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Here, a = 2, d = 7 – 2 = 5 and n = 10.

We have a_n = a + (n – 1) d

So, a₁₀ = 2 + (10 – 1) × 5 = 2 + 45 = 47

Therefore, the 10^th term of the given AP is 47.

1^st no. is 105 and last no. is 994.

Find n 994 = 105 + (n + 1)7

So, n = 128 

So, Sum, S₁₂₈=128/2[105 994]

11, 8, 5, 2, ……….. -150

a = 11, d = 8 11 = -3. a_n = -150.

a + (n – 1) d = a_n

11 + (n – 1) (-3) = -150

11 – 3n + 3 = -150

-3n + 14 = -150

-3n = -150 – 14

-3n = -164

3n = 164

∴ n = \(\frac{164}{3}\)

Here value of ‘n’ is not perfect. 

Hence -150 is not a term of the A.P.

C.Gauss

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

Given, A.P. 11, 8, 5, 2, …

Here, a = 11 and d = a₂ – a₁ = 8 – 11 = -3

We know that, n^th term a_n = a + (n – 1)d

Required to check n^th term a_n = -150

a + (n – 1)d = -150

11 + (n – 1)(-3) = -150

11 – 3n + 3 = -150

3n = 150 + 14

3n = 164

⇒ n = 164/3, which is not a whole number.

Therefore, -150 is not a term in the A.P.

Given, A.P. 7, 10, 13,…

Here, a = 7 and d = a₂ – a₁ = 10 – 7 = 3

We know that, n^th term a_n = a + (n – 1)d

Required to check n^th term a_n = 68

a + (n – 1)d = 68

7 + (n – 1)3 = 68

7 + 3n – 3 = 68

3n + 4 = 68

3n = 64

⇒ n = 64/3, which is not a whole number.

Therefore, 68 is not a term in the A.P.

a = 11, d = - 3

A_n = a + (n – 1) d

-150 = 11 + (n – 1) (- 3)

-150 = 11 – 3n + 3

-150 = 14 – 3n

-164 = 3n

Since, -164 is not divisible by 3

Therefore, -150 isn’t the term of the given A.P

a = 7, d = 3

A_n = a + (n – 1) d

68 = 7 + (n – 1) 3

68 = 4 + 3n

3n = 64

Since, 64 is not divisible by 3

Therefore, 68 isn’t the term of the given A.P.

Given,

A.P is 7, 10, 13,… 

Here, 

a₁ = a = 7, 

a₂ = 10 

Common difference, 

d = a₂ – a₁ 

= 10 – 7 

= 3 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a₁ and d is common difference and n is any natural number 

∴ a_n = 7 + (n – 1)3 

⇒ a_n = 7 + 3n – 3 

⇒ a_n = 3n + 4 

Now,

To find whether 68 is a term of this A.P. or not 

Put a_n = 68 

∴ 3n + 4 = 68 

⇒ 3n = 68 – 4 

⇒ 3n = 64

⇒ n = \(\frac{64}{3}\)

\(\frac{64}{3}\) is not a natural number

Hence, 

68 is not a term of given A.P.

a = 3, d = 5

A_n = a + (n – 1) d

302 = 3 + (n – 1) 5

302 = - 2 + 5n

304 = 5n

Since, 304 is not divisible by 5

Therefore, 302 isn’t the term of the given A.P.

Given, A.P. 3, 8, 13,…

Here, a = 3 and d = a₂ – a₁ = 8 – 3 = 5

We know that, n^th term a_n = a + (n – 1)d

Required to check n^th term a_n = 302

a + (n – 1)d = 302

3 + (n – 1)5 = 302

3 + 5n – 5 = 302

5n – 2 = 302

5n = 304

⇒ n = 304/5, which is not a whole number.

Therefore, 302 is not a term in the A.P.

(B) 3.5

Explanation:

We know that 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

(Since, d = 0, it’s a constant A.P)

Given,

A.P is 3, 8, 13,… 

Here, 

a₁ = a = 3, 

a₂ = 8 

Common difference, 

d = a₂ – a₁ 

= 8 – 3 

= 5

 We know, 

a_n = a + (n – 1)d 

Where a is first term or a₁ and d is common difference and n is any natural number 

∴ a_n = 3 + (n – 1)5 

⇒ a_n = 3 + 5n – 5 

⇒ a_n = 5n – 2 

To find : 

Whether 302 is a term of this A.P. or not 

Put a_n = 302 

∴ 5n – 2 = 302 

⇒ 5n = 302 + 2 

⇒ 5n = 304

⇒ n = \(\frac{304}{5}\)

\(\frac{304}{5}\) is not a natural number

Hence,

304 is not a term of given A.P.

(B) an AP with d = 4

Explanation:

According to the question,

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, it’s an A.P with d = 4

Correct option is (B) 104

Given sequence is 5, 14, 23, ......

\(\therefore a_1=5,a_2=14,a_3=23\)

Now \(a_2-a_1=14-5=9,\)

\(a_3-a_2=23-5=9\)

\(\because\) \(a_2-a_1\) \(=a_3-a_2\)

\(\therefore\) Given sequence is an arithmetic progression whose first term is 5 & common difference is 9.

\(\because\) \(a_{12}=a+11d\)        \((\because a_n=a+(n-1)d)\)

\(=5+11\times9\)             \((\because a=a_1=5\;\&\;d=a_2-a_1=9)\)

\(=5+99=104\)

Correct option is B) 104

4, 7, 10, 13, 16, …… 

Subtracting ‘2’ from the each term of A.P in given series, we get 

4 – 2, 7 – 2, 10- 2, 13 – 2, 16 – 2, …… 

2, 5, 8, 11, 14, …… 

In the list obtained, the first term 

a₁ = 2 , a₂ = 5, a₃ = 8, a₄ = 11, 

Also a₂ – a₁ = 5 – 2 = 3 

a₃ – a₂ = 8 – 5 = 3 

a₄ – a₃ = 11 – 8 = 3 …………………………………… 

i.e., d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = 3 

∴ The resulting list forms an A.P.

4, 7, 10, 13, 16, …… 

Multiplying each term by 3, we get 

4 × 3, 7 × 3, 10 × 3, 13 × 3, 16 × 3, …… 

12, 21, 30, 39, 48, ……. 

In the list obtained the first term a₁ = 12 and a₂ = 21, a₃ = 30, …. 

Also a₂ – a₁ = a₃ – a₂ = …… = 9 

∴ The resulting list also forms an A.P. 

Now divide every term by 7, we get

\(\frac{4}{7}\), \(\frac{7}{7}\), \(\frac{10}{7}\), \(\frac{13}{7}\), \(\frac{16}{7}\), …… is the resulting list.

Since (x + 2), 2x and (2x + 3) are in AP, we have

2x - (x + 2) = (2x + 3) - 2x

\(\Rightarrow\) x - 2 = 3

\(\Rightarrow\) x = 5

∴ x = 5

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0 

a₁ = 3, a₂ = -1, a₃ = -5, a₄ = -9 

Now, 

a₂ – a₁ = -1 – 3 = -4 

a₃ – a₂ = -5 – (-1) 

= -5 + 1 = -4 

a₄ – a₃ = -9 – (-5) 

= -9 + 5 = -4 

As, 

a₂ – a₁ = a₃ – a₂ = a₄ – a₃ 

The given sequence is A.P 

Common difference, 

d = a₂ – a₁ = -4 

To find next three more terms of A.P, firstly find a_n 

We know, 

a_n = a + (n-1)d where a is first term or a₁ and d is common difference 

∴ a_n = 3 + (n-1) -4 

⇒ a_n = 3 – 4n + 4 

⇒ a_n = 7 – 4_n 

When n = 5 : 

a₅ = 7 – 4(5) 

⇒ a₅ = 7 – 20 

⇒ a₅ = -13 

When n = 6 : 

a₆ = 7 – 4(6) 

⇒ a₆ = 7 – 24 

⇒ a₆ = -17 

When n = 7 : 

a₇ = 7 – 4(7) 

⇒ a₇ = 7 – 28 

⇒ a₇ = -21 

Hence, 

A.P is 3, -1, -5, -9, -13, -17, -21,…

9, 7, 5, 3 …

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0 

a₁ = 9, a₂ = 7, a₃ = 5, a₄ = 3 

Now, 

a₂ – a₁ = 7 – 9 = -2 

a₃ – a₂ = 5 – 7 = -2 

a₄ – a₃ = 3 – 5 = -2 

As, 

a₂ – a₁ = a₃ – a₂ = a₄ – a₃ 

The given sequence is A.P Common difference, 

d = a₂ – a₁ = - 2 

To find the next three more terms of A.P, firstly find a_n 

We know, 

a_n = a + (n-1) d 

Where a is first term or a1 and d is common difference 

∴ a_n = 9 + (n-1) -2 

⇒ a_n = 9 – 2n + 2 

⇒ a_n = 11 – 2n 

When n = 5 : 

a₅ = 11 – 2(5) 

⇒ a₅ = 11 – 10 

⇒ a₅ = 1 

When n = 6 : 

a₆ = 11 – 2(6) 

⇒ a₆ = 11 – 12 

⇒ a₆ = -1 

When n = 7 : 

a₇ = 11 – 2(7) 

⇒ a₇ = 11 – 14 

⇒ a₇ = -3 

Hence, 

A.P is 9, 7, 5, 3, 1, -1, -3,….

Differences between a Sequence and Set:

There are two types of sequences:

a. Finite Sequence: If the number of terms in a sequence is finite (countable) i.e. if there is an end of terms in the sequence then it is called a Finite Sequence.

Examples:

i. 1, 2, 3, … 20.

ii. 4, 6, 8, … 50.

iii. 1, 4, 9, 16, … 100.

b. Infinite Sequence: If the number of terms in a sequence is infinite (uncountable) i.e. there is no end of terms in the sequence then it is called an Infinite Sequence.

Examples:

i. 1, 3, 5, 7, …

ii. 5, 10, 15, …

iii. 2, 4, 6, 8, …

Each number in the sequence is called a term of the sequence.

The number in the first position is called the first term and is denoted by t₁.

The number in the second position is called the second term and is denoted by t₂.

Similarly, the number in the ‘n^th’ position of the sequence is called the nth term and is denoted by t_n.

If t_n is given, then a sequence can be formed.

A sequence is a collection of numbers arranged in a definite order according to some definite rule.

Examples: i. 1, 4, 9, 16, … (Collection of perfect squares of natural numbers)

ii. -2, -4, -6, … (Collection of negative even integers)

The correct option is: (C) 10 

Explanation:

Given A.P. is 5, 2, -1 , ......

=> a = 5, d = 2 - 5 = -3

T_n = -22 = a + (n-1)d = -22

=> 5 + (n- 1) (-3) = -22+ n = 10

Hence, 10^th term of the given A.P is -22.

Given,

a₁ = 3 and a_n = 3a_n–1 + 2, for all n > 1 

We can find the first four terms of a sequence by putting values of n from 1 to 4 

When n = 1 : 

a₁ = 3 

When n = 2 : 

a₂ = 3a_2–1 + 2 

⇒ a₂ = 3a₁ + 2 

⇒ a₂ = 3(3) + 2 

⇒ a₂ = 9 + 2 

⇒ a₂ = 11 

When n = 3 : 

a₃ = 3a_3–1 + 2 

⇒ a₃ = 3a₂ + 2 

⇒ a₃ = 3(11) + 2 

⇒ a₃ = 33 + 2 

⇒ a₃ = 35 

When n = 4 : 

a₄ = 3a_4–1 + 2 

⇒ a₄ = 3a_{3 + 2} 

⇒ a₄ = 3(35) + 2 

⇒ a₄ = 105 + 2 

⇒ a₄ = 107 

∴ First four terms of sequence are 3, 11, 35, 107.

Given as

a₁ = 3 and a_n = 3a_n–1 + 2, for all n > 1

By using the values n = 1, 2, 3, 4 we can find the first four terms.

Now, when n = 1:

a₁ = 3

When n = 2:

a₂ = 3a_2–1 + 2

= 3a₁ + 2

= 3(3) + 2

= 9 + 2

= 11

When n = 3:

a₃ = 3a_3–1 + 2

= 3a₂ + 2

= 3(11) + 2

= 33 + 2

= 35

When n = 4:

a₄ = 3a_4–1 + 2

= 3a₃ + 2

= 3(35) + 2

= 105 + 2

= 107

Thus, first four terms of sequence are 3, 11, 35, 107.

The correct option is: (A) p + q - n

Explanation:

We have given that a_p = q and a_q = p

=> q = a+ (p - 1)d and      ....(i)

p = a + (q - 1)d       ....(ii)

Subtracting (ii) from (i), we get

q - p = d(p - q) = d = -1

Now, q = a + 1 - p         [From (i)]

=> a = q + p - 1

. ^.. a_n = a + (n - 1) d = q + p -1 +(n - 1) (- 1)

= q + p - 1 + 1 - n = q + p - n

Given, 

a₁ = 2 = a₂, a_n = a_n–1 – 1, n>2 

We can find the first five terms of a sequence by putting values of n 

From 1 to 5 

When n = 1 : 

a₁ = 2 

When n = 2 : 

a₁ = 2 

When n = 3 : 

a₃ = a_3–1 – 1 

⇒ a₃ = a₂ – 1 

⇒ a₃ = 2 – 1 

⇒ a₃ = 1 

When n = 4 : 

a₄ = a_4–1 – 1 

⇒ a₄ = a_{3 – 1} 

⇒ a₄ = 1 – 1 

⇒ a₄ = 0 

When n = 5 : 

a₅ = a_5–1 – 1 

⇒ a₅ = a₄ – 1 

⇒ a₅ = 0 – 1 

⇒ a₅ = -1 

∴ First five terms of the sequence are 2, 2, 1, 0, -1.

Given, 

a₁ = 1 = a₂, a_n = a_n–1 + a_n–2, n>2 

We can find the first five terms of a sequence by putting values of n from 1 to 5 

When n = 1 : 

a₁ = 1 

When n = 2 : 

a₁ = 1 

When n = 3 : 

a₃ = a_3–1 + a_3–2 

⇒ a₃ = a₂ + a₁ 

⇒ a₃ = 1 + 1 

⇒ a₃ = 2 

When n = 4 : 

a₄ = a_4–1 + a_4–2 

⇒ a₄ = a₃ + a₂ 

⇒ a₄ = 2 + 1 

⇒ a₄ = 3 

When n = 5 : 

a₅ = a_5–1 + a_5–2 

⇒ a₅ = a₄ + a₃ 

⇒ a₅ = 3 + 2 

⇒ a₅ = 5 

∴ First five terms of the sequence are 1, 1, 2, 3, 5.

(i) a₁ = 1, a_n = a_n–1 + 2, n > 1

By using the values n = 1, 2, 3, 4, 5 we can find that the first five terms.

Given as

a₁ = 1

Now, when n = 2:

a₂ = a_2–1 + 2

= a₁ + 2

= 1 + 2

= 3

When n = 3:

a₃ = a_3–1 + 2

= a₂ + 2

= 3 + 2

= 5

When n = 4:

a₄ = a_4–1 + 2

= a₃ + 2

= 5 + 2

= 7

When n = 5:

a₅ = a_5–1 + 2

= a₄ + 2

= 7 + 2

= 9

Thus, first five terms of the sequence are 1, 3, 5, 7, 9.

(ii) a₁ = 1 = a₂, a_n = a_n–1 + a_n–2, n > 2

By using the values n = 1, 2, 3, 4, 5 we can find the first five terms.

Given as

a₁ = 1

a₂ = 1

When n = 3:

a₃ = a_3–1 + a_3–2

= a₂ + a₁

= 1 + 1

= 2

When n = 4:

a₄ = a_4–1 + a_4–2

= a₃ + a₂

= 2 + 1

= 3

When n = 5:

a₅ = a_5–1 + a_5–2

= a₄ + a₃

= 3 + 2

= 5

Thus, first five terms of the sequence are 1, 1, 2, 3, 5.

(iii) a₁ = a₂ =2, a_n = a_n–1 – 1, n > 2

By using the values n = 1, 2, 3, 4, 5 we can find the first five terms.

Given as

a₁ = 2

a₂ = 2

When n = 3:

a₃ = a_3–1 – 1

= a₂ – 1

= 2 – 1

= 1

When n = 4:

a₄ = a_4–1 – 1

= a₃ – 1

= 1 – 1

= 0

When n = 5:

a₅ = a_5–1 – 1

= a₄ – 1

= 0 – 1

= -1

Thus, first five terms of the sequence are 2, 2, 1, 0, -1.

Given, 

a₁ = 1, 

a_n = a_n–1 + 2, n > 1 

We can find the first five terms of a sequence by putting values of n 

From 1 to 5 

When n = 1 : 

a₁ = 1 

When n = 2 : 

a₂ = a_2–1 + 2 

⇒ a₂ = a₁ + 2 

⇒ a₂ = 1 + 2 

⇒ a₂ = 3 

When n = 3 : 

a₃ = a_3–1 + 2 

⇒ a₃ = a₂ + 2 

⇒ a₃ = 3 + 2 

⇒ a₃ = 5 

When n = 4 : 

a₄ = a_4–1 + 2 

⇒ a₄ = a₃ + 2 

⇒ a₄ = 5 + 2 

⇒ a₄ = 7 

When n = 5 : 

a₅ = a_5–1 + 2 

⇒ a₅ = a₄ + 2 

⇒ a₅ = 7 + 2 

⇒ a₅ = 9 

∴ First five terms of the sequence are 1, 3, 5, 7, 9.

To prove : 

Sum of the n^th term from the beginning and the nth term from the end is a + l 

We know, 

a_n = a + (n – 1)d 

Where a is first term or a₁ and d is the common difference, and n is any natural number, and n th term from the end is an'= l – (n – 1)d 

Now, 

a_n + a_n' = a + (n – 1)d + l – (n – 1)d 

⇒ a_n + a_n' = a + nd – d + l – nd + d 

⇒ a_n + an' = a + l 

Hence Proved.

Let the three terms be a - d, a, a + d

a – d + a + a – d = 21 (given)

3a = 21

a = 7

According to question,

(a – d) (a + d) = a + 6

a ² – d² = a + 6

7 ² – d² = 7 + 6

49 – d² = 13

d ² = 36

d = 6

Therefore,

a – d = 7 – 6 = 1

a + d = 7 + 6 = 13

Thus, the three terms are 1, 7 and 13.

To Find: 2 - digit numbers divisible by 3.

First 2 - digit number divisible by 3 is 12

Second 2 - digit number divisible by 3 is 15 and

Last 2 - digit number divisible by is 99.

Given: The AP is 12, 15, 18,…………..,99

a₁ = 12, a₂ = 15, d = 15 –12 = 3 and a_n = 99

(Where a = a₁ is First term, a₂ is Second term, a_n is n^th term and d is common difference of given AP)

Formula Used: a_n = a + (n – 1)d

99 = 12 + (n - 1)3

87 = (n – 1)3

29 = (n – 1)

n = 30

So, There are total of 30 two - digit number which is divisible by 3.

Given : 

The sum of first three terms is 21 

To find : 

The first three terms of AP 

Assume the first three terms are a - d, a, a + d where a is the first term and d is the common difference 

So, 

Sum of first three terms is a - d + a + a + d = 21 

3a = 21 

a = 7 

It is also given that product of first and third term exceeds the second by 6.

So,

(a - d)(a + d) - a = 6 

a² - d² - a = 6

Substituting a = 7,

7² - d² - 7 = 6 

d² = 36 

d = 6 

or d = - 6 

Hence,

The terms of AP are a - d, a, a + d which is 1, 7, 13 or 13, 7, 1.

Correct option is (B) 13

Given sequence is -8, -6, -4, .......

\(\because\) \(a_2-a_1=-6-(-8)=2\)

\(a_3-a_2=-4-(-6)=2\)

\(\because\) \(a_3-a_2\) = \(a_2-a_1\)

\(\therefore\) Sequence -8, -6, -4, ....... is an A.P.

whose common difference is d = 2 & first term is \(a=a_1=-8.\)

Let n terms are to be added to make the sum 52 in the given series.

i.e., \(S_n=52\)

\(\Rightarrow\frac n2[2a+(n-1)d]=52\)

\(\Rightarrow\frac n2[2\times-8+(n-1)2]=52\)      \((\because a=-8\;\&\;d=2)\)

\(\Rightarrow n[-16+2n-2]=52\times2=104\)

\(\Rightarrow2n^2-18n-104=0\)

\(\Rightarrow n^2-9n-52=0\)

\(\Rightarrow n^2-13n+4n-52=0\)

\(\Rightarrow n(n-13)+4(n-13)=0\)

\(\Rightarrow(n-13)(n+4)=0\)

\(\Rightarrow n-13=0\;or\;n+4=0\)

\(\Rightarrow n=13\;or\;n=-4\)

\(\because\) Number of terms never be negative.

\(\therefore n\neq-4\)

Therefore n = 13

Hence, 13 terms are to be added to make the sum 52 in the given series.

Correct option is B) 13

Given: 

The n^th term of an A.P. (-4n+15).

To find: the sum of first 20 terms of this A.P. 

Solution: We have, 

T_n = (-4n + 15) 

T₁ = -4 + 15 = 11 

T₂ =( -4 × 2 )+ 15 = 7

 T₃ =( -4 × 3 ) + 15 = 3

Hence the A.P is 11,7,3,.........

The first term is 11 and the common difference is, d = T₂ – T₁ = 7 – 11 = -4 

We calculate the sum of terms of A.P by using the formula:S_n= \(\frac{n}{2}\)[2a + (n – 1) d] 

Substitute the known values to get the sum.

The sum of first 20 terms, S₂₀ = \(\frac{20}{2}\)[(2 x 11) + (20 - 1)(-4)]

S₂₀ = \(\frac{20}{2}[(22 + (19)(-4)]\)

S₂₀ = 10 (22 – 76) 

S₂₀ = 10(-54)

S₂₀ = -540

We have sum of n terms, 

S_n = 3n² + 4n 

Put n = 1 

S₁ = T₁ = 3(1)² + 4 (1) = 7 

Put n = 2 

S₂ = 3(2)² + 4 (2) = 20 

T₂ = S₂ – S₁ = 20 – 7 = 13 

Put n = 3 

S₃ = 3(3)² + 4(3) = 39

T₃ = S₃ – S₂ = 39 – 20 = 19 

Therefore, first term is 7 and common difference, 

d = 13 – 7 = 6 

The 25^th term is, T_n= a + (n – 1) d 

T₂₅ = 7 + (25 – 1) x 6 

= 7 + 24 x 6 = 151

The given A.P. is -12, -9, -6,…21 

Here, a = -12, d = -9 – (-12) = 3 

Let the number of terms be n 

T_n = a + (n – 1) d 

21 = -12 + (n – 1) (3) 

21 = -15 + 3n 

36 = 3n

n = 12 

If 1 is added to each term than A.P. is -11, -8, -5…20 

Here, a = -11, d = 3 

S₁₂ = \(\frac{n}{2}\)[2(-11) + (12 -1) x 3] = 6 (-22 + 33)

= 66

AP is 21, 18, 15,…

a = 21,

d = 18 – 21 = -3

Sum of terms = S_n = 0

(n/2) [2a + (n – 1)d] = 0

(n/2) [2(21) + (n – 1)(-3)] = 0

(n/2) [45 – 3n] = 0

[45 – 3n] = 0

n = 15 (number of terms)

Thus, 15 terms of the given AP sums to zero.

We know that,

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (2, 5)

Multiples of 2 or 5 = Multiple of 2 + Multiple of 5 – Multiple of LCM (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)

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

Consider the first 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

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_1,

S₁ = S₂₅₀ = (250/2) ×[2+500]

S₁ = 125(502)

S₁ = 62750 … (1)

Consider the second 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

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_2,

S₂ = S₁₀₀ = (100/2) ×[5+500]

S₂ = 50(505)

S₂ = 25250 … (2)

Consider the third 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

Sum of an AP, S_n = (n/2) [ a + a_n]

Let the sum of this AP be S_3,

S₃ = S₅₀ = (50/2) × [2+510]

S₃ = 25(510)

S₃ = 12750 … (3)

Therefore, the required Sum, S = S₁ + S₂ – S₃

S = 62750 + 25250 – 12750

= 75250

Show that: the sum of the interior angles of polygons with 3, 4, 5, 6, …. sides form an arithmetic progression.

To Find: The sum of the interior angles for a 21 - sided polygon.

Given: That the sum of the interior angles of a triangle is 180°.

NOTE: We know that sum of interior angles of a polygon of side n is (n – 2) x 180°.

Let a_n = (n – 2) x 180°

⇒ Since a_n is linear in n.

So it forms AP with 3, 4, 5, 6,……sides

{a_n is the sum of interior angles of a polygon of side n}

By using the above formula, we have

a₂₁ = (21 – 2) x 180°

a₂₁ = 3420°

So, the Sum of the interior angles for a 21 - sided polygon is equal to 3420°.

We know the arithmetic mean of ‘a’ and ‘b’ is (a+b)/2
Hence (a+b)/2 = (aⁿ⁺¹+bⁿ⁺¹)/(aⁿ+bⁿ)
(a+b)/2 = (aⁿ x a + bⁿ x b)/(aⁿ+bⁿ)  
(aⁿa + aⁿb + bⁿa + bⁿb) = (2aⁿa+2bⁿb)
(aⁿb + bⁿa) = (aⁿa + bⁿb)
aⁿ(b – a)=bⁿ(b – a)
aⁿ = bⁿ
This is possible only when n = 0

Given,

A.P is 12 + 8i, 11 + 6i, 10 + 4i, … 

Here, 

a₁ = a = 12 + 8i, 

a₂ = 11 + 6i 

Common difference, 

d = a₂ – a₁ 

= 11 + 6i – (12 + 8i) 

= 11 – 12 + 6i – 8i 

= -1 – 2i 

We know,

a_n = a + (n – 1)d 

Where a is first term or a₁ and d is common difference and n is any natural number.

∴ a_n = 12 + 8i + (n – 1) -1 – 2i 

⇒ a_n = 12 + 8i – n – 2ni + 1 + 2i 

⇒ a_n = 13 + 10i – n – 2ni 

⇒ a_n = (13 – n) + (10 – 2n)i 

To find purely real term of this A.P., 

Imaginary part have to be zero 

∴ 10 – 2n = 0 

⇒ 2n = 10

⇒ n = \(\frac{10}{2}\)

⇒ n = 5 

Hence, 

5^th term is purely real 

To find purely imaginary term of this A.P., 

Real part have to be zero 

∴ 13 – n = 0 

⇒ n = 13 

Hence, 

13^th term is purely imaginary.

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.

Correct option is (B) \(\sqrt{mn} \)

Let first term & common ratio of the G.P. is a & r respectively.

Given that \(a_{P+Q}=m,a_{P-Q}=n\)

\(\Rightarrow\) \(ar^{P+Q-1} = m\)       _____________(1)

and \(ar^{P-Q-1}=n\)     _____________(2)    \((\because a_n=ar^{n-1})\)

Divide equation (1) by (2), we get

\(\frac{ar^{P+Q-1}}{ar^{P-Q-1}}=\frac mn\)

\(\Rightarrow r^{P+Q-1-(P-Q-1)}=\frac mn\)

\(\Rightarrow r^{2Q}=\frac mn\)

\(\Rightarrow r^Q=\sqrt{\frac mn}\)       _____________(3)

From (1), we have

\(ar^{P+Q-1}=m\)

\(\Rightarrow ar^{P-1}.r^Q=m\)    \((\because a^{m+n}=a^m.a^n)\)

\(\Rightarrow a_P.\sqrt{\frac mn}=m\)       (From (3) \(\&\;a_P=ar^{P-1})\)

\(\Rightarrow a_P=m.\sqrt{\frac nm}=\sqrt{mn}\)

Correct option is  B) \(\sqrt{mn}\)

Correct option is (D) 91

\(\because\) \(a_n = \frac{n(n+1)(2n+1)}{6}\)

\(\therefore\) \(a_6=\frac{6(6+1)(12+1)}{6}\)     (By putting n = 6)

\(=7\times13=91\)

Correct option is D) 91

Correct option is (C) \(\frac{340}{19}\)

We have \(a_n = \frac{n(n+3)}{n+2}\)

\(\therefore\) \(a_{17}=\frac{17(17+3)}{17+2}\)       (Putting n = 17)

\(=\frac{17\times20}{19}=\frac{340}{19}\)

Correct option is C) \(\cfrac{340}{19}\)