Arithmetic Sequences

The number of dots in first 3 triangles are 3, 6 and 10 respectively.

The difference between the first two terms is 6-3 = 3

And 10-6 = 4

The series continues 3, 4, 5…

Hence, this forms an AP with first term as 3 and common difference as 1

Next two terms are 4 + 1 = 5

And 5 + 1 = 6

So 4^th pattern has 10 + 5 = 15 dots

5^th pattern has 15 + 6 = 21 dots

For the following sequences, let

a = First term of sequence

d = common difference if the sequence is an AP

a) Number of sides – 3,4,5,6,7…. And so on is an Arithmetic Progression.

Where a = 3 and d = 1

b) Sum of interior angles of a regular polygon with n sides = (n-2)×180

Putting n = 3, 4, 5, 6… and so on

The required sequence is 180,360,540,720… and so on.

An AP with a = 180 and d = 180

c) Sum of exterior angles of any closed polygon is 360

Hence the sequence is 360,360,360… and so on i.e. AP with a = 360 and d = 0

d) One Interior angle = (Sum of all interior angles)/(No. of sides)

=

Putting n = 3, 4, 5… and so on…

The sequence is 60, 90, 108, 120… and so on. Here, the sequence is not an AP

e) One exterior angle =

(∵ no. of exterior angles = no. of sides)

=

Putting n = 3, 4, 5… and so on

The sequence is 120, 90, 72, 60… and so on which is clearly not an AP.

Dividend = ( Divisor × Quotient) + Remainder

Here, remainder = 1

Divisor = 3

Dividend = ?

On taking quotient as 1, 2, 3 and so on…

Dividend = 4, 7, 10… and so on which is an AP with

a = 4 and d = 3

Now, for 2^nd case –

remainder = 2

Divisor = 3

Dividend = ?

On taking quotient as 1,2,3 and so on…

Dividend = 5,8,11… and so on which is an AP with

a = 5 and d = 3

Sequence of natural numbers ending in 1 or 6 is,

1, 6, 11, 16, 21, 26, 31, 36, …

It can also be described as,

Arithmetic sequence with first term as 1 and common difference as 5

Or

Sequences of natural numbers, which leave 1 as the remainder when divided by 5.

Volume of cube = side × side × side

For side = 1, Volume = 1 cc.

For side = 2, Volume = 8 cc.

For side = 3, Volume = 27 cc.

Hence, the sequence is 1, 8, 27… and so on which is clearly not an AP

i) Sequence of odd numbers-

Odd numbers are those which are not divisible by 2

i.e. 1, 3, 5… and so on

i.e. (2-1),(4-1),(6-1)…and so on

Thus Sn = 2n-1, where n = 1, 2, 3… and so on

ii) Sequence of natural numbers which leave remainder 1 on division by 3. -

Dividend = (Divisor × Quotient) + (Remainder)

Here, Divisor = 3

Remainder = 1

For getting the required sequence, putting quotient as 1, 2, 3… and so on

Dividend = 4,7,10…

i.e. Sn= 3n + 1

iii) The sequence of natural numbers ending in 1.

Natural numbers ending in 1 are 1, 11, 21… and so on

These can be written as (10-9),(20-9),(30-9),… and so on

Hence, the sequence Sn = 10n-9 where n = 1, 2, 3… and so on

iv) The sequence of natural numbers ending in 1 or 6.

The natural numbers ending in 1 or 6 are 1, 6, 11, 16, 21…and so on

Observing the pattern, these numbers can be written as (5-4), (10-4), (15-4),…and so on

Hence, the sequence Sn = 5n-4, where n = 1, 2, 3… and so on

a) Sum of interior angles of any n sided regular polygon = (n-2) × 180 (for n>2)

Hence, Sn = (n-2) × 180

b) Sum of exterior angles of any regular polygon is 360

Hence, Sn= 360 (independent of n)

c) Measure of an interior angle = (Sum of all interior angles) ÷ (No. of sides) (∵ no. of angles = no. of sides)

Sn = ((n-2) × 180) ÷ n

d) Measure of an exterior angle = (Sum of all exterior angles) ÷ (no. of sides)

= 360÷n

S_n = 360/n; where n = 1, 2, 3… and so on

i) How many red triangles are there in each picture?

→ In 1^st picture, 1 red triangle

In 2^nd picture, 3 red triangles

In 3^rd picture, 9 red triangles

S₁ = 1, 3, 9…

ii) Taking the area of the first triangles as 1, compute the area of a small triangle in each

picture.

→ In 2^nd picture, each triangle is divided into 4 parts in which there are 3 red and 1`white triangle.

Hence, area of each small triangle = (1÷ 4) = 0.25 sq. units

In 3^rd picture, each smaller triangle from the 2^nd figure, except the middle triangle is further divided into further 4 equal parts.

Hence, area of each small triangle = (0.25/4) = 0.625 sq. units

S₂ = 1, 0.25, 0.625…

iii) What is the total area of all the red triangles in each picture?

→ In 1^st triangle, only red triangle is present.

In 2^nd triangle, 3 red and 1 white triangle is present each of equal areas = 0.25 sq. units

∴ Area of red triangles = 3 × 0.25 = 0.75 sq. units

In 3^rd picture, each smaller triangle from the 2^nd figure, except the middle triangle is further divided into further 4 equal parts.

Area of all red triangles = 9 × 0.0625 = 0.5625

S₃ = 1, 0.75, 0.5625…

iv) Write the algebraic expressions for these three sequences obtained by continuing this process.

For 1^st question, the sequence is 1, 3, 9…

Hence, the expression is 3^n-1 , where n = 1,2,3…and so on

For 2^nd question, the sequence is 1, 0.25, 0.625…

Hence, the expression is (1÷4) ^(n-1) = (0.25)^n-1 ,where n = 1, 2, 3… and so on

For 3^rd question, the sequence is 1,0.75,0.675…

Here, the expression is product of above 2 expressions

i.e. 3^n-1 × 0.25^n-1 = 0.75^n-1

i) Sequence of odd numbers

→ Odd numbers are those which are not divisible by 2

i.e. 1,3,5,7…

As, can be seen from the above sequence, it is an Arithmetic Progression.

Common difference = 3-1 = 5-3 = 2

ii) Sequence of even numbers

→ Even numbers are those which are divisible by 2

I.e. 2,4,6…

As, can be seen from the above sequence, it is an Arithmetic Progression.

Common difference = 4-2 = 6-4 = 2

iii) Sequence of fractions got as half the odd numbers

→ Odd numbers are those which are not divisible by 2

i.e. 1,3,5,7…

The sequence mentioned is half the odd numbers…

I.e. 1/2, 3/2, 5/2, 7/2,…

As, can be seen from the above sequence, it is an Arithmetic Progression.

Common difference = = = 1

iv) Sequence of powers of 2

→ The sequence of powers of 2 is as follows

2¹,2²,2³,… and so on

i.e. 2,4,8,…

Difference of second and first term = 4-2 =2

Difference of third and second term = 8-4 =4

As the difference is not same, it is not an AP.

v) Sequence of reciprocals of natural numbers

→ Natural numbers are those starting from 1 with common difference as 1

Sequence of reciprocals of natural numbers is as follows –

1/1, 1/2, 1/3 , 1/4 , 1/5… and so on

Difference of second and first term = (1/2)-(1/1) = -1/2

Difference of third and second term = (1/3) – (1/2) = -1/6

As the difference is not same, it is not an AP.

No. of coloured squares in

Figure 1 – 8

Figure 2 – 12

Figure 3 – 16

As can be seen from the figures, the side of each coloured square increases linearly from 3,4,5… and so on. Hence, it is an AP

With common difference = 12-8 = 16-12 = 4

i) How many small squares are there in each rectangle?

→ No. of small squares in

Rectangle 1 - 2

Rectangle 2 - 4

Rectangle 3 - 6

Rectangle 4 - 8

As is evident, it forms an AP with common difference = 4-2 =2

ii) How many large squares?

→ No. of large squares in

Rectangle 1 - 0

Rectangle 2 - 1 (intersection of all 4 small squares)

Rectangle 3 - 2 (2 same size overlapping squares as in rectangle 2)

Rectangle 4 - 3

As is evident, it forms an AP with common difference = 1-0 =1

iii) How many squares in all?

→ All squares = small squares + large squares

No. of all squares in

Rectangle 1 - 2 + 0 = 2

Rectangle 2 - 4 + 1 = 5

Rectangle 3 - 6 + 2 = 8

Rectangle 4 - 8 + 3 = 11

As is evident, it forms an AP with common difference = 2+1 = 3

In Δ fab , Δ gac , and so on…

The subtended angle θ is same

∴ tanθ = constant

∴

Now l(ab)=l(bc)=l(cd)… given

∴ l(ac)=2× l(ab)

l(ad)= 3× l(ab)

For, tanθ = constant

l(gc)=2× l(fb)

l(hd)=3× l(fb)

Hence, lengths fb,gc,hd…are in AP

i.e. the length of perpendiculars are in AP.

For it to be an AP, x_n+1 – x_n = constant

x_n+1-x_n = {(n+1)³-6(n+1)²+13(n+1)-7}-{ n³ – 6n²+ 13n – 7}

= (n+1)³-n³-6{(n+1)²-n²}+13(n+1-n)-7+7

= (n+1-n)(n²+2n+1+n²+n+n²)-6(2n+1)+13

= 1× (3n²+3n+1)-12n+7

= 3n²-9n+8

Which is not independent of n. Hence, the sequence is not an AP.

Let a = first term of arithmetic sequence

a_{2 ,} a_{3 ,} a_4… are second, third, fourth terms and so on

d = common difference

i) Here, a = 24

a₂ = 42

d = 42-24 = 18

a₃ = 42 + 18 = 60

a₄ = 60 + 18 = 78

ii) Here, a₂ = 24

a₃ = 42

d = 42-24 = 18

a = a₂ – d = 24-18 = 6

a₄ = 42 + 18 = 60

iii) Here, a₃ = 24

a₄ = 42

d = a₄ – a₃ = 42-24 = 18

a₂ = 24 – 18 = 6

a = 6-18 = -12

iv) Here, a = 24

a₃ = 42

a₃ = a + 2d

Solving, d = 9

∴ a₂ = 24 + 9 = 33

a₄ = 42 + 9 = 51

v) Here, a₂ = 24

a₄ = 42

a₄ = a + 3d …(eq)1

and,a₂ = a + d …(eq)2

Eliminating a;

d = 9

a = 24-9 = 15

a₃ = 24 + 9 = 33

vi) Here, a = 24

a₄ = 42

a_{4 =} a + 3d

Solving, d = 6

a₂ = 24 + 6 = 30

a₃ = 30 + 6 = 36

Hence, all the answers are found.

i) a₃ = 34

a₆ = 67

a₃ = a + 2d and a₆ = a + 5d

Solving,a₆ = a₃ + 3d

∴ d = 11

a₂ = 34-11 = 23

a = 23-11 = 12

a₄ = 34 + 11 = 45

a₅ = 45 + 11 = 56

First 5 terms are 12,23,34,45 and 56

ii) a₃ = 43

a₆ = 76

a₃ = a + 2d and a₆ = a + 5d

Solving,a₆ = a₃ + 3d

∴ d = 11

a₂ = 43-11 = 32

a = 32-11 = 21

a₄ = 43 + 11 = 54

a₅ = 54 + 11 = 65

First 5 terms are 21,32,43,54 and 65

iii) a₃ = 2

a₅ = 3

a₃ = a + 2d and a₅ = a + 4d

Solving, a₅ = a₃ + 2d

d = 0.5

a₂ = 2-0.5 = 1.5

a = 1.5-0.5 = 1

a₄ = 2 + 0.5 = 2.5

First 5 terms are 1,1.5,2,2.5 and 3

iv) a₄ = 2

a₇ = 3

a₄ = a + 3d

a₇ = a + 6d

Solving,a₇ = a₄ + 3d

d =

a₃ = =

a₂ = =

a = = 1

a₅ = 2 + =

First 5 terms are 1,4/3,5/3,2 and 7/3

v) a₂ = 5

a₅ = 2

a₂ = a + d

a₅ = a + 4d

Solving,a₅ = a₂ + 3d

d = -1

a = 5-(-1) = 6

a₃ = 5-1 = 4

a₄ = 4-1 = 3

a₅ = 3-1 = 2

First 5 terms are 6,5,4,3,2.

Hence solved.

Given:

a₅ = 38 and,

a₉ = 66

We know, nth term is given as,

[a_n = a + (n-1)× d]

⇒ a₅ = a + 4d ..(1)

And, a₉ = a + 8d …(2)

Solving eq(1) and eq(2), we get,

a₉ = a₅ + 4d

⇒ 66 = 38 + 4d

⇒ .

Now, from eq (1),

a₅ = a + 4d

⇒ 38 = a + 4(7)

⇒ a = (38-(4×7))

⇒ a = 38 – 28

⇒ a = 10

And,

25^th term, a₂₅ = a + 24d

= 10 + (24×7)

= 178

Hence, 25^th term is 178

a = 13

a₂ = 24

a₃ = 35

d = 35-24 = 11

Case i)

a_n = 101

a_n = 13 + (n-1)d

Solving, n = 9

∴ 101 is the term of this AP.

Case ii)

a_n = 1001

a_n = 13 + (n-1)d

Solving, n = 90.8

∴ 1001 is not the term of this AP.

Dividend = (divisor× quotient) + remainder

Here, divisor = 7

Remainder = 3

First 3 digit no. is 100

For, dividend = 100, quotient = 13.6…

Hence, first 3 digit no. divisible by 7 with remainder 3 comes with quotient>13.6… = 14

When quotient = 14, dividend = 101

Last 3 digit no. divisible by 7 with remainder 3 = 997

a_n = 997

a = 101

d = 7

a_n = a + (n-1)× d

∴ n-1 = 128

∴ n = 129

Hence,129 numbers are present.

For 1^st row,

a = 1 a₄ = 4

a₄ = a + 3d

∴ d = 1

a₂ = 1 + 1 = 2

∴ A₁₂ = 2

a₃ = 2 + 1 = 3

∴ A₁₃ = 3

For 1^st column,

a = 1 a₄ = 7

a₄ = a + 3d

∴ d = 2

a₂ = 1 + 2 = 3

∴ A₂₁ = 3

a₃ = 3 + 2 = 5

∴ A₃₁ = 5

For 4^th row,

a = 7 a₄ = 28

a₄ = a + 3d

∴ d = 7

a₂ = 7 + 7 = 14

∴ A₄₂ = 14

a₃ = 14 + 7 = 21

∴ A₄₃ = 21

For 4^th column,

a = 4 a₄ = 28

a₄ = a + 3d

∴ d = 8

a₂ = 4 + 8 = 12 ∴ A₂₄ = 12

a₃ = 12 + 8 = 20 ∴ A₃₄ = 20

For 2^nd row,

a = 3 a₄ = 12

a₄ = a + 3d

∴ d = 3

a₂ = 3 + 3 = 6 ∴ A₄₂ = 6

a₃ = 6 + 3 = 9 ∴A₂₃ = 9

For 3^rd row,

a = 5 a₄ = 20

a₄ = a + 3d

∴ d = 5

a₂ = 5 + 5 = 10

∴ A₃₂ = 10

a₃ = 10 + 5 = 15

∴ A₃₃ = 15

If we use numbers other than 1,4,7,and 28 it is not guaranteed that the grid will be completely filled such that each row and column is in AP.

a) a = 11

d = 22-11 = 11

i) let a_n = 123

a_n = a + (n-1)×d

∴ n = 11.18

Since, n is not an integer, 123 is not a part of sequence

ii) let a_n = 132

a_n = a + (n-1)×d

∴ n = 12

Since, n is an integer, 132 is a part of sequence

b) a = 12

d = 23-12 = 11

i) let a_n = 100

a_n = a + (n-1)×d

∴ n = 9

Since, n is an integer, 100 is a part of sequence

ii) let a_n = 1000

a_n = a + (n-1)×d

∴ n = 90.81

Since, n is not an integer, 1000 is not a part of sequence

c) a = 21

d = 32-21 = 11

i) let a_n = 100

a_n = a + (n-1)×d

∴ n = 8.18

Since, n is not an integer, 100 is not a part of sequence

ii) let a_n = 1000

a_n = a + (n-1)×d

∴ n = 90

Since, n is an integer, 1000 is a part of sequence

d) a = 1/4

d = (1/2)-(1/4) = 1/4

i) let a_n = 3

a_n = a + (n-1)×d

∴ n = 12

Since, n is an integer, 3 is a part of sequence

ii) let a_n = 4

a_n = a + (n-1)×d

∴ n = 16

Since, n is an integer, 4 is a part of sequence

e) a = 3/4

d = (3/2)-(3/4) = 3/4

i) let a_n = 3

a_n = a + (n-1)×d

∴ n = 4

Since, n is an integer, 3 is a part of sequence

ii) let a_n = 4

a_n = a + (n-1)×d

∴ n = 5.33

Since, n is not an integer, 4 is not a part of sequence

S_n = 30 where n = 5

For an AP with a = first term

d = common difference

S_n =

Substituting in above equation…

∴ 2a + (5-1) × d = 12

∴ 2a + 4d = 12

∴ a + 2d = 6

i) Let a = 2, hence d = 2

Sequence is 2,4,6,8,10…

ii) Let a = 4, hence d = 1

Sequence is 4,5,6,7,8…

iii) Let a = 6, hence d = 0

Sequence is 6,6,6,6,6…

Hence, 3 sequences are found.

a = 1 S₄ = 100 n = 4

S_n =

Substituting everything in above equation,

2a + (4-1) × d = 50

∴ 3d = 48

∴ d = 16

First four terms are 1,17,33,49.

Hence, the answer is written above.

Let a,b,c and e are four consecutive terms of an AP with d = common difference.

b = a + d

c = a + 2d

e = a + 3d

First term + Last term = a + e = a + (a + 3d) = 2a + 3d …(1)

Second term + Third term = b + c = (a + d) + (a + 2d) = 2a + 3d …(2)

Hence,

First term + Last term = Second term + Third term

S_n = 100 where n = 4

For an AP with a = first term

d = common difference

S_n =

∴ 2a + (4-1) × d = 50

∴ 2a + 3d = 50

Let a = 1, hence d = 16

Sequence is 1,17,33,49…

Let a = 4, hence d = 14

Sequence is 4,18,32,46…

Let a = 7 hence d = 12

Sequence is 7,19,31,43…

a₈ = 12,

a₁₂ = 8

a_n = a + (n-1) × d

a₈ = a + 7d = 12 …(eq)1

a₁₂ = a + 11d = 8 …(eq)2

(eq)2-(eq)1

4d = -4

d = -1

a = 19

So, sequence is 19,18,17,16…

Let the possible no of birds be x,

Therefore, according to given condition

Where N is any natural number.

⇒ 11x = 4N

Also, no of birds should also be a natural number, therefore N should be a multiple of 11, i.e. N = 11n for any natural no 'n'

Hence possible values of N = 11, 22, 33…

And possible values of x = 4, 8, 12, …

Algebraic expression for N is 11n

And

For x is 4n.

For the sequence,

a =

d =

a_n = a + (n-1) × d

Let p be any natural number

p = + (n-1) ×

= +

=

6p = n + 1

Thus, for every natural number p, a value of n exists.

Hence, this arithmetic sequence contains all natural numbers.

For the sequence,

a =

d =

a_n = a + (n-1) × d

= + (n-1) ×

=

= k(2n-1) (where k = constant)

As a_n is a multiple of (2n-1) and not (2n) it contains all odd numbers but no even number..

a = first term = 4

d = common difference = 7-4 = 3

a_n = a + (n-1) × d

a_n = 4 + 3n-3 = 3n + 1

(a_n)² = (3n + 1)² = 9n² + 6n + 1 = 3n(3n + 2) + 1 = 3{n(3n + 2)} + 1

As, (a_n)² is also of the form 3n + 1, the squares of all the terms of the arithmetic sequence 4, 7, 10, … belong to the sequence.

a = 5

d = 8-5 = 3

a_n = a + (n-1) × d

a_n = 5 + (n-1) × 3

∴ a_n = 3n + 2

Let p be a natural number

p² = 3n + 2

n =

Now, for all integers from 0 to 9(i.e. p from 0 to 9), n does not come out to be an integer.

Hence, the arithmetic sequence 5, 8, 11, … contains no perfect squares.

Sum of all angles of a regular polygon with n sides = (n-2) × 180

Let a,a + d,a + 2d,a + 3d,a + 4d are 5 terms of an AP.

Sum of all angles = 5a + 10d = (5-2) × 180 = 540

∴ a + 2d = 108

The minimum angle will be obtained when a = d.

a + 2a = 3a = 108

∴ a = 36 is the minimum angle.

Let a =

a₂ =

a₃ =

a₂-a = -

a₃-a₂ = -

As a₃-a₂ = a₂-a

Hence, the common difference is same, i.e. the sequence forms an AP.

a = 11 = first term

d = 22-11 = 11 = common difference

S_n =

Here, n = 25

∴ S₂₅ =

= 25× 11× 13

= 3575

a = 12

d = 23-12 = 11

S_n =

Here, n = 25

∴ S₂₅ =

= 25× 24× 6

= 3600

a = 21

d = 32-21 = 11

S_n =

Here, n = 25

∴ S₂₅ = 3825

a = 19

d = 28-19 = 9

S_n =

Here, n = 25

∴ S₂₅ =

= 3175

a = 1

d = 6-1 = 5

S_n =

Here, n = 25

∴ S₂₅ =

= 1525

Hence, the sum of first 5 terms for each sequence is found out.

For the arithmetic sequence,

a = 6

d = 10-6 = 4

S_n = …(eq)1

Sum of first 20 terms = S₂₀

Sum of next 20 terms = S₄₀-s₂₀

Substituting values in eq1

S₂₀ =

= 10× 88

= 880

S₄₀ =

= 20× 168

= 3360

Sum of first 20 terms = S₂₀ = 880

Sum of next 20 terms = S₄₀-s₂₀ = 3360-880 = 2480

∴ Difference between sum of first 20 terms = 2480-880 = 1600

And next 20 terms

Hence, answer is found out.

Case i)

a = 6

d = 10-6 = 4

n = 20

S_n1 =

= 10×88

= 880

Case ii)

a = 15

d = 19-15 = 4

n = 20

S_n2 =

= 10×106

= 1060

The difference between the sum of the first 20 terms of the arithmetic sequences 6, 10, 14, … and 15, 19, 23, … = S_n1 – S_n2

= 1060- 880

= 180

Hence, answer is found out.

Three digit numbers which are multiples of 9 are in AP

The sequence is 108,117..999

a = 108

a_n = 999

d = 9

a_n = a + (n-1)×d

999 = 108 + (n-1)× 9

∴ n = 100

S_n =

=

= 50× 1107

= 55350

a^m× aⁿ = a^{m + n}

LHS = 5² × 5⁴ × 5⁶ × … × 5²ⁿ

= 5^{2 + 4 + 6 + …2n}

= 5^{2(1 + 2 + 3 + …n)}

{1 + 2 + 3 + …n = }

=

= 5^{n(n + 1)}

RHS = (0.008)^-30 = (8× 10^-3)^-30

= (2³× (2×5)^-3)^-30

= (2³× 2^-3× 5^-3)^-30

= 5⁹⁰

∵ (a^m)ⁿ = a^mn

LHS = RHS

∴ 5^{n(n + 1)} = 5⁹⁰

∴ n(n + 1) = 90

∴n² + n-90 = 0

Solving for n,

n = 9 and n = -10

But, n<0 is not possible

Hence, n = 9

S_n = n² + 2n

a_{n + 1} = S_{n + 1} - S_n

= {(n + 1)² + 2(n + 1)}-{n² + 2n}

= n² + 4n + 4-n²-2n

= 2n + 4

= 2(n + 1) + 2

∴ a_n = 2n + 2

S_n = 2n² + n

a_{n + 1} = S_{n + 1} - S_n

= {2(n + 1)² + (n + 1)}- {2n² + n}

= (2n² + 4n + 4 + n + 1)-(2n² + n)

= 4n + 5

= 4(n + 1) + 1

∴ a_n = 4n + 1

S_n = n² – 2n

a_{n + 1} = S_{n + 1}-S_n

= {(n + 1)²-2(n + 1)}- (n² – 2n)

= n² + 2n + 1-2n-2-n² + 2n

= 2n-1

= 2(n + 1)-3

a_n = 2n-3

S_n = 2n² - n

a_{n + 1} = S_{n + 1} - S_n

= {2(n + 1)²-(n + 1)}- {2n² - n}

= (2n² + 4n + 2-n-1)-(2n²-n)

= 4n + 1

= 4(n + 1)-3

a_n = 4n-3

S_n = n²-n

a_{n + 1} = S_{n + 1}-S_n

= {(n + 1)²-(n + 1)}-(n²–n)

= n² + 2n + 1-n-1-n² + n

= 2n

= 2(n + 1)-2

a_n = 2n-2

(i) Here, we will use formula in which we can calculate sum of AP by first term(a) and last term(a_n) i.e.

Where n is no of terms

Clearly first term, a = 51

Last term, a_n = 70

And no of terms, n = 70 - 51 + 1 = 20

Hence, sum is

= 10(121)

= 1210

(ii)

Here, we will use formula in which we can calculate sum of AP by first term(a) and last term(a_n) i.e.

Where n is no of terms

Clearly first term,

Last term,

Common difference,

Therefore, we can calculate no of terms usinh,

⇒ 11 = n - 1

⇒ n = 12

Hence, we have sum as

⇒ S_n = 6(14) = 84

(iii)

Here, we will use formula in which we can calculate sum of AP by first term(a) and last term(a_n) i.e.

Where n is no of terms

Clearly first term,

Last term,

Common difference,

Therefore, we can calculate no of terms usinh,

⇒ n = 25

Hence, we have sum as

S₁₀ = 350

S₅ = 100

S_n =

When n = 10,

S₁₀ = 350 =

∴ 2a + 9d = 70 …(eq)1

S₅ = 100 =

∴ a + 2d = 20 …(eq)2

Solving equations 1 and 2…

d = 6

a = 8

Hence, the sequence is 8,14,20…

a_n = a + (n-1)×d = 8 + (n-1)×6 = 6n + 2

a = 16

d = 24-16 = 8

Let number of terms = n

S_n =

=

= n(12 + 4n)

S_n + 9 = 4n² + 12n + 9 = (2n + 3)²

Hence, the sum of any number of terms of the arithmetic sequence 16, 24, 32, …. starting from the first, added to 9 gives a perfect square.

The first terms of each line are 4,7,13,22…

The difference between each consecutive term is

7-4 = 3

13-7 = 6

22-13 = 9

i.e. 3,6,9… in AP

For 4^th line,

First term = 22 + 12 = 34 , d = 3

For 5^th line

First term = 34 + 15 = 49 , d = 3

Also from observation, nth line has n terms

Hence, 4^th line

34,37,40,43

And 5^th line

49,52,55,58,61

The sequence of common difference is

The difference between the first terms of each line forms an AP which is

3,6,9…

For first term of 20^th line, sum of 19 terms of above sequence to be added to first term of first line.

Sum of 19 terms =

= 570

For 20^th line,

First term = 4 + 570 = 574

It has 20 terms

Last term = 574 + 19×3 = 574 + 57 = 631