Arithmetic Progression

Comparing the given arithmetic progression with general form,
We get, a = 6
and common difference = difference of two consecutive term
i.e. 9 – 6 = 3, 12 – 9 = 3.

Hence, First term a = 6 and common arithmetic difference = 3.

Comparing the given arithmetic progression with general form,
We get, a = -7
and common difference = difference of two consecutive term
i.e. (–9) – (-7) = 2, (-11) – (-9) = -2.

Hence, First term a = -7 and common arithmetic difference = -2.

Comparing the given arithmetic progression with general form,
We get, a =
and common difference = difference of two consecutive term
i.e.

Hence, First term a = and common arithmetic difference = -1.

Comparing the given arithmetic progression with general form,
We get, a = 1
and common difference = difference of two consecutive term
i.e. (–2) – (1) = -3, (-5) – (-2) = -3.

Hence, First term a = 1 and common arithmetic difference = -3.

Comparing the given arithmetic progression with general form,
We get, a = -1
and common difference = difference of two consecutive term
i.e.

Hence, First term a = -1 and common arithmetic difference = .

Comparing the given arithmetic progression with general form,
We get, a = 3
and common difference = difference of two consecutive term
i.e. (1) – (3) = -2, (-1) – (1) = -2.

Hence, First term a = 3 and common arithmetic difference = -2.

Comparing the given arithmetic progression with general form,
We get, a = 3
and common difference = difference of two consecutive term
i.e. (–2) – (3) = -5, (-7) – (-2) = -5.

Hence, First term a = 3 and common arithmetic difference = -5.

Let the first four terms of arithmetic progression be a₁, a₂, a₃, a₄
Given, First term a = -1 and common difference d = .
⇒ a₁ = a = -1.
⇒ a₂ = a₁ + d =
⇒ a₃ = a₂ + d =
⇒ a₄ = a₃ + d =

Hence, First four terms are .

Let the first four terms of arithmetic progression be a₁, a₂, a₃, a₄
Given, First term a = and common difference d = .
⇒ a₁ = a = .
⇒ a₂ = a₁ + d =
⇒ a₃ = a₂ + d =
⇒ a₄ = a₃ + d =

Hence, First four terms are

Let the first four terms of arithmetic progression be a₁, a₂, a₃, a₄
Given, First term a = 0.6 and common difference d =1.1 .
⇒ a₁ = a = 0.6
⇒ a₂ = a₁ + d = 0.6 + 1.1 = 1.7
⇒ a₃ = a₂ + d = 1.7 + 1.1 = 2.8
⇒ a₄ = a₃ + d = 2.8 + 1.1 = 3.9

Hence, First four terms are 0.6, 1.7, 2.8, 3.9.

Let the first four terms of arithmetic progression be a₁, a₂, a₃, a₄
Given, First term a = 4 and common difference d = -3.
⇒ a₁ = a = 4
⇒ a₂ = a₁ + d = 4 + (-3) = 1
⇒ a₃ = a₂ + d = 1 + (-3) = -2
⇒ a₄ = a₃ + d = (-2) + (-3) = -5

Hence, First four terms are 4, 1, -2, -5.

Let the first four terms of arithmetic progression be a₁, a₂, a₃, a₄
Given, First term a = 11 and common difference d = -4.
⇒ a₁ = a = 11
⇒ a₂ = a₁ + d = 11 + (-4) = 7
⇒ a₃ = a₂ + d = 7 + (-4) = 3
⇒ a₄ = a₃ + d = 3 + (-4) = -1

Hence, First four terms are 11, 7, 3, -1.

Let the first four terms of arithmetic progression be a₁, a₂, a₃, a₄
Given, First term a = -1.25 and common difference d = -0.25.
⇒ a₁ = a = -1.25
⇒ a₂ = a₁ + d = -1.25 + (-0.25) = -1.5
⇒ a₃ = a₂ + d = -1.5 + (-0.25) = -1.75
⇒ a₄ = a₃ + d = -1.75 + (-0.25) = -2.00

Hence, First four terms are -1.25, -1.5, -1.75, -2.00.

Let the first four terms of arithmetic progression be a₁, a₂, a₃, a₄
Given, First term a = 20 and common difference d = .
⇒ a₁ = a = 20
⇒ a₂ = a₁ + d =
⇒ a₃ = a₂ + d =
⇒ a₄ = a₃ + d =

Hence, First four terms are .

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ =
a₃ – a₂ =
a₄ – a₃ =
For each term difference is same . So this sequence is an arithmetic progression and its common difference is d = . Next four terms are :
⇒ a₅ = a₄ + d =
⇒ a₆ = a₅ + d =
⇒ a₇ = a₆ + d =
⇒ a₈ = a₇ + d =

Hence, Yes, Next four terms are .

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ =
a₃ – a₂ =
a₄ – a₃ =

For each term difference is same 0. So this sequence is an arithmetic progression and its common difference is d = 0. Next four terms are :
⇒ a₅ = a₄ + d =
⇒ a₆ = a₅ + d =
⇒ a₇ = a₆ + d =
⇒ a₈ = a₇ + d =

Hence, Yes, d = 0, Next four terms are .

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ = a² – a = a(a - 1)
a₃ – a₂ = a³ – a² = a²(a - 1)
a₄ – a₃ = a⁴ – a³ = a³(a - 1)
Here common difference in not same. So this sequence is not an arithmetic progression.

Hence, No.

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ = √6 - √3 = √3(√2) - √3 = √3(√2 - 1)
a₃ – a₂ = √9 - √6 = √3(√3) - √3(√2) = √3(√3 - √2)
a₄ – a₃ = √12 - √9 =√3(√4) - √3(√3) = √3(√4 - √3) = √3(2 - √3)

Here common difference in not same. So this sequence is not an arithmetic progression.

Hence, No.

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ = √8 - √2 = √2(√4) - √2 = 2√2 - √2 = √2
a₃ – a₂ = √18 - √8 = √2(√9) - √2(√4) = 3√2 - 2√2 = √2
a₄ – a₃ = √32 - √18 = √2(√16) - √2(√9) = 4√2 - 3√2 = √2

For each term difference is same √2. So this sequence is an arithmetic progression and its common difference is d =√2 . Next four terms are :
⇒ a₅ = a₄ + d = √32 + √2 = 4√2 + √2 = 5√2 = √25(√2) = √50
⇒ a₆ = a₅ + d = √50 + √2 = 5√2 + √2 = 6√2 = √36(√2) = √72
⇒ a₇ = a₆ + d = √72 + √2 = 6√2 + √2 = 7√2 = √49(√2) = √98
⇒ a₈ = a₇ + d = √98 + √2 = 7√2 + √2 = 8√2 = √64(√2) = √128

Hence, Yes, d = √2, Next four terms are √50, √72, √98, √128.

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
For each term difference is same i.e., a. So this sequence is an
arithmetic progression and its common difference is d = a. Next four terms are :
⇒ a₅ = a₄ + d = 4a + a = 5a
⇒ a₆ = a₅ + d = 5a + a = 6a
⇒ a₇ = a₆ + d = 6a + a = 7a
⇒ a₈ = a₇ + d = 7a + a = 8a

Hence, Yes, d = a, Next four terms are 5a, 6a, 7a, 8a.

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002

For each term common difference is not same. So this sequence is not an arithmetic progression.

Hence, No

Let the arithmetic progression be a₁, a₂, a₃, a₄ ……. So here
a₂ – a₁ = (3+√2) – (3) = √2
a₃ – a₂ = (3+2√2) – (3+√2) = √2
a₄ – a₃ = (3+3√2) – (3+2√2) = √2

For each term difference is same √2. So this sequence is an arithmetic progression and its common difference is d =√2 . Next four terms are :
⇒ a₅ = a₄ + d = 3+3√2 + √2 = 3+4√2
⇒ a₆ = a₅ + d = 3+4√2 + √2 = 3+5√2
⇒ a₇ = a₆ + d = 3+5√2 + √2 = 3+6√2
⇒ a₈ = a₇ + d = 3+6√2 + √2 = 3+7√2

Hence, Yes, d = √2, Next four terms are 3+4√2, 3+5√2, 3+6√2, 3+7√2.

The given arithmetic progression is 2, 7, 12, …
Its first term a = 2
Common difference = d = 7 – 2 = 5.
So n^th term a_n of the given arithmetic progression is given by
a_n = a + (n -1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Thus 10^th term a₁₀ = 2 + (10 - 1) × 5.
⇒ a₁₀ = 2 + 9 × 5 = 2 + 45 = 47

Hence, 10^th term is 47.

The given arithmetic progression is √2, 3√2, 5√2, …
Its first term a = √2
Common difference = d = 3√2 – √2 = 2√2.
So n^th term a_n of the given arithmetic progression is given by
a_n = a + (n -1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Thus 18^th term a₁₈ = √2 + (18 - 1) × 2√2.
⇒ a₁₈ = √2 + 17 × 2√2 = √2 + 34√2 = 35√2

Hence, 18^th term is 35√2.

The given arithmetic progression is 9, 13, 17, 21, …
Its first term a = 9
Common difference = d = 13 – 9 = 4.
So n^th term a_n of the given arithmetic progression is given by
a_n = a + (n -1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Thus 24^th term a₂₄ = 9 + (24 - 1) × 4.
⇒ a₂₄ = 9 + 23 × 4 = 9 + 92 = 101

Hence, 24^th term is 101.

Let the n^th term of the progression be -81.
First term a = 21 and common difference = 18 – 21 = -3
Since, a_n = a + (n -1)d
∴ -81 = 21 + (n – 1) × (-3)
⇒ -81 = 21 -3n -3(-1)
⇒ -81 = 21 -3n + 3
⇒ -81- 24 = -3n
⇒ -105 = -3n

⇒ n = 35

Hence, -81 is the 35^th term of the progression.

Let the n^th term of the progression be 0.
First term a = 84 and common difference = 80 – 84 = -4
Since, a_n = a + (n -1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

∴ 0 = 84 + (n – 1) × (-4)
⇒ 0 = 84 -4n -4(-1)
⇒ 0 = 84 -4n + 4
⇒ -88 = -4n
⇒ n = 22

Hence, Zero is the 22^nd term of the progression.

The arithmetic progression is 5, 11, 17, 23, ….
Here, First term a = 5 and common difference = 11 – 5 = 6
Let us assume that n^th term a_n of the progression is 301.
Since, a_n = a + (n -1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

∴ 301 = 5 + (n – 1) × (6)
⇒ 301 = 5 + 6n -6
⇒ 302 = 6n
⇒ n = 50.33
Here n is a fraction, But any term n should be an integer.
∴ 301 is not any term of the arithmetic progression.

Hence, 301 is not any term of the progression.

The arithmetic progression is 11, 8, 5, 2, ….
Here, First term a = 11 and common difference = 8 – 11 = -3
Let us assume that n^th term a_n of the progression is -150.
Since, a_n = a + (n -1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

∴ -150 = 11 + (n – 1) × (-3)
⇒ -150 = 11 - 3n + 3
⇒ -150 -14 = -3n
⇒ -164 = -3n
⇒ n = 54.667
Here n is a fraction, But any term n should be an integer.
∴ -150 is not any term of the arithmetic progression.

Hence, -150 is not any term of the progression.

Let first term be a and common difference be d.
We know that, a_n = a + (n - 1)d
∴ a₆ = a + (6 - 1)d
⇒ 19 = a + 5d …(i)
and a₁₇ = a + (17 - 1)d
⇒ 41 = a + 16d …(ii)

On solving eq. (i) and (ii), we get,
a = 9 and d = 2

⇒ a₄₀ = a + (40 - 1)d
⇒ a₄₀ = 9 + (40 - 1) × 2
⇒ a₄₀ = 9 + 39× 2 = 9 + 78 = 87

Hence, 40^th term is 87.

Let first term be a and common difference be d.
We know that, a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

∴ a₃ = a + (3 - 1)d
⇒ 4 = a + 2d …(i)
and a₉ = a + (9 - 1)d
⇒ -8 = a + 8d …(ii)

On solving eq. (i) and (ii), we get,
a = 8 and d = -2

Let the n^th term be zero.
⇒ a_n = a + (n - 1)d
⇒ 0 = 8 + (n - 1) × (-2)
⇒ 0 = 8 – 2n -2(-1)
⇒ 0 = 8 – 2n + 2
⇒ -10 = – 2n
⇒ n = 5

Hence, 5^th will be zero.

Let first term be a and common difference be d.
We know that, a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

∴ a₃ = a + (3 - 1)d
⇒ 16 = a + 2d …(i)
and a₇ = a + (7 - 1)d
⇒ a₇ = a + 6d …(ii)
and a₅ = a + (5 - 1)d
⇒ a₅ = a + 4d …(iii)

Given, a₇ = a₅ + 12
⇒ a + 6d = a + 4d + 12 (From (ii) and (iii))

⇒ 6d – 4d = 12
⇒ 2d = 12
⇒ d = 6
Substituting the value of d in eq. (i), we get,
16 = a + 2(6)
⇒ a = 16 – 12 = 4

Hence first term a = 4 and common difference d = 6
∴ a₁ = 4
⇒ a₂ = a₁ + d = 4 + 6 = 10
⇒ a₃ = a₂ + d = 10 + 6 = 16
⇒ a₄ = a₃ + d = 16 + 6 = 22

Hence, The arithmetic progression is 4, 10, 16, 2.

Smallest 3-digit number that is divisible by 7 is 105 and greatest 3-digit number that is divisible by 7 is 994.
Let the no. of three digits numbers that are divisible by 7 be n.

Thus, a_n = 994, first term a = 105 and common difference d = 7.

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

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ 994 = 105 + (n - 1) × 7
⇒ 994 – 105 = 7 × (n - 1)
⇒ 889 = 7 × (n - 1)
⇒ 127 = n – 1
⇒ n = 128

Hence, Total 3-digit numbers that are divisible by 7 are 128.

Here last term of the progression l = -62.
First term a = 10 and common difference = 7 – 10 = -3
We know, In an AP, the nth term from last is

a_n = l – (n – 1)d

where a_n is nth term from last, l is last term and ‘d’ is common difference.

Thus 11^th term from the end
= l – (11 -1)d
= -62 – (10) × (-3)
= -62 + 30
= -32

Hence, 11^th term from the last is -32.

Here last term of the progression l = 88.
First term a = 1 and common difference = 4 – 1 = 3
We know, In an AP, the nth term from last is

a_n = l – (n – 1)d

where a_n is nth term from last, l is last term and ‘d’ is common difference.

Thus 8^th term from the end
= l – (8 -1)d
= 88 – (7) × (3)
= 88 - 21
= 67

Hence, 8^th term from the last is 67.

Given, a = 7 and a₆₀ = 125.
∴ a_n = a + (n – 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ a₆₀ = 7 + (60 - 1)d
⇒ 125 = 7 + (59)d
⇒ 118 = 59 × d
⇒ d = 2

Now 32^th term, a₃₂ = a + (32 - 1)d

= 7 + 31 × 2
= 7 + 62
= 69

Hence, 32^th term is 69.

Let the first number be a and common difference be d, Then arithmetic progression be a, a+d, a+2d, a+3d.

Given, a + a+d + a+2d + a+3d = 50
⇒ 4a + 6d = 50 …(i)
Also, a + 3d = 4(a)
⇒ 3d = 3a
⇒ a = d …(ii)

Solving eq. (i) and (ii), we get,
a = 5 and d = 5.

Hence the numbers are :
a₁ = 5
a₂ = a + d = 5 + 5 = 10
a₃ = a + 2d = 5 + 2(5) = 5 + 10 = 15
a₄ = a + 3d = 5 + 3(5) = 5 + 15 = 20.

Hence, The numbers are 5, 10, 15, 20.

The arithmetic progression 1, 3, 5, 7, …. is given.
Here first term is 1 and common difference is 3 – 1 = 2.
Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of first 12 terms S₁₂
= 6[ 2 + 11×2]
= 6[2 + 22]
= 6 × 24
= 144

Hence, Sum of first 12 terms is 144.

The arithmetic progression 8, 3, -2, …. is given.
Here first term is 8 and common difference is 3 – 8 = -5.
Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of first 22 terms S₂₂
]
= 11[ 16 + 21×(-5)]
= 11[16 - 105]
= 11 × (-89)
= -979

Hence, Sum of first 22 terms is -979.

The arithmetic progression …. is given.
Here first term is and common difference is .
Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of first 11 terms S₁₁

Hence, Sum of first 11 terms is .

The arithmetic progression 3, 11, 19, …., 803 is given.
Here first term is 3 and common difference is 11 – 3 = 8.
Last term l = 803

Let no. of terms be n.
So, a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ 803 = 3 + (n - 1) × 8
⇒ 800 = (n - 1) × 8
⇒ 100 = n – 1
⇒ n = 101

Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of 101 terms S₁₀₁
]
= 50.5[ 6 + 100×(8)]
= 50.5[ 6 + 800]
= 50.5 × (806)
= 40703

Hence, Sum is 40703.

The arithmetic progression is given.
Here first term is 7 and common difference is
Last term l = 84

Let no. of terms be n.
Since, a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒
⇒
⇒ 22 = n – 1
⇒ n = 23

Since the sum of n terms is
S_{n =}
So sum of 23 terms S₂₃
]
= 11.5 [14 + 22 × 3.5]
= 11.5[ 14 + 77]
= 11.5 × (91)
= 1046.5

Hence, Sum is 1046.5 = .

Here first term a = 9 and common difference d = 17 – 9 = 8.
Also, S_n = 636
Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

636 = n[ 9 + (n - 1)4]
636 = 9n + 4n(n - 1)
636 = 9n + 4n² – 4n
636 = 4n² + 5n
4n² + 5n – 636 = 0
⇒ n = 12, n = -13.25 (which is not applicable ∵ n can neither be negative nor be in decimal)
∴ n = 12

Hence, 12 terms of the arithmetic progression must be taken to get sum as 636.

Here first term a = 63 and common difference d = 60 – 63 = -3.
Also, S_n = 693
Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

693 × 2 = n[ 126 + (n - 1)(-3)]
1386 = 126n - 3n(n - 1)
1386 = 126n - 3n² + 3n
1386 = -3n² + 129n
3n² - 129n + 1386 = 0
n² - 43n + 462 = 0
⇒ n = 21, n = 22
∴ n = 21 or n = 22

Hence, Either 21 or 22 terms of the arithmetic progression must be taken to get sum as 693.

General n^th term a_n is given as:
a_n = a + (n - 1)d … (i)

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

and a_n = 3 + 4n
⇒ a_n = (7-4) + 4n
⇒ a_n = 7 + 4n – 4
⇒ a_n = 7 + 4(n – 1)
⇒ a_n = 7+ (n – 1)× 4 …(ii)

On comparing eq. (i) and (ii), we get,
a = 7 and d = 4

Here first term is 7 and common difference is 4.
Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of first 25 terms S₂₅
= 12.5 × [ 14 + 24×4]
= 12.5 × [ 14 + 96]
= 12.5 × 110
= 1375

Hence, Sum of first 25 terms is 1375.

General n^th term a_n is given as:
a_n = a + (n - 1)d … (i)

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

and a_n = 7 - 3n
⇒ a_n = (4 + 3) - 3n
⇒ a_n = 4 - 3n + 3
⇒ a_n = 4 + (-3) × (n – 1)
⇒ a_n = 4+ (n – 1)× (-3) …(ii)

On comparing eq. (i) and (ii), we get,
a = 4 and d = -3

Here first term is 4 and common difference is -3.
Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of first 25 terms S₂₅
= 12.5 × [ 8 + 24×(-3)]
= 12.5 × [ 8 - 72]
= 12.5 × (- 64)
= -800

Hence, Sum of first 25 terms is -800.

Common difference = 18 – 14 = 4
∴ a = a₂ – d = 14 – 4 = 10
First term a = 10 and common difference d = 4

Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of first 51 terms S₅₁
= 25.5 × [ 20 + 50×4]
= 25.5 × [ 20 + 200]
= 25.5 × 220
= 5610

Hence, Sum of first 51 terms is 5610.

Here first term is 17 and common difference is 9
Last term l = 350

Let no. of terms be n.
Since, a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ 350 = 17 + (n - 1) × 9
⇒ 333 = (n - 1) × 9
⇒ 37 = n – 1
⇒ n = 38

Since the sum of n terms is
S_{n =}
So sum of 38 terms S₃₈
= 19 [34 + 37 × 9]
= 19 × 367
= 6973

Hence, Number of terms is 38 and Sum is 6973.

Starting from 1 the first odd number divisible by 3 is 3 then second odd number divisible by 3 is 9.
Greatest odd number till 1000 i.e. divisible by 3 is 999
Thus arithmetic progression is 3, 9, 15, …, 999
Here first term a = 3 and last term l = 999
and common difference = 9 – 3 = 6.

Let number of all odd number divisible by 3 be n.
Since, a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ 999 = 3 + (n - 1) × 6
⇒ 996 = (n - 1) × 6
⇒ 166 = n – 1
⇒ n = 167

Since the sum of n terms is
S_{n =}
So sum of 167 terms S₁₆₇
= 83.5 × [6 + 166 × 6]
=n
= 83.5 × 1002
= 83667

Hence, Sum is 83667.

Here first term a = 8, n^th term = 33 and s_n = 123
General n^th term a_n is given as:
a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

33 = 8 + (n -1) × d
25 = (n - 1) × d … (i)

Since the sum of n terms is
S_n=

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of first 25 terms S₂₅
]
246 = n [ 16 + (n - 1) × d]
246 – 16 n = (n) × (n - 1) × d … (ii)

On dividing eq. (ii) from (i), we get,
⇒
⇒ (246 – 16n) = 25n
⇒ 25n + 16n = 246
⇒ 41n = 246
⇒ n = 6

Substituting the value of n in eq. (i), we get,
d = 5

Hence, Value of n is 6 and common difference is 5.

Given,
S_n = 280, n = 4 and common difference is -20.

Let the first term be a.

Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

So sum of 4 terms S₄
]
280 = 2× [2a + 3 × (-20)]
140 = 2a – 60
2a = 140 + 60
2a = 200
a = 100

a₁ = a = 100
a₂ = a₁ + d = 100 - 20 = 80
a₃ = a₂ + d = 80 - 20 = 60
a₄ = a₃ + d = 60 - 20 = 40

Hence, The value of prizes are 100, 80, 60 and 40.

Given, a₃ = 600 and a₇ = 700

Let the first term be a and common difference be d.
Since, nth term an AP is given by :
a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ a₃ = a + (3 - 1)d
⇒ 600 = a + 2d … (i)

⇒ a₇ = a + (7 - 1)d
⇒ 700 = a + 6d …(ii)

On solving eq. (i) and (ii), we get :
a = 550 and d =25

∴ production in first year a₁ = a 550
and production in tenth year a₁₀
= a + (10 - 1)d
= 550 + 9 × 25
= 550 + 225 = 775
∴ a₁₀ = 775

Since the sum of n terms is
S_{n =}
Total production in first seven year S₇
S_{7 =}
S₇ = 3.5 × [ (2× 550) + (6 × 25)]
S₇ = 3.5 × [ 1100 + 150]
S₇ = 3.5 × 1250
S₇ = 4375

Hence, (i) the production in the 1^st year is 550
(ii) the production in the 10^th year is 775
(iii) the total production in first 7 years is 4375.

Since the common difference of both the term is same.
∴ each term will change by same number.
⇒ Difference of their 30^th term is same as difference of their 1^st term.
⇒ Difference of 30^th term = Difference of 1^st term = 8 – 3 = 5.

Since they are in arithmetic progression their common difference is same.
⇒ a – 18 = b – a = -3 – b
⇒ a – 18 = -3 – b
⇒ a + b = 18 – 3
⇒ a + b = 15.

Let the first term be a and common difference be d.
Since, nth term an is given by :
a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ a₇ = a + (7 - 1)d
⇒ 34 = a + 6d … (i)

⇒ a₁₃ = a + (13 - 1)d
⇒ 64 = a + 12d …(ii)

On solving eq. (i) and (ii), we get :
a = 4 and d = 5

⇒ a₁₈ = a + (18 - 1)d
⇒ a₁₈ = 4 + 17 × 5
⇒ a₁₈ = 4 + 85
⇒ a₁₈ = 89

Given, first term a = 2 and common difference d = 8 and S_n = 90.

Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

90 ₌
⇒ 180 = n [4 + 8(n - 1)]
⇒ 180 = 4n + 8n(n – 1)
⇒ 180 = 4n + 8n² – 8n
⇒ 180 = 8n² – 4n
⇒ 45 = 2n² – n
⇒ 2n² – n – 45 = 0
⇒ n = 5, n = -4.5
Since, n is a number of terms, it cannot be negative.
∴ n = 5.

Since the sum of n terms is
S_{n =}…(i)

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Also S_n = 3n² + 5n
⇒ S_n = n(3n + 5)
⇒ S_n = n(3n + 8 - 3)
⇒ S_n = n[8 + 3(n -1)]
⇒ S_n = n[8 + (n - 1)3]
⇒ S_n …(ii)

On comparing eq. (i) and (ii), we get :
a = 8 and d = 6

Since, n^th term a_n is given by :

a_n= a + (n – 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

a_n = 8 + (n - 1)6
⇒ 164 = 8 + (n - 1) × 6
⇒ 156 = (n - 1) × 6
⇒ (n - 1) = 26
⇒ n = 27.

Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ S_2n = 3S_n
⇒ 2[2a + (2n - 1)d] = 3[2a + (n - 1)d]

⇒ 4a + (2n - 1)2d] = 6a + (n - 1)3d
⇒ 4nd – 2d = 2a + 3nd – 3d
⇒ nd + d = 2a
⇒ (n + 1)d = 2a

S_3n:S_{n =}
=
=
=
=
= 6
S_3n : S_n = 6:1

Given, first term a = 1, last term l = 11 and S_n = 36.
Since, n^th term a_n is given by :
a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

⇒ 11 = 1 + (n - 1)d
⇒ 10 = (n -1)d … (i)

Since the sum of n terms is
S_{n =}

⇒ 72 = n[ 2 + (n-1)d] …(ii)

From eq. (i) and (ii), we get,
⇒ 72 = n[ 2 + 10]
⇒ 72 = 12n
⇒ n = 6

Here last term of the progression l = 201.
First term a = 3 and common difference = 5 – 3 = 2
We know, In an AP, the nth term from last is

a_n = l – (n – 1)d

where a_n is nth term from last, l is last term and ‘d’ is common difference.

Thus 5^th term from the end
= l – (5 -1)d
= 201 – (4) × (2)
= 201 - 8
= 193

Hence, 5^th term from the last is 193.

We know that if three terms A, B and C are in AP then
2B = A + C

Hence, Value of a is .

Required arithmetic progression is 1, 2, 3, …, 1000
Here, first term a= 1 and common difference d = 1.

Since the sum of n terms is
S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

S_{1000 =}
= 500 [ 2 + 999]

= 500 × 1001
= 500500

Hence, Sum of first 1000 positive integers is 500500.

The arithmetic progression is 5, 11, 17, 23, ….
Here, First term a = 5 and common difference = 11 – 5 = 6
Let us assume that n^th term a_n of the progression is 301.
we know, a_n = a + (n -1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

∴ 299 = 5 + (n – 1) × (6)
⇒ 299 = 5 + 6n -6
⇒ 300 = 6n
⇒ n = 5

Hence, Yes, 299 is the 5^th term of the progression.

Here, first term a = 20 and common difference d We know that general term a_n = a + (n - 1)d

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Since the common difference is negative we get the first negative term when -(n - 1)d > a


⇒ (n - 1) > 26.66
⇒ n > 26.66 + 1
⇒ n > 27.66
⇒ n = 28

Hence, 28^th term of the arithmetic progression is first negative term.

Let the first term be a – 3d and common difference be 2d.
Thus arithmetic progression will be a - 3d, a - d, a + d, a+3d.

Given, Sum of terms = 20
⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20
⇒ a = 5 …(i)

Also,

Sum of their square = 120

(a – 3d)² + (a – d)² + (a + d)² + (a + 3d)² = 120

⇒ (5 – 3d)² + (5 – d)² + (5 + d)² + (5 + 3d)² = 120

⇒ (25 + 9d² – 30d) + (25 + d² – 10d) + (25 + d² + 10d) + (25 + 9d² + 30d) = 120

⇒ 100 + 20d² = 120

⇒ 20d² = 20

⇒ d² = 1

⇒ d = ± 1

If d = 1

First term, a – 3d = 5 – 3 = 2 and common difference, 2d = 2 and four terms will be 2, 4, 6, 8

If d = -1

First term, a – 3d = 5 –(– 3) = 8 and common difference, 2d = -2 and four terms will be 8, 6, 4 and 2.

Since the sum of n terms is
S_{n =}…(i)

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Also S_n =
⇒ S_n =
⇒ S_n =
⇒ S_n =
⇒ S_n = …(ii)

On comparing eq. (i) and (ii), we get:
a = 4 and d = 3

Also, n^th term a_n is given by:
a_n = a + (n – 1)d

For given AP, we have

a_n = 4 + (n - 1)6
⇒ a₂₅ = 4 + (25 - 1) × 3
⇒ a₂₅ = 4 + 24 × 3
⇒ a₂₅ = 4 + 72
⇒ a₂₅ = 76

Hence, 25^th term is 76.

Clearly, No of houses are in AP and

Given,

First term, a = 1

common difference d = 1

According to given statement: S_{1 to x- 1} = S_{(x+1) to 49}
⇒ S_{x+1 to 49} = S₄₉ – S_x and S_{1 to x- 1} = S_{x- 1}

Since the sum of n terms is S_{n =}

Where,

a = First term of AP
d = Common difference of AP
and no of terms is ‘n’

Applying given condition:

S_{1 to x- 1} = S_{(x+1) to 49}
⇒ S_{x- 1} = S₄₉ – S_x
⇒ S_{x- 1} = S₄₉ – S_x


⇒ (x - 1)(x) = 49(50) - x(x+1)
⇒ x² – x = 49(50) – x² -x
⇒ 2x² = 49 × 50
⇒ x² = 49 × 25
⇒ x = 35 and x = -35
Here since the x is a number between 1 to 49, ∴ x = 35.

Hence, The value of x is 35.