ a + 25 = 26

∴ a = 26 - 25

∴ a = 1

∴ t₁ = a = 1

t₂ = t₁ + d = 1 + 5 = 6

t₃ = t₂ + d = 6 + 5 = 11

t₄ = t₂ + d = 11 + 5 = 116

∴ The required A.P. is 1, 6, 11, 16, …

Given: t₄ = -15, t₉ = -30

Now, t_n = a + (n - 1)d

∴ t₄ = a + (4 - 1)d

∴ -15 = a + 3d

∴ a + 3d = -15 … (I)

And, t₉ = a + (9 - 1)d

∴ -30 = a + 8d

∴ a + 8d = -30 … (II)

Subtracting equation (I) from (II), we get

Substituting d = -3 in equation (I), we get

a + 3×(-3) = -15

∴ a - 9 = -15

∴ a = -15 + 9

∴ a = -6

Since,

∴ The sum of the first 10 numbers is -195.

The first A.P. is 9, 7, 5, …

Here, a = 9, d = 7 - 9 = -2

Now, n^th term = t_n = a + (n - 1)d

∴ n^th term = 9 + (n - 1) (-2)

= 9 - 2n + 2

= 11 - 2n

The second A.P. is 24, 21, 18, …

Here, a = 24, d = 21 - 24 = - 3

∴ n^th term = a + (n - 1)d

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

= 24 - 3n + 3

= 27 - 3n

Since, the n^th terms of the two A.P.'s are equal.

∴ 11 - 2n = 27 - 3n

∴ 3n - 2n = 27 - 11

∴ n = 16

∴ t₁₆ = 9 + (16 - 1) (-2)

= 9 + 15 × (-2)

= 9 - 30

∴ t₁₆ = -21

∴ The values of n and n^th term are 16 and -21 respectively.

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

As per the first condition, we have

t₃ + t₈ = 7

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

∴ a + (3 - 1) d + a + (8 - 1)d = 7

∴ a + 2d + a + 7d = 7

∴ 2a + 9d = 7 … (I)

As per the second condition, we have

t₇ + t₁₄ = -3

∴ a + (7 - 1)d + a + (14 - 1 )d = -3

∴ a + 6d + a + 13d = -3

∴ 2a + 19 d = - 3 … (II)

Subtracting equation (I) from (II), we get

Substituting d = -1 in equation (I), we get

2a + 9×(-1) = 7

∴ 2a - 9 = 7

∴ 2a = 7 + 9

∴ 2a = 16

∴ a = 8

∴ t₁₀ = 8 + (10 - 1)(-1)

= 8 + 9(-1)

= 8 - 9

= -1

∴ The 10^th term of the A.P. is -1.

Let the number of terms in the A.P. be n and the common difference be 'd'.

∴ a = -5, t_n = 45, S_n = 120 … (Given)

Now, t_n = a + (n - 1)d

∴ 45 = -5 + (n - 1)d

∴ 45 + 5 = (n - 1)d

∴ (n - 1)d = 50 … (I)

Since,

Substituting n = 6 in equation (I), we get

(6 - 1)d = 50

∴ 5d = 50

∴ d = 10

∴ The number of terms is 6 and the common difference is 10.

The natural numbers from 1 to n are 1, 2, 3, …, n.

The above sequence is an A.P. as the difference between every two consecutive terms is constant.

∴ a = 1, d = 2 - 1 = 1

Also, S_n = 36 …[Given]

Since,

∴ 72 = n² + n

∴ n² + n - 72 = 0

∴ n² + 9n - 8n - 72 = 0

∴ n(n + 9) - 8 (n + 9) = 0

∴ (n + 9) (n - 8) = 0

∴ n + 9 = 0 or n - 8 = 0

∴ n = -9 or n = 8

But, number of terms cannot be negative.

∴ n = 8

∴ The value of n is 8.

Let the three parts of 207 be a - d, a, a + d.

As per the first condition, we have

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

∴ 3a = 207

∴ a = 69 … (I)

As per the second condition, we have

(a - d) × a = 4623

∴ (69 - d) × 69 = 4623 … [From (I)]

∴ 69 - d = 67

∴ d = 69 - 67

∴ d = 2

∴ a - d = 69 - 2 = 67

a = 69

a + d = 69 + 2 = 71

∴ The three parts of 207 that are in A.P. are 67, 69 and 71.

The number of terms is 37.

∴ Middle term

∴ The terms 18^th, 19^th and 20^th are placed exactly at the middle.

As per the first condition, we have

t₁₈ + t₁₉ + t₂₀ = 225

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

∴ a + (18 - 1)d + a + (19 - 1)d + a + (20 - 1)d = 225

∴ 3a + 17d + 18d + 19d = 225

∴ 3a + 54d = 225 … (I)

As per the second condition, we have

t₃₅ + t₃₆ + t₃₇ = 429

∴ a + (35 - 1)d + a + (36 - 1)d + a + (37 - 1)d = 429

∴ 3a + 34d + 35d + 36d = 429

∴ 3a + 105d = 429 … (II)

Subtracting equation (I) from (II), we get

Substituting d = 4 in equation (I), we get

3a + 54 × 4 = 225

∴ 3a + 216 = 225

∴ 3a = 225 - 216

∴ 3a = 9

∴ a = 3

∴ The required A.P. is 3, 3 + 4, 3 + 2×4, 3 + 3×4, …, 3 + (37 - 1)4

i.e. 3, 7, 11, 15, …, 144.

Given: t₁ = a, t₂ = b, t_n = c

∴ d = t₂ - t₁ = b - a

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

∴ c = a + (n - 1)(b - a)

Now,

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

The sum of first n terms of an A.P. is given by

As per the given condition, we have

S_p = S_q

∴ 2ap + p(p - 1)d = 2aq + q(q - 1)d

∴ 2ap + p²d - pd = 2aq + q²d - qd

∴ 2ap - 2aq + p²d - q²d - pd + qd = 0

∴ 2a(p - q) + d(p² - q²) - d(p - q) =

Since, a² - b² = (a + b)(a - b)

∴ 2a(p - q) + d(p + q)(p - q) - d(p - q) = 0

∴ (p - q)[2a + d(p + q) - d] = 0

∴ (p - q)[2a + d(p + q - 1)] = 0

∴ (p - q) = 0 or [2a + d(p + q - 1)] = 0

But p ≠ q

∴ 2a + d(p + q - 1) = 0 … (I)

Now, sum of (p + q) terms is

∴ The sum of first (p + q) terms is 0.

As per the given condition, we have

m × t_m = n × t_n

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

∴ m[a + (m - 1)d] = n[a + (n - 1)d]

∴ ma + md(m - 1) = na + nd(n - 1)

∴ ma + m²d - md = na + n²d - nd

∴ ma + m²d - md - na - n²d + nd = 0

∴ (ma - na) + (m²d - n²d) - (md - nd) = 0

∴ a(m - n) + d(m²- n²) - d(m - n) = 0

Since, a² - b² = (a + b)(a - b)

∴ a(m - n) + d(m + n) (m - n) - d(m - n) = 0

∴ (m - n)[a + (m + n - 1) d] = 0

∴ m - n = 0 or [a+ (m + n - 1)d] = 0

But m ≠ n

∴ a + (m + n - 1)d = 0

∴ t_(m+n) = 0

∴ The (m + n)^th term of the A.P. is 0.

Simple interest =

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,, respectively.

From this d = , and a =

Amount of simple interest after 20 years

tn= a + (n - 1)d

t₂0 =+ (20 - 1)

t₂0 =

Amount of simple interest after 20 years is =