Arithmetic Progressions

(i)

Put n = 1

A₁ = 3(1) + 2 = 3+2=5

Put n = 2

A₂ = 3(2) + 2=8

Put n = 3

A₃ = 3(3) + 2 = 9+2=11

Put n = 4

A₄ = 3(4) +2 = 12+2=14

Put n =5

A₅ = 3(5) + 2 = 15+2=17

(ii)

Put n = 1

A₁= =

Put n = 2

A₂= = 0

Put n = 3

A₃= =

Put n = 4

A₄= =

Put n = 5

A₅= = = 1

(iii)

Put n = 1

A₁= 31= 3

Put n = 2

A₂= 32= 9

Put n = 3

A₃= 33= 27

Put n = 4

A₄= 34= 81

Put n = 5

A₅= 35= 243

(iv)

Put n = 1

A₁= = =

Put n = 2

A₂= = =

Put n = 3

A₃= = =

Put n = 4

A₄= = = = 2

Put n = 5

A₅ = = =

(v)

Put n = 1

A₁= (-1)1.21 = (-1).2 = -2

Put n = 2

A₂= (-1)2.22 = (1).4 = 4

Put n = 3

A₃= (-1)3.23 = (-1).8 = -8

Put n = 4

A₄= (-1)4.24 = (1).16 = 16

Put n = 5

A₅= (-1)5.25 = (-1).32 = -32

(vi)

Put n = 1

A₁= =

Put n = 2

A₂= = 0

Put n = 3

A₃= =

Put n = 4

A₄= = = 4

Put n = 5

A₅= =

(vii)

Put n = 1

A₁= (1)² – 1+1 = 1

Put n = 2

A₂= (2)² -2+1 = 3

Put n = 3

A₃= (3)² -3+1 = 9-2 = 7

Put n = 4

A₄ = (4)² -4+1 = 16-3 = 13

Put n = 5

A₅= (5)² – 5+1 = 25-4 = 21

(viii)

Put n = 1

A₁= 2(1)²– 3(1) + 1 = 2-3+1 = 0

Put n = 2

A₂= 2(2)² – 3(2) + 1 = 8-6+1 = 3

Put n = 3

A₃= 2(3)² – 3(3) + 1 = 18-9+1 = 10

Put n = 4

A₄= 2(4)² – 3(4) + 1 = 32-12+1 = 21

Put n = 5

A₅= 2(5)² – 3(5) + 1 = 50-15+1 = 36

(ix)

Put n = 1

A₁= =

Put n = 2

A₂= =

Put n = 3

A₃= = =

Put n = 4

A₄= =

Put n = 5

A₅= =

(i)

Put n = 12

A₁₂= 5(12) – 4

= 60-4 = 56

Put n = 15

A₁₅= 5(15) – 4

= 75-4 = 71

(ii)

Put n = 7

A₇= = =

Put n = 8

A₈= = =

(iii)

Put n = 5

A₅= 5(5-1) (5-2)

= 5(4)(3) = 60

Put n = 8

A₈= 8(8-1) (8-2)

= 8(7)(6) = 336

(iv)

Put n = 1

A₁= 1(1-1) (2-1) (2-2) (3+2)

= 0

Put n = 2

A₂= 2(2-1) (2-2) (3+2)

= 0

Put n = 3

A₃= 3(3-1) (2-3) (3+3)

= 3 (2) (-1) (6)

= 36

(v)

Put n = 3

A₃= (-1)³ (3) = -3

Put n = 5

A₅= (-1)⁵ (5) = -5

Put n = 8

A₈= (-1)⁸ (8) = 8

(i)

Put n = 2

A₂= a_2-1 + 2

A₂= a₁ +2

= 1+2 = 3

Put n = 3

A₃= a_3-1 +2

= a₂ + 2

= 3+2 = 5

Put n = 4

A₄= a_4-1 + 2

= a3 + 2

= 5+2 = 7

Put n = 5

A₅= a_5-1 + 2

= a₄ + 2

= 7+2 = 9

Put n = 6

A₆ = a_6-1 +2

= a₅ +2

= 9+2 = 11

(ii)

Put n = 3

A₃= a_3-1 – 3

= a₂ – 3

= 2-3 = -1

Put n = 4

A₄= a_4-1 – 3

= a₃ – 3

= -1 – 3 = -4

Put n = 5

A₅ = a_5-1 -3

= a₄ – 3

= -4 – 3 = -7

Put n = 6

A₆= a_6-1 – 3

= a₅ – 3

= -7 – 3 = -10

Put n = 7

A₇= a_7-1 – 3

= a₆ – 3

= -10 – 3 = -13

(iii)

Put n = 2

A2= a_2-1 / 2

= -1/2

Put n = 3

A₃= a_3-1 / 3

= -1/6

Put n = 4

A₄ = a_4-1 / 3

= -1/24

Put n = 5

A₅= a_5-1 / 3

= -1/120

Put n = 6

A₆= a_6-1 / 3

= -1/720

(iv)

Put n = 2

A₂= 4a_2-1 + 3

= 4a₁ + 3 = 19

Put n = 3

A₃= 4a_3-1 +3

= 4a₂ + 3 = 79

Put n = 4

A₄ = 4a_4-1 + 3

= 4a₃ + 3

= 316 + 3

= 319

Put n = 5

A₅ = 4a_5-1 + 3

= 4a₄ + 3

= 1276 + 3

= 1279

Put n = 6

A₆= 4a_6-1 + 3

= 4a₅ + 3

= 5116 + 3

= 5119

(i)

First term, a = -5

Common difference, d = -1 – (-5)

= -1 + 5 = 4

(ii)

First term, a =

Common difference, d = - =

(iii)

First term, a = 0.3

Common difference, d = 0.55 – 0.30 = 0.25

(iv)

First term, a = -1.1

Common difference, d = -3.1 – (-1.1)

= -2

(i)

a₁= 4, d= -3

A₂= a₁+ d = 4-3 =1

A₃= a₁+ 2d = 4+ 2(-3) = 4-6 = -2

A₄= a₁ =3d = 4+ 3(-3) = 4-9 = -5

Series = 4, 1, -2, -5,…

(ii)

a₁= -1, d =

A₂= a₁ + d = -1 + =

A₃= a₁ + 2d = -1 + 1 = 0

A₄= a₁ + 3d = -1 + =

Series = -1, , 0, ,…

(iii)

a₁= -1.5, d= -0.5

A₂= a₁+ d = -1.5 – 0.5 = -2

A₃= a₁+ 2d = -1.5 + 2(-0.5) = -1.5 – 1 = -2.5

A₄= a₁+ 3d = -1.5 + 3(-0.5) = -1.5 – 1.5 = -3

Series =

(i) The cost of digging a well for the first metre = Rs. 150

The cost of digging a well for the second metre = 150 + 20 = Rs. 170

The cost of digging a well for the third metre = 170 + 20 = Rs. 190

The cost of digging a well for the fourth metre = 190 + 20 = Rs. 210

Difference between first and second metre = 170 – 150 = 20

Difference between second and third metre = 190 – 170 = 20

Difference between third and fourth metre = 210 – 190 = 20

Since, all the differences are equal. Therefore, it’s an A.P.

(ii) Let the amount of air present in the cylinder first time = x

Amount of air present in the cylinder second time = x - =

Amount of air present in the cylinder third time = - =

Difference in the amount of air present in the cylinder between first and the second time = – x =

Difference in the amount of air present in the cylinder between second and the third time = -

=

Since, the differences are unequal. Therefore, it’s not an A.P.

Put n = 1

A₁= 5(1) – 7

= -2

Put n = 2

A₂= 5(2) – 7

= 10 – 7 = 3

Put n = 3

A₃= 5(3) – 7

= 15 – 7 = 8

Common difference, d₁= a₂ – a₁= 3 – (-2) = 5

Common difference, d₂= a₃ – a₂= 8 – 3 = 5

Since, d₁ = d₂ and Common difference = 5

Therefore, it’s an A.P.

Given: The sequence .

To prove: the sequence defined by is not A.P.

Proof:

Consider the sequence ,


Put n = 1

A₁= 3(1)² – 5 = 3 – 5 = -2

Put n = 2

A₂= 3(2)² – 5 = 12- 5 = 7

Put n = 3

A₃= 3(3)² – 5 = 27 – 5 = 22

In an A.P the difference of consecutive terms should be same.

Common difference, d₁= a₂ – a₁ =7 – (-2) = 9

Common difference, d_{2 =} a₃ – a₂ = 22 – 9 = 13

Since, d₁ ≠ d₂

Therefore, it’s not an A.P.

Put n = 1

A₁= -4(1)² + 15 = 11

Put n = 2

A₂= -4(2)² + 15 = -1

Put n = 3

A₃= -4(3)² + 15 = -21

Common difference, d₁= a₂ – a₁ = -1 – 11 = -12

Common difference, d₂= a₃ – a₂ = -21 + 1 = 20

Since, d₁ ≠ d₂

Therefore, it’s not an A.P.

(i)

Common difference, d = -2 – 1 = -3

A₅= -8 – 3 = -11

A₆= -11 – 3 = -14

A₇= -14 – 3 = -17

A₈= -17 – 3 = -20

(ii)

Common difference, d = -3 – 0 = -3

A₅= -9 – 3 = -12

A₆= -12 – 3 = -15

A₇= -15 – 3 = -18

A₈= -18 – 3 = -21

(iii)

Common difference, d = + 1 =

A₄= + =

A₅= + = = 4

A₆= 4 + =

A₇= + = =

(iv)

Common difference, d = + 1 =

A₄= + =

A₅= + =

A₆= + =

A₇= + = 0

Put n = 1

A₁= a + b

Put n = 2

A₂= a + 2b

Put n = 3

A₃= a + 3b

Put n = 4

A₄= a + 4b

Common difference, d₁ = a₂ – a₁ = a + 2b – a – b = b

Common difference, d₂= a₃ – a₂ = a+ 3b – a – 2b = b

Common difference, d₃ = a₄ – a₃ = a+ 4b – a – 3b = b

Since, d₁ = d₂ = d₃

Therefore, it’s an A.P. with common difference ‘b’

(i) Put n = 1

A₁= 3 + 4(1) = 7

Put n = 2

A₂= 3 + 4(2) = 11

Put n = 3

A₃= 3 + 4(3) = 15

Common difference, d₁= a₂ – a₁ = 11 – 7 = 4

Common difference, d₂= a₃ – a₂ = 15 – 11 = 4

Since, d₁ = d₂

Therefore, it’s an A.P. with sequence 7, 11, 15,…

(ii) Put n = 1

A₁= 5 + 2(1) = 7

Put n = 2

A₂= 5 + 2(2) = 9

Put n = 3

A₃= 5 + 2(3) = 11

Common difference, d₁= a₂ – a₁= 9 – 7 = 2

Common difference, d₂= a₃ – a₂= 11 – 9 = 2

Since, d₁ = d₂

Therefore, it’s an A.P. with sequence 7, 9, 11,…

(iii) Put n = 1

A₁= 6 – 1 = 5

Put n = 2

A₂= 6 – 2 = 4

Put n = 3

A₃= 6 – 3 = 3

Common difference, d₁= a₂ – a₁= 4 – 5 = -1

Common difference, d₂ = a₃- a₂= 3 – 4 = -1

Since, d₁=d₂

Therefore, it’s an A.P. with sequence 5, 4 , 3,…

(iv) Put n = 1

A₁= 9 – 5(1) = 4

Put n = 2

A₂= 9 – 5(2) = -1

Put n = 3

A₃= 9 – 5(3) = -6

Common difference, d₁= a₂ – a₁ = -1 – 4 = -5

Common difference,d₂= a₃ – a₂ = -6 – (-1) = -5

Since, d₁=d₂

Therefore, it’s an A.P. with sequence 4, -1, -6,…

(i)

Common difference, d₁ = 6 – 3 = 3

Common difference, d₂= 12 – 6 = 6

Since, d₁ ≠ d₂

Therefore, it’s not an A.P.

(ii)

Common difference, d₁ = -4 – 0 = -4

Common difference, d₂= -8 – (-4) = - 4

Since, d₁ = d₂

Therefore, it’s an A.P. with common difference, d = -4

(iii)

Common difference, d₁= - =

Common difference, d₂ = - =

Since, d₁ ≠ d₂

Therefore, it’s not an A.P.

(iv)

Common difference, d₁= 2 – 12 = -10

Common difference, d₂= -8 -2 = -10

Since,d₁ = d₂

Therefore, it’s an A.P. with common difference, d = -10

(v)

Common difference, d₁= 3 – 3 = 0

Common difference, d₂= 3 – 3 = 0

Since, d₁=d₂

Therefore, it’s an A.P. with common difference, d = 0

(vi)

Common difference, d₁= p + 90 – p = 90

Common difference, d₂= p + 180 – p – 90 = 90

Since, d₁=d₂

Therefore, it’s an A.P. with common difference, d = 90

(vii)

Common difference, d₁= 1.7 – 1.0 = 0.7

Common difference, d₂= 2.4 – 1.7 = 0.7

Since, d₁=d₂

Therefore, it’s an A.P. with common difference, d = 0.7

(viii)

Common difference, d₁= -425 + 225 = -200

Common difference, d₂= -625 + 425 = -200

Since, d₁=d₂

Therefore, it’s an A.P. with common difference, d = -200

(ix)

Common difference, d₁= 10 + 2⁶ – 10 = 2⁶ = 64

Common difference, d₂ = 10 + 2⁷ – 10 – 2⁶ = 2⁶ (2 – 1) = 64

Since, d₁=d₂

Therefore, it’s an A.P. with common difference, d = 64

(x)

Common difference, d₁ = (a + 1) + b – a – b = 1

Common difference, d₂ = (a + 1) + (b + 1) – (a + 1) – b = 1

Since, d₁ = d₂

Therefore, it’s an A.P. with common difference, d = 1

(xi)

Common difference, d₁= 3² – 1² = 8

Common difference, d₂ = 5² – 3² = 25 – 9 = 16

Since, d₁≠d₂

Therefore, it’s not an A.P.

(xii)

Common difference, d₁ = 5² – 1² = 24

Common difference, d₂ = 7² – 5² = 24

Since, d₁ = d₂

Therefore, it’s an A.P. with common difference, d = 24

(i)

Common difference, d = 59 – 51 = 8

A₅ = 75 + 8 = 83

A₆ = 83 + 8 = 91

(ii)

Common difference, d = 67 – 75 = -8

A₅= 51 - 8 = 43

A₆= 43 – 8 = 35

(iii)

Common difference, d = 2.0 – 1.8 = 0.2

A₅= 2.4 + 0.2 = 2.6

A₆= 2.6 + 0.2 = 2.8

(iv)

Common difference, d = – 0 =

A₅= + = 4

A₆= 4 + =

(v)

Common difference, d = 136 – 119 = 17

A₅= 170 + 17 = 187

A₆= 187 + 17 = 204

a_n= 6n + 2

Put n = 1

A₁= 6(1) + 2 = 8

Put n = 2

A₂= 6(2) + 2 = 14

Common difference, d = a₂ – a₁ = 14 – 8 = 6

(i) 10^th term of the A.P.

a = 1, d = 4 – 1 = 3

A₁₀= a + (10 – 1) d

= 1 + (9)3 = 28

(ii) 18^th term of the A.P.

a = √2, d = 3√2 - √2 = 2√2

A₁₈= a + (18 – 1) d

= √2 + (17) 2√2 = 35√2

(iii) n^th term of the A.P.

a = 13, d = -5

A_n= a + (n-1)d = 13 + (n-1) (-5)

= 13 -5n + 5 = 18 – 5n

(iv) 10^th term of the A.P.

a = -40, d = 25

A₁₀= a + (10-1) d = -40 + 9 (25)

= -40 + 225 = 185

(v) 8^th term of the A.P.

a = 117, d = -13

A₈= a + (8 – 1) d = 117 + 7 (-13)

= 117 – 91 = 26

(vi) 11^th term of the A.P.

a = 10.0, d = 10.5 – 10.0 = 0.5

A₁₁= a + (11 – 1) d = 10.0 + 10 (0.5)

= 10.0 + 5 = 15.0

(vii) 9^th term of the A.P.

a = , d = - =

A₉= a + (9 – 1) d = + 8 () =

a = 3, d = 5

A_n= a + (n – 1) d

248 = 3 + (n – 1) 5 = 3 + 5n – 5

248 = -2 + 5n

250 = 5n

n = 50

a = 84, d = -4,A_n = 0

We know,

A_n= a + (n – 1) d

0 = 84 + (n – 1) -4

0 = 84 – 4n + 4

0 = 88 – 4n

n = 22

a = 4, d = 5

A_n= a + (n – 1) d

254 = 4 + (n – 1) 5

254 = 4 + 5n – 5

254 = -1 + 5n

255 = 5n

n = 51

a = 21, d = 21

A_n= a + (n – 1) d

420 = 21 + (n – 1) 21

420 = 21 + 21n – 21

420 = 21n

n = 20

a = 121, d = -4

A_n = -ve

A_n= a + (n – 1) d

= 121 + (n – 1) -4

= 125 – 4n

Since, we want –ve term

Therefore, 4n should be a number greater than 125

Hence, 128 is the number greater than 125 and divisible by 4

4n = 128

n = 32

Therefore, 32^nd term of the given A.P. is –ve

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.

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.

(i)

a = 7, d = 3

A_n = 43

a + (n – 1) d = 43

7 + (n – 1) 3 = 43

3n = 39

n = 13

Therefore, there are total 13 terms in the A.P.

(ii)

a = -1, d =

A_n =

a + (n – 1) d =

-1 + (n -1) =

n = 26

(iii)

a = 7, d = 6

A_n = 205

a + (n – 1) d = 205

7 + (n – 1) 6 = 205

6n = 204

n = 34

(iv)

a = 18, d =

A_n = -47

a + (n – 1) d = -47

18 + (n – 1) = -47

18 - + = -47

n = 27

Given: The first term of an A.P. is 5, the common difference is 3 and the last term is 80

To find: the number of terms.


Solution :


We have a = 5, d = 3

a_n = 80

From the formula a_n = a + (n-1) d


Number of terms is :

= + 1

= + 1

= 25 + 1

= 26

Hence the number of terms in a given A.P. is 26.

a₆= 19 = a + (n – 1) d

=19 = a + 5d …(i)

A₁₇ = 41 = a + (n – 1) d

= 41 = a + 16d …(ii)

Subtracting (i) from (ii), we get

22 = 11d

d = 2

Now substituting the value of d in (i): 19 = a + 10

= a = 9

A₄₀= a + (40 – 1) 2

= 9 + 78

= 87

a₉ = 0

a + (9 – 1)d =0

a + 8d = 0

a = -8d …(i)

To Prove: a₂₉ = 2a₁₉

Proof: LHS= a₂₉ = a + 28d = -8d + 28d = 20d

RHS= 2a₁₉ = 2[a + (18)d] = 2(-8d + 18d) = 2(10d) = 20d

Since, LHS=RHS

Hence, proved

10a₁₀ = 15a₁₅

10[a + (10 – 1)d] = 15[a + (15 – 1)d]

2a + 2(9)d = 3a + 3(14)d

-a = 42d – 18d

-a = 24d

a = -24d

a₂₅ = a + 24d

= -24d + 24d

= 0

a₁₀ = 41

a + (10 – 1)d = 41

a + 9d = 41 …(i)

a₁₈ = 73

a + (18 – 1)d = 73

Substituting the value of a from (i),

41 – 9d + 17d = 73

8d = 32

d =4

a = 41 – 9(4)

a =41 – 36

a =5

a₂₆ =a + (26 – 1)d

=5 + 25(4)

=5 + 100

=105

Given: a₂₄ = 2a₁₀

To Prove: a₇₂ = 2a₃₄

Proof: a₂₄ = 2a₁₀

a + (24 –1)d = 2a + 2(10 –1)d

23d – 18d = a

a = 5d

LHS= a₇₂ =a + 71d =5d +71d =76d

RHS= a₃₄ =2a + 2(33)d =10d + 66d =76d

Since, LHS=RHS

Hence, proved

Given:
(m+1)^th term of an A.P. is twice the (n+1)^th term
a_{(m +1)} =2a_{(n +1)}

To Prove:
(3m+1)^th term is twice (m+n+1)^th term
a_{(3m + 1)} =2a_{(m +n +1)}

Proof:
a_{(m +1)} =2a_{(n +1)}

⇒ a + (m + 1 – 1) d = 2a + 2(n +1 -1)d

⇒ -a = 2nd – md

⇒ a = md- 2nd (i)

LHS:
a_3m+1= a + (3m +1 -1)d
= md – 2nd + 3md
= 2d(2m - n)

RHS:
2a_{(m + n + 1)} = 2[a +(m +n +1 -1)d]
= 2[md -2nd +md +nd]
= 2d(2m - n)

LHS = RHS

Hence, proved

A=9, D=7-9=-2

a=15, d=-3

A_n=a_n

A+(n -1)D=a +(n -1)d

9 – (n -1)2=15 – (n -1)3

(n -1)(3 -2)=6

n -1=6

n=7

(i)

a=3, d=5 -3=2, a_n =201

a_n= a +(n -1)d

201 =3 +2n – 2

N =100

Now, we have to find 12^th term from the last that means,

100^th – 11=89^th term

Then,

a₈₉ =a + (89 – 1)d

=3 +88

=179

Hence, the 12^th term from the end of the A.P. is 179.

(ii)

a=3, d= 8 -3 =5

a_n= 253

a + (n -1)d =253

3 + (n -1)5 =253

n=51

Now, we have to find 12^th term from the last that means,

51th -11 = 40^th term

Then,

a₄₀ = a + (n -1)d

= 3 + 39(5)

=198

Hence, the 12^th term from the end of the A.P. is 198

(iii)

a=1, d= 4 -1 =3

a_n= 88

a + (n -1)d =88

1 + (n -1)3 =88

n =30

Now, we have to find 12^th term from the last that means,

30^th -11 = 19^th term

a₁₉ = a + 18d

= 1 + 18(3)

=55

Hence, the 12^th term from the end of the A.P. is 198.

Given:

a₄= 3a₁ …(i)

a₇= 2a₃ +1 …(ii)

We know a_n = a + (n-1) d

a₃ = a +2d

a₄ = a +3d

a₇ = a +6d

Put all the values of a₁ and a₄ in (i),

a + 3d = 3a

⇒ 3d = 3a - a

⇒ 3d = 2a

⇒ d=

Put all the values of a₇ and a₃ in (ii),

⇒ a + 6d= 2(a +2d) +1

⇒ a + 6d = 2a + 4d + 1

⇒ 2d = a +1

Put d = ,

⇒ 2() = a +1

⇒ a= 3

Now, difference, d=

⇒ d= 2

Hence, first term is 3 and the common difference is 2.

a₆= a + 5d

12= a + 5d …(i)

a₈= a + 7d

22= a + 7d …(ii)

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

10= 2d

d= 5

from (i)

12 = a + 5(5)

12= a + 25

a= -13

a₂= a +d

=-13 +5

=-8

a_n= a + (n-1)d

= -13 + (n-1)5

=-18 +5n

Two digit numbers divisible by 3 are 12, 15, 18,… ,99

Hence, a= 12, ,d=3

a_n= a + (n -1)d

99= 12(n -1)3

99= 9 = 3n

n=30

Hence, the total numbers of two digit divisible by 3 are 30.

n =60, a=7, a₆₀ =l =125

a₆₀ = a + 59d

125 = 7 + 59d

d=

d=2

a₃₂= a + 31d

= 7 + 31(2)

=69

Hence, 32th term is 69 in the given A.P.

Given: a₄ + a₈ = 24 …(i)

a₆ + a₁₀ = 34 …(ii)

a₄ = a + 3d

a₈ =a + 7d

a₆ =a + 5d

a₁₀=a + 9d

Put the value of a₄ and a₈ in (i)

(a + 3d) + (a + 7d) = 24

a + 5d =12 …(iii)

Put the value of a₆ and a₁₀ in (ii)

(a +5d) + (a +9d) =34

a +7d =17 …(iv)

Subtracting (iii) from (iv), we get

2d = 5

d =

Putting the value of d in (iii), we get

a = 12 - =

Hence, first term is and common difference is

Given, a = 5

a₁₀₀= -292

5 – 99d = -292

d = = 3

Now, 50^th term, a₅₀ = a + (50 – 1) d

= 5 + (49) 3 = 5 + 147 = 152

Hence, 50^th term of given A.P. is 152

(i)

a = -9, d = -14 + 9 = -5

a₃₀= a + 29d

= -9 + 29(-5) = -154

a₂₀= a + 19d

= -9 + 19(-5) = -104

a₃₀ – a₂₀= -154 – (-104)

= -154 + 104

= -50

(ii)

a = a, d = a + d – a = d

a₃₀= a + 29d

a₂₀= a+ 19d

a₃₀ – a₂₀ = a + 29d – a – 19d

= 10d

Let n^th term a_n = a + (n – 1) d

= a + nd – d

k^th term, a_k = a + (k – 1) d

= a + kd – d

Now,

a_n - a_k = (a + nd – d) – (a + kd – d)

= nd – kd = d (n – k)

(i) a₁₁ = 5

a + 10d = 5 (i)

a₁₃ = 79

a + 12d = 79 (ii)

By subtracting (i) from (ii), we get

2d = 74

d = 37

Hence, the common difference is 37

(ii)a₁₀ = a + 9d (i)

a₅ = a + 4d (ii)

a₁₀ – a₅ = a + 9d – a – 4d

200 = 5d

d = 40

Hence, common difference is 40

(iii) a₂₀ = a + 19d (i)

a₁₈ = a + 17d (ii)

Given that a₂₀ = a₁₈ + 10

a + 19d = a + 17d + 10

2d = 10

d = 5

Hence, common difference is 5

(i)

a = 25, d = 50 – 25 = 25

Last term, l = 1000

Number of terms, n = + 1

= + 1 = 40

Hence, the value of n is 40

(ii)

a = -1, d = -2

Last term, l = -151

Number of terms, n = + 1

= + 1 = 76

Hence, the value of n is 76

(iii)

a = , d =

Last term, l = 550

l = a + (n – 1) d

550 = + (n – 1)

550 =

1100 = 11n

n = 100

Hence, number of terms, n = 100

(iv)

a = 1, d = – 1 =

Last term, l =

a + (n – 1) d =

1 + (n – 1) =

= – 1 +

=

10n = 170

n = 17

Hence, value of n is 17

Given, a = a

Last term, l = l

We have to prove that the sum of the m^th term from the beginning and m^th term from the end is (a + 1)

Now, m^th term from the beginning,

a_m(b) = a + (m – 1) l

= a + ml – l (i)

Again m^th term from the end

a_m(e) = l – (m – 1)l

= l – ml + l = 2l – ml (ii)

By adding (i) and (ii), we get

a + ml – l + 2l – ml = a + l

Hence, proved

Given, a₃ = 16

a + 2d = 16 (i)

a₅ = a + (5 – 1) d

= a + 4d

a₇ = a + (7 -1)d

= a + 6d

Acc. To question,

a₇ = 12 + a₅ (ii)

Put the value of a₅ and a₇ in (ii), we get

a + 6d = 12 + a + 4d

2d = 12 + a – a

2d = 12

d = 6

From equation (i),

16 = a + 2(6)

16 = a + 12

a = 4

We have to find A.P.

a₁ = 4

a₂ = a + d = 4 + 6 = 10

a₃ = a + 2d = 4 + 2(6) = 16

a₄ = a + 3d = 4 + 3(6) = 22

Hence, the A.P. is 4 , 10, 16 , 22,…

a₇ = 32

a + 6d = 32 (i)

a₁₃ = a + 12d = 62 (ii)

By subtracting (i) from (ii), we get

6d = 30

d= 5

From (i), a = 32 – 6(5) = 2

We have to find A.P.

a₁ = 2

a₂ = a + d = 2 + 5 = 7

a₃ = a + 2d = 2 + 2(5) = 12

a₄ = a + 3d = 2 + 3(5) = 17

Hence, A.P. is 2, 7, 12, 17,…

a = 3, d = 7

a₁₃ = 3 + (13 – 1)7

= 3 + 84 = 87

Now, n^th term is 84 more than the 13^th term

a_n = 84 + a₁₃

= 84 + 87

= 171

Now we have to find n,

a_n = a + (n – 1) d

171 = 3 + (n – 1) 7

7n = 175

n = 25

Hence, 25^th term of the given A,.P. is 84 more than its 13^th term

Two arithmetic progressions have the same common difference.

Let, common difference = d

First term of first A.P. = a

First term of second A.P. = a’

Given that difference between their 100^th term is 100

100^th term of first A.P., a₁₀₀ = a + 99d

100^th term of second A.P., a’₁₀₀ = a’ + 99d

Acc. To question,

a₁₀₀ – a’₁₀₀ = 100

a + 99d – a’ – 99d = 100

a – a’ = 100 (i)

1000^th term of first A.P., a₁₀₀₀ = a + 999d

1000^th term of second A.P., a’₁₀₀₀ = a’ + 999d

Acc. To question,

a₁₀₀₀ – a’₁₀₀₀ = a + 999d – a’ – 999d

= a – a’

a₁₀₀₀ – a’₁₀₀₀ = 100 [from (i)]

Hence, difference between 1000^th term of two A.P. is 100

Given, A.P. = 63, 65, 67,…

A.P.’ = 3, 10, 17,…

Let for A.P.,

First term, a = 63, d = 65 – 63 = 2

And n^th term = a_n

a_n = 63 + (n – 1) 2

= 63 + 2n – 2

= 61 + 2n

For A.P.’

First term, a = 3, d = 10 – 3 = 7

n^th term = a_n

Now n^th term of both A.P. are equal

61 + 2n = -4 + 7n

7n – 2n = 61 + 4

n = 13

Hence, 13^th term of both the A.P. are equal

Let, multiple of 4 lies between 10 and 250

12, 16, 20, 24,…248

We know, a_n = a + (n – 1)d

First term, a = 12

C0mmon difference, d = 16 – 12 = 4

Last term, a_n = 248

a_n = a+ (n – 1)d

248 = 12 + (n – 1) 4

248 = 12 + 4n – 4

4n = 248 – 12 + 4

4n = 240

n = 60

Hence, multiple of 4 lies between 10 and 250 is 60

Let, three digit number divisible by 7 are

105, 112, 119,….994

Here, First term, a = 105

Common difference, d = 112 – 105 = 7

And last term, a_n = 994

a_n = a+ (n – 1) d

994 = 105 + (n – 1) 7

994 = 105 + 7n – 7

994 = 98 + 7n

7n = 896

n = 128

Hence, three digit number that are divisible by 7 are 128

a = 8, d = 6,

Last term = a_n

a_n= a + (n – 1) d

= 8 + (n – 1) 6

= 2 + 6n (i)

Let, 41^st term, a₄₁ = 8 + (41 – 1) 6

= 8 + 40 * 6 = 248

Now the term is 72 more than its 41^st term

a_n = 72 + a₄₁

= 72 + 248 = 320

Putting this value in (i), we get

320 = 2 + 6n

6n = 318

n = 53

Hence, 53^rd term of the given A.P. is 72 more than its 41^st term

First term, a = 9, d = 12 – 9 = 3

Let, last term be a_n

a_n = a + (n – 1) d

= 9 + (n – 1)3

= 9 + 3n – 3 = 6 + 3n (i)

Let, 36^th term, a₃₆ = a + 35d

= 9 + 35 (3) = 114

Now the term is 39 more than its 36^th term

a_n = 39 + a₃₆

= 39 + 114 = 153

Putting the value in (i), we get

153 = 6 + 3N

3n = 153 – 6

3n = 147

n = 49

Hence, 49^th term of the given A.P. is 39 more than its 36^th term

a = 7, d = 3

Last term, a_n = 184

a_n = a + (n – 1) d

184 = 7 + (n - 1) 3

184 = 7 + 3n – 3

180 = 3n

n = 60

Now we have to find last 8^th term, it means = 60^th – 7 = 53^rd term

a₅₃ = 7 + (53 – 1)3

= 7 + 52 * 3

= 7 + 156 = 163

Hence, 8^th term from the end of given is 163

Given,
First term, a = 8
Common difference, d = 2

Let the number of terms be 'n'
We know, nth term of an AP is
a_n = a + (n – 1)d
where ‘a’ and ‘d’ are first term and common difference of AP respectively
Last term, a_n = 126

⇒ a_n = a + (n – 1) d

⇒ 126 = 8 + (n – 1) 2

⇒ 120 = 2n

⇒ n = 60

Now we have to find last 10^th term, means = 60^th term – 10 + 1 = 51^th term

Now, a₅₁ = 8 + (51 – 1)2

= 8 + 50(2)
= 8 + 100
= 108

Hence, 10^th term from the last of given A.P. is 108

Given, a₄ + a₈ = 24 (i)

a₆ + a₁₀ = 44 (ii)

We know, a_n = a + (n – 1) d

4^th term, a₄ = a + 3d

6^th term, a₆ = a + 5d

8^th term, a₈ = a + 7d

10^th term, a₁₀ = a + 9d

Putting the value of a₄ and a₈ in (i), we get

a + 3d + a + 7d = 24

2a + 10d = 24 (iii)

Put the value of a₆ and a₁₀ in (ii), we get

a + 5d + a + 9d = 44

2a + 14d = 44 (iv)

By subtracting (iii) from (iv), we get

4d = 20

d = 5

Now putting value of d in (iii), we get

2a = 24 – 10(5)

a = -13

a₁ = -13

a₂ = a + d = -13 + 5 = - 8

a₃ = a + 2d = -13 + 10 = -3

Hence, the A.P. is -13, -8, -3,…

a = 3, d = 15 – 3 = 12

Let last term be a_n

a_n = a + (n – 1) d

= 3 + (n – 1) 12

= 12n – 9 (i)

a₂₁ = a + 20d

= 3 + 20(12) = 243

Now, the term is 120 more than the 21^st term

a_n = 120 + a₂₁

= 120 + 243

= 363

Putting this value in (i), we get

363 = 12n – 9

12n = 363 + 9

n = 31

Hence, 31^st term of given A.P. is 120 more than its 21^st term

Given: The 17^th term of an A.P. is 5 more than twice its 8^th term and the 11^th term of the A.P. is 43.

To find: the n^th term.

Solution:

Given,

a₁₇ = 5 + 2(a₈) ......(i)

And

a₁₁ = 43

As a_n = a + (n-1) d

For n=11,

a + 10d = 43

a = 43 - 10d ......(ii)


For n = 8,

a₈ = a + 7d


For n=17,

a₁₇ = a + 16d


Put the value of a from (ii)

⇒ a₁₇= 43 – 10d + 16d

⇒ a₁₇ = 43 + 6d

Putting the value of a₈ and a₁₇ in (i), we get

⇒ 43 + 6d = 5 + 2(a + 7d)


⇒ 43 + 6d = 5 + 2a + 14d

⇒ 43 – 5 = 2a + 14d – 6d

⇒ 38 = 2a + 8d

⇒ 38 = 2(43 – 10d) + 8d [from (ii)]

⇒ 38 = 86 – 20d + 8d

⇒ 38 = 86 – 12d

⇒ 12d = 86 – 38

⇒ 12d = 48

⇒ d = 4

From (ii), a = 43 – 10d

⇒ 43 – (10 × 4) = 43 – 40 = 3

We know, n^th term of A.P., a_n = a + (n -1) d

a_n = 3 + (n – 1) 4

a_n = 3 + 4n – 4


a_n = 4n – 1

Hence, n^th term is 4n – 1.

Let the three digit numbers divisible by 9 are

108, 117, 126,… ,999

a=108, d=9, last term, l= 999

Number of terms, n = +1

= + 1

= +1

=100

Hence, the number of all three digit natural numbers which are divisible by 9 are 100

Given: a₉ =19

a + 8d= 19 (i)

Acc. to question,

a₁₉= 3a₆ (ii)

a₁₉=a + 18d

a₆=a + 5d

Putting the value of a₁₉ and a₆ in (ii)

a + 18d =3(a +5d)

3d= 2a

3d=2(19 -8d) (from (i))

19d=38

d=2

Now, putting the value of d in (i)

a=19 -8(2) =3

a₁ = a=3

a₂ =a +d=3 +2=5

a₃ =a +2d=3 + 2(2)=7

Hence, the A.P. is

Given: a₅ = 22

a + 4d=22 (i)

Acc. to question,

a₉ = 6a₂ (ii)

a₉= a +8d

a₂=a+d

Putting the value of a₉ and a₂ in (ii)

a +8d= 6(a +d)

2d=5a

2d=5(22 -4d)

22d=110

d=5

Now, putting the value of d in (i)

a=22 -4(5)=2

a₁ = a=2

a₂ =a +d=2 +5=7

a₃ =a +2d=2 + 2(5)=12

Hence, the A.P. is

Given: a₂₄=2a₁₀

a +23d= 2(a +9d)

5d=a (i)

To Prove:a₇₂ =4a₁₅

Proof: L.H.S. =a₇₂

= a +71d

=5d +71d [from (i)]

=76d

R.H.S= 4a₁₅

=4(a +14d)

=4a +56d

=4(5d) +56d [from (i)]

=20d +56d

=76d

Since, L.H.S=R.H.S.

Hence proved

We have to find the numbers between 101 and 999 which are divisible by both 2 and 5

That means the numbers divisible by 10 between 101 and 999.

Let, the required divisible numbers be,

110, 120, 130, …, 990

a=110, d=10, last term, l=990

Number of terms, n = +1

= + 1

= +1

=89

Hence, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 are 89.

We know, nth term of an AP is
a_n = a + (n – 1)d
where ‘a’ and ‘d’ are first term and common difference of AP respectively
Given,

Subtracting equation [1] from [2], we get

Putting this value of d in equation [1], we get

Now, 63rd term = a + 62d

Hence Proved!

a₅ +a₉=30

(a +4d) + (a +8d)=30

2a +12d=30

a + 6d=15 (i)

Acc. To question,

a₂₅ =3a₈

a+ 24d = 3(a +7d)

3d=2a

3d=2(15 -6d) [from(i)]

15d=30

d=2

Putting the value of d in (i),

a= 15 – 6(2)=3

a₁ = a=3

a₂ =a +d=3 +2=5

a₃ =a +2d=3 + 2(2)=7

Hence, the A.P. is

a=40, d=37 -40= -3

Let, a_n =0

a + (n -1) d =0

40 + (n -1) (-3) =0

-3n = -43

n=43/3

Since, ‘n’ can’t be a fraction

Hence, the answer is no

a=213, l=37

Middle term of A.P. =

=

=

=125

a₅=31

a +4d =31 (i)

a₂₅= 140 + a₅

a + 24d= 140 + 31

31 -4d +24d=171 [ from (i) ]

20d=140

d=7

Putting the value of d in (i)

a= 31 – 4(7) = 3

a₁ = a=3

a₂ =a +d=3 +7=10

a₃ =a +2d=3 + 2(7) =17

Hence, the A.P. is 3, 10, 17,…

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

a² – d² = a + 6

7² – d² = 7 + 6

49 – d² = 13

d² = 36

d = 6

Therefore, a – d = 7 – 6 = 1

a + d = 7 + 6 = 13

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

Let the three numbers be a – d, a, a + d

a – d + a + a + d = 27

3a = 27

a = 9

According to question,

(a – d) (a) (a + d) = 648

(a² – d²) (a) = 648

(9² – d²) 9 = 648

(81 – d²) = 72

d² = 9

d = 3

Therefore, a – d = 9 – 3 = 6

A + d = 9 + 3 = 12

Let the four numbers be (a – 3d), (a – d), (a + d) and (a + 3d)

a – 3d + a – d + a + d + a + 3d = 50

4a = 50

a =

According to question,

(a + 3d) = 4(a – 3d)

a + 3d = 4a – 12d

3a = 15d

5d =

d =

Therefore,

a – 3d = - = 5

a – d = - = 10

a + d = + = 15

a + 3d = + = 20

Let the angles be (a – 3d), (a – d), (a + d) and (a + 3d)

a – 3d + a – d + a + d + a + 3d = 360^o (sum of angles of quadrilateral)

4a = 360^o

a = 90^o

According to question,

(a + d) – (a – d) = 10^o

2d = 10^o

d = 5^o

Therefore, (a – 3d) = 90^o – 15^o = 75^o

(a – d) = 90^o – 5^o = 85^o

(a + d) = 90^o + 5^o = 95^o

(a + 3d) = 90^o + 15^o = 105^o

Let the numbers be (a – d), a, (a+ d)

A – d + a + a+ d = 12

3a = 12

a = 4

According to question,

(a – d)³ + a³ + (a + d)³ = 288

a³ – d³ – 3ad (a – d) + a³ + a³ + d³ + 3ad (a + d) = 288

3a³ – 3a²d + 3ad² + 3a²d + 3ad² = 288

3(4)³ + 6(4) d² = 288

24d² = 96

d² = 4

d = �4

Therefore, when a = 4 and d = 2

a – d = 4 – 2 = 2

a + d = 4 + 2 = 6

When a = 4 and d = -2

a – d = 4 + 2 = 6

a + d = 4 – 2 = 2

We know that, if three numbers are a,b,c are in AP then, 2b = a + c

According to question, (8x+4), (6x-2), (2x+7) are in AP.

Hence,

2 (6x – 2) = 8x + 4 + 2x + 7

12x – 4 = 10x + 11

2x = 15

x = 15/2 = 7.5

Given terms: x+1, 3x and 4x+2

Since, the given terms are in A.P.

Therefore, their common difference will be same

3x – (x + 1) = (4x + 2) – 3x

3x – x - 1 = 4x + 2 – 3x

2x - 1 = x + 2

x = 3

The terms given below are : (a–b)², (a²+b²) and (a + b)²

Common difference, d₁ = a² + b² – (a – b)²

d₁ = a² + b² – (a² + b² - 2ab)

d₁ = a² + b² – a² - b² + 2ab

d₁= 2ab

Common difference, d₂ = (a + b)² – (a² + b²)

d₂ = a² + b² + 2ab – a² –b²

d₂ = 2ab

Since, d₁ = d₂ i.e. the common difference is same.

Therefore, the given terms are in A.P.

(i)to 10 terms

S_n = [2a + (n – 1) d]

S₁₀ = [100 + 9(-4)]

= 5 [100 – 36]

= 5(64) = 320

(ii)to 12 terms

S_n = [2a + (n – 1) d]

= [2 + (12 – 1) 2]

= 6 (2 + 22)

= 144

(iii)to 25 terms

S_n = [2a + (n – 1) d]

= [2 (3) + 24()]

= [6 + 36]

= 525

(iv)to 12 terms

S_n = [2a + (n – 1) d]

= [82 - 55]

= 6 [27] = 162

(v)to 22 terms

S_n = [2a + (n – 1) d]

= [2(a + b) + (21) (-2b)]

= 11 [2a + 2b – 42b]

= 11 [2a – 40b]

= 22a – 440b

(vi)to n terms

S_n = [2a + (n – 1) d]

= [2(x – y)² + (n – 1) (2xy)]

= (2) [(x – y)² + (n – 1) xy]

= n [(x – y)² + (n – 1) xy]

(vii)to n terms

S_n = [2a + (n – 1) d]

= [2() + (n – 1) xy]

= {2(x – y) + (n – 1) (2x – y)}

= {n (2x – y) – y}

(viii)to 36 terms

S_n = [2a + (n – 1) d]

= [-52 + (35) 2]

= 18 [-52 + 70]

= 18 [18] = 324

S_n = [2a + (n – 1) d]

= [10 + (n – 1) -3]

= [13 – 3n]

Put n = 1

a₁ = 5 – 6(1) = 5 – 6 = -1

Put n = 2

a₂= 5 – 6(2) = 5 – 12 = -7

a = -1, d = -7 + 1 = -6

S_n = [2a + (n – 1) d]

= [2(-1) + (n -1) (-6)]

= [-1 + 3n + 3]

= n [2 -3n]

a = 25, d = -3

S_n = [2a + (n – 1) d]

116 = [50 – 3n +3]

116 = [53 – 3n]

232 = 53n – 3n²

3n² – 53n + 232 = 0

n = 8 or n = , which isn’t possible as n must be a natural number.

Therefore, n = 8

[a + l] = 116

[25 + l] = 116

l = 4

Hence, the last term is 4

S_n = [2a + (n – 1) d]

0 = [2(18) + (n – 1) (-2)]

38n – 2n² = 0

n = 0 (which is not possible)

n = 19

Therefore, n = 19 terms

a₅= a + (n -1) d

2 = -14 + 4d

16 = 4d

d = 4

S_n = [2a + (n – 1) d]

40 = [2(-14) + (n -1) (4)]

40 = 12n² – 16n

2n² – 16n -40 = 0

n² – 8n -20 = 0

n² – 10n + 2n – 20 = 0

(n – 10) (n + 2) = 0

n = 10 and n = -2 (not possible)

S_n = [2a + (n – 1) d]

636 = [2(9) + (n – 1) 8]

636 = n [9 + 4n -4]

636 = 5n + 4n²

4n² + 5n – 636 = 0

n = 12 or n = (not possible)

S_n = [2a + (n – 1) d]

693 = [2(63) + (n – 1) (-3)]

693 = [126 -3n + 3]

462 = n [43 – n]

n² – 43n + 462 = 0

n = 22 or n = 21

a = 17, l = 350, d = 9

Number of terms, n = + 1

= + 1

= = = 38

S_n= (a + l)

= (350 + 17)

= 19 (367) = 6973

We know, nth term of an AP is
a_n = a + (n – 1)d
where ‘a’ and ‘d’ are first term and common difference of AP respectively
Given, third term of an A.P. is 7
a₃ = 7
⇒ a + 2d = 7 (i)
the seventh term exceeds three times the third term by 2

⇒ a₇ = 3a₃ + 2
⇒ a + 6d = 3(7) + 2
⇒ 7 – 2d +6d = 21 + 2
⇒ 4d = 16
⇒ d = 4

From (i),
a = 7 – 2(4)

= -1

Also, we know sum of first 'n' terms of an AP is

Sum of first 20 terms is

= 10[-2 + 76]
= 740

a = 2, l = 50

S_n= [a + l]

442 = [50 +2]

n = = 17

n = + 1

17 = + 1

16d = 48

d = 3

a₁₂= -13

a + (12 – 1) d = -13

a + 11d = -13

S₄ = [2a + (4 – 1) d]

24 = 2 [2a + 3d]

12 = 2a + 3d

12 = 2 (-11d – 13) + 3d

12 = -22d – 26 + 3d

38 = -19d

d = -2

a = -13 + 11(2)

= -13 + 22

= 9

S₁₀= [2(9) + (10 – 1) (-2)]

= 5 [18 – 18]

= 0

a₂₂= a + 21d

149 = a + 21(22)

a = 149 – 462

a = -313

S₂₂= [2a + (22 – 1) d]

= 11 [-626 + 462]

= 11 (-164)

= -1804

a = 3, l = 99, n = 33

S₃₃= (a + l)

= (3 + 99)

= (102)

= 33 (51) = 1683

a = 1, d = 2

Sum of first n odd numbers, S_n = [2a + (n – 1) d]

= [2(1) + (n – 1) 2]

= n [1 + n – 1]

= n²

(i) 0 and 50

a = 1, l = 49, n = 25

S₂₅= [a + l]

= [1 + 49]

= (50) = 625

(ii) 100 and 200

a = 101, l = 199, n = 50

S₅₀= [a + l]

= 25 (101 + 199)

= 25 (300) = 7500

a = 3, l = 999, n = = 167

S_n= (a + l)

= (3 + 999)

= (1002)

= 167 (501) = 83667

To find: the sum of all integers between 84 and 719, which are multiples of 5.

Solution:


The smallest and largest digits between 84 and 719 which are divisible by 5 is 85 and 715.

So the sequence will be 85,90,........... 715.

a = 85, d = 5, l = 715


From the formula l=a+(n-1)d

The number of terms are:

n = + 1

n= + 1

n=


n= 127

⇒ S₁₂₇= (a + l)

⇒ S₁₂₇= (85 + 715)

⇒ S₁₂₇= (800) = 50800

So the sum of the terms between 84 and 719 is 50800.

a = 56, d = 7, l = 497

n = + 1

= + 1

= + 1 =

= 64

S₆₄= (a + l)

= 32 (56 + 497)

= 32 (553) = 17696

a = 102, l = 998

n = = 449

S₄₄₉= (a + l)

= (102 + 998)

= = 246950

a = 108

L = 549

D = 9

549 = 108 + (n - 1)9

549 = 99 + 9n

9n = 450

n = 50

S_n= [216 + 49 (9)]

= 16425

a = 22, d = -4

S_n = 64

64 = (2a + (n – 1) d)

n [44 – (n – 1)4] = 128

n (44 – 4n + 4) = 128

48n – 4n² = 128

n² – 12n + 32 = 0

n² – 8n – 4n + 32 = 0

(n – 8) (n – 4) = 0

n = 8 or n = 4

According to question,

a₅= 30

a + 4d = 30

a = 30 – 4d (i)

a₁₂= 65

a + 11d = 65

30 – 4d + 11d = 65 [from (i)]

7d = 35

d = 5

Put d in (i)

a = 30 – 4(5)

= 10

S₂₀= [2(a) + (20 – 1) d]

= 10 [2(10) + 19(5)]

= 10 {20 + 95}

= 1150

(i) 11 terms of the A.P :

a = 2, d = 6 – 2 = 4

S₁₁= [2(a) + 10d]

= [2(2) + 10(4)]

= 11 [2 + 20]

= 242

(ii) 13 terms of the A.P :

a = -6, d = 0 + 6 = 6

S₁₃= [2(a) + 12d]

= 13 [-6 + 6(6)]

= 13 [-6 + 36]

= 13 (30) = 390

(iii) 51 terms of the A.P. whose second term is 2 and fourth term is 8.

a₂= 2

a + d = 2 (i)

a₄ = 8

a + 3d = 8

2 – d + 3d = 8

2 + 2d = 8

d = 3

a = -1

S₅₁= [2(a) + 50d]

= 51 [-1 + 25(3)]

= 51 (74)

= 3774

(i) The first 15 multiples of 8

a = 8, d = 8

S₁₅= [2(8) + 14(8)]

= 960

(ii) The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(a) a = 3, d = 3

S₄₀= [2(a) + 39(d)]

= 20 [2(3) + 39(3)]

= 60 (41) = 2460

(b) a = 6, d = 5

S₄₀= [2(5) + 39(5)]

= 100 [41]

= 4100

(iii) All 3 – digit natural numbers which are divisible by 13.

a = 6, d = 6

S₄₀= [2(6) + 39(6)]

= 120 (41)

= 4920

(iv) All 3 – digit natural numbers, which are multiples of 11.

a = 110, d = 11, l = 990

n = + 1

= + 1

=

=

= 81

S₈₁= [2(110) + 80(11)]

= 8910 [1 + 4]

= 44550

(i)

a = 2, d = 4 – 2 = 2, l = 200

n = + 1

= + 1 = 100

S₁₀₀= [2(2) + 99(2)]

= 100 [101] = 10100

(ii)

a = 3, d = 11 – 3 = 8, l = 803

S_n= + 1 = + 1

= 101

S₁₀₁= [2(3) + 100(8)]

= 101 (403) = 40703

(iii)

a = -5, d = -8 + 5 = -3, l = -230

n = + 1

= = 76

S_n= 38(5) [-2 – 45]

= 190 (-47) = -8930

(iv)

a = 1, d = 3 – 1 = 2, l = 199

n = + 1

= 100

S₁₀₀= [2(1) + 99(2)]

= 100 (100) = 10000

(v)

a = 7, d = - 7 =

Last term, a_n = 84

a_n = a + (n – 1)d

84 = 7 + (n – 1)

84 =

84 * 2 = 7 + 7n

168 = 7 + 7n

7n = 168 – 7

7n = 161

n = 23

Sum of n^th term, S_n = [2a + (n – 1)d]

S₂₃ = [2(7) + (23 – 1)]

= [14 + 22 * ]

= [14 + 77]

= * 91 =

Hence, sum of given A.P. is

(vi)

a = 34, d = 32 – 34 = -2

Last term, a_n = 10

a_n = a + (n – 1)d

10 = 34 + (n – 1) (-2)

10 = 34 – 2n + 2

2n = 34 – 10 + 2

2n = 26

n = 13

Sum of n terms,

S_n = [2a + (n – 1) d]

S₁₃ = [2(34) + (13 – 1) (-2)]

= [68 + 12(-2)]

= [68 – 24]

= * 44 = 286

Hence, Sum of given A.P. is 286

(vii)

a = 25, d = 3

Last term, a_n = 100

A_n = a + (n -1) d

100 = 25 (n – 1)3

75 = 3n – 3

78 = 3n

n = 26

Sum of n terms, S_n = [2a + (n – 1) d]

= [2(25) + (26 – 1) 3]

= 13 [50 + 25 * 3]

= 13 * 125 = 1625

Hence, Sum of given A.P. is 1625

(viii)

a = 18, d = - 18 =

Last term, a_n =

a_n = a+ (n – 1)d

= 18 + (n – 1) ()

-99 = 36 + 5 – 5n

-140 = -5n

n = 28

S₂₈= [a + l]

= 14 (18 - )

= 7 (36 – 99)

= 7 (-63) = -441

(i)

Put n = 1

a₁ = 3 + 4(1) = 7

Put n = 15

a₁₅ = 3 + 4(15) = 63 = l

Sum of 15 terms, S₁₅ = [a + l]

= [7 + 63] = 525

(ii)

Put n = 1

b₁ = 5 + 2(1) = 7

Put n = 15

b₁₅ = 5 + 2(15) = 35 = l

Sum of 15 terms, S₁₅= [a + l]

= [7 + 35] = 315

(iii)

Put n = 1

x₁ = 6 – 1 = 5

Put n = 15

x₁₅ = 6 – 15 = -9

Sum of 15 terms, S₁₅= [a + l]

= [5 - 9] = -30

(iv)

Put n = 1

y₁= 9 – 5(1) = 4

Put n = 15

y₁₅= 9 – 5(15) = -66

Sum of 15 terms, S₁₅= [a + l]

= [4 - 66] = -465

Given, a_n= An + B

Put n = 1, a₁= A + B

Put n = 20, a₂₀= 20A + B

S₂₀= [a + l]

= 10 [A + B + 20A + B]

= 10 [21A + 2B]

= 210A + 20B

Given, n^th term, a_n = 2 – 3n

Put n = 1, a₁ = 2 – 3(1) = -1

Put n = 25, a₁₅= 2 – 3(15) = -43

Therefore, S₂₅ = (-1 – 43)

= (-44) = -925

Given, a_n= 7 - 3n

Put n = 1

a₁= 7 – 3(1) = 4

Put n = 25

a₂₅= 7 – 3(25) = -68 = l

S₂₅= [4 – 68]

= 25 [-32] = -800

a₂= 14

a + d = 14 (i)

a₃= 18

a + 2d = 18 (ii)

Subtracting (i) from (ii), we get

d = 4

Putting the value of d in (i), we get

a = 14 – 4 = 10

S₂₅= [2(a) + 24(d)]

= 25 [58]

= 1450

S₇= 49

[2a + 6d] = 49

a + 3d = 7 (i)

S₁₇= 289

[2a + 16d] = 289

a + 8d = 17 (ii)

Subtract (i) from (ii), we get

5d = 10

d = 2

Put d = 2 in (i), we get

a = 7 – 6 = 1

S_n= [2(1) + (n – 1)2]

= n [1 + n – 1]

= n²

a = 5, l = 45

S_n= 400

[5 + 45] = 400

[50] = 400

n = 16

16^th term is 45

a₁₆= 45

5 + 15d = 45

15d = 40

d =

Given,

_{Let an} be the n^th term of the A.P.

_an= S_n – S_{n – 1}

= 3n²/2 + 13n/2 – 3(n – 1)²/2 – 13(n – 1)/2

= {n² – (n – 1)²} + {n – (n – 1)}

= 3n - +

= 3n + = 3n + 5

Put n = 25

a₂₅= 3(25) + 5

= 75 + 5 = 80

Therefore, 25^th term is a₂₅ = 80

(i)

Given, a_n= 50

a + (n – 1) d = 50

5 + (n – 1) 3 = 50

(n – 1)3 = 45

(n -1) = 15

n = 16

S_n= [a + l]

= 8 [5 + 50] = 8 [55] = 440

(ii)

a_n= 4, d = 2, S_n= -14

a + (n – 1) 2 = 4

a + 2n = 6, and

[2a + (n – 1)2] = -14

n [ 2a + 2n – 2] = -14, or

[a + a_n] = -14

[a + 4] = -14

n [6 -2n +4] = -28

n [10 -2n] = -28

2n² – 10n – 28 = 0

2(n² – 5n – 14) = 0

(n + 2) (n – 7) = 0

n = -2, n = 7

Therefore, n = -2 is not a natural number. So, n = 7

(iii)

Given, a = 3, n = 8, S_n= 192

S_n = {2a + (n – 1) d]

192 * 2 = 8 [6 + (8 – 1)d]

= 6 + 7d

48 = 6 + 7d

7d = 42

d = 6

(iv)

a_n= 28, S_n = 144, n = 9

S_n= [a + l]

144 = [a + 28]

= a = 28

a + 28 = 32

a = 4

(v)

Given, a = 8, a_n= 62 and S_n= 210

S_n= [a + l]

210 = [8 + 62]

210 * 2 = n (70)

n = = 6

a + (n – 1) d = 62

8 + (6 – 1) d = 62

5d = 54

d = 10.8

(vi)

Given, a = 2, d = 8 and S_n = 90

90 = [4 + (n – 1)d] {Therefore, S_n= [2a + (n – 1)d]}

180 = n [4 + 8n – 8]

8n² – 4n – 180 = 0

4 (2n² – n – 45) = 0

2n² – n – 45 = 0

(2n + 1) (n – 5) = 0

Therefore, n = is not a natural number

n = 5

a_n= a + (n – 1)d

= 2 + 4(8) = 32

Given,

Man saved in 10 years, S₁₀= 16500

In each year after first he saved, d = 100

We know, sum of n terms, S_n= [2a + (n – 1) d]

For 10 years

S₁₀= [2a + (10 – 1) 100]

16500 = 5 [2a + 9 * 100]

= 2a + 900

3300 = 2a + 900

2a = 3300 – 900

2a = 2400

a = 1200

Hence, Rs. 1200 saved by him in first year

Given, A man saved in first year, a = 32

A man saved in second year, a₂ = 36

Increase saving, d = 4

In n years his saving will be 200, S_n = 200

We know,

S_n = [2a + (n – 1) d]

200 = [2(32) + (n – 1) 4]

400 = n [64 + 4n – 4]

400 = n [60 + 4n]

400 = 4n [15 + n]

100 = 15n + n²

n² + 15n – 100 = 0

n² + 20n – 5n – 100 = 0

n (n + 20) – 5 (n + 20) = 0

(n – 5) (n + 20) = 0

Here, n – 5 = 0, n = 5

n + 20 = 0, n = -20

The term can never be negative. So, we consider n = 5

Hence, in 5 years his saving will be Rs. 200

Given, Total amount of payable in 40 annual installments, S₄₀ = 3600

After 30 installment he died and leaving of the debt unpaid

Which means, total in 30 installment, S₃₀= of the debt

S₃₀= * 3600 = 2400

We know sum of n terms, S_n = [2a + (n – 1) d]

For 30 installments, S₃₀ = [2a + (30 – 1) d]

2400 = 15 [2a + 29d]

160 = 2a + 29d

2a = 160 – 29d

a = (i)

For 40 installments, S₄₀ = [2a + (n – 1) d]

3600 = [2a + (40 – 1) d]

180 = 2a + 39d

2a = 180 – 39d

a = (ii)

From (i) and (ii), we get

=

160 – 29d = 180 – 39d

39d – 29d = 180 – 160

10d = 20

d = 2

Putting the value of d in (i), we get

a =

= = = 51

Hence, value of first installment is 51

Total numbers of trees = 25

Distance between trees = 5

Total distance between well and first tree is = 10 m

Now, gardener return back to well after watering each tree then,

Distance between well and second tree is = 10 + 10 + 5 = 25

Distance between well and third tree = 25 + 5 + 5 = 35

Distance between well and fourth tree = 35 + 5 + 5 = 45

Distance between well and last tree without return back = 10 + 24(5) = 10 + 120 = 130

So, common difference = 35 – 25 = 10

Now, Total distance covered by 24 trees,

We knew, S_n= [2a + (n – 1) d]

= [2(25) + (24 – 1) 10]

= 12 [50 + 23 * 10]

= 12 [50 + 230]

= 12 * 280

= 3360

Now, distance covered by 25 trees = 10 + 3360 = 3370

Total distance covered by gardener with return back = 3370 + 130 = 3500 m

Hence, total distance will covered by gardener is 3500.

Total amount for counting = Rs. 10710

In 1 min he counts =Rs. 180

In half an hour (30 min) he count = Rs.180×30 =Rs.5400

Amount left count after half an hour, S_n= Rs.10710 –Rs. 5400 = 5310

Now, in 31^st min he count 3 less than preceding minute = Rs.180 – Rs.3 = Rs.177

In 32^nd min he count 3 less than preceding minute = Rs. 177 –Rs. 3 =Rs. 174

Then, Arithmetic progression formed is 177, 174,…

Here, difference between minutes = 174 – 177 = -3

We know, sum of n terms, S_n= (n/2) [2a + (n – 1) d]

5310 = (n/2) [2(177) + (n – 1) -3]

5310 × 2 = n [354 – 3n + 3]

10620 = 354n – 3n² + 3n

10620 = 357n – 3n²

⇒ 3n² – 357n + 10620 = 0

On taking 3 common from the complete equation, we get,

⇒ 3 (n² – 119n + 3540) = 0

⇒ n² – 119n + 3540 = 0

Now, from factoring by splitting the middle term method, we get,

⇒ n² – 60n – 59n + 3540 = 0

⇒ n (n – 60) – 59 (n -60) = 0

⇒ (n – 59) (n – 60) = 0

Therefore, either n = 59 or n = 60

We will use 59 as these are minutes

So The total time to calculate whole amount is 59 + 30 = 89 min
= 1 hr 29 min

Given: A piece of equipment cost a certain factory Rs.600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and soon.

To find: The value at the end of 10 years if all percentages applying to the original cost.

Solution:

Cost of equipment = 6, 00, 000

In 1 year the value depreciate by 15%


⇒ The value of the equipment after first year= 6, 00, 000 * = 90, 000

In 2 year depreciate by 13.5%



⇒ The value of the equipment after second year= 6, 00, 000 * = 81, 000

In 3 year depreciate by 12%



⇒ The value of the equipment after third year= 6, 00, 000 * = 72, 000

Now A.P. is 90000, 81000, 72000,…

So, common difference = 81000 – 90000 = -9, 000

We have to find total depreciation for 10 years,


Since the value of depreciation is constant in any consecutive years i.e its value is -9000.



So to find the depreciation after ten years we will use the formula:


S_n= [2a + (n – 1) d]

S₁₀= [2(90, 000) + (10 – 1) (-9000)]

S₁₀= 5 [180000 + 9 * (-9000)]

S₁₀= 5 [180000 – 81000]

S₁₀= 5 [99000] = 495000

Hence,


cost of equipment at the end of 10 years = original cost – depreciation

= 6, 00, 000 – 4, 95, 000


= Rs. 1, 05, 000

Amount of money = 700

Total number of prize = 7

Each prize is Rs. 20 less than its preceding prize

Let, Prize of first prize = a

Prize of second prize = a – 20

Prize of third prize = a – 20 – 20 = a – 40

Here, A.P. is a, a – 20, a – 40,…

So, common difference, d = a – 20 – a = -20

We know, S_n = [2a + (n – 1) d]

700 = [2a + (7 – 1) -20]

= 2a + 6 (-20)

2a = 320

a = 160

So, value of first prize = 160

Value of second prize, 160 – 20 = 140

Value of third prize, 140 – 20 = 120

Value of fourth prize, 120 – 20 = 100

Value of fifth prize, 100 – 20 =80

Value of sixth prize, 80 – 20 = 60

And value of seventh prize, 60 – 20 = 40

Hence, value of prizes, Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, Rs 40.

Given, First term, a = 8

N^th term, a_n = 33 = l

And sum of n terms, S_n = [a + l]

123 = [8 + 33]

123 * 2 = n * 41

n = = = 6

We know,

a_n= a + (n – 1) d

33 = 8 + (6 – 1) d

33 – 8 = 5d

25 = 5d

d = 5

Hence, number of terms n = 6 and common difference, d = 5

Given, First term, a = 22

n^th term, a_n= -11 = l

and sum of n terms, S_n= 66

We know sum of n terms, S_n= [a + l]

66 = [22 – 11]

66 * 2 = n * 11

n = = 12

we know, a_n= a + (n – 1) d

-11 = 22 + (12 – 1) d

-11 – 22 = 11d

-33 = 11d

d = -3

Hence, number of terms, n = 12 and common difference, d = -3

Given, S_n = 4n – n²

Putting n = 1, 2

S₁ = 4(1) – (1)² = 3

S₂ = 4(2) – (2)² = 4

We know, a_n = S_n – S_{n – 1}

For first term, a₁ = S₁ – S₀ = 3 – 0 = 3

For term a₂ = S₂ – S_{1 =}4 – 3 = 1

Now fore term a₃,

S₂ = 4(2) – (2)² = 8 – 4 = 4

S₃ = 4(3) – (3)² = 12 – 9 = 3

a₃ = S₃ – S₂ = 3 – 4 = -1

Given,

First term, a₁ = 17

Last term, a_n = 350 = l

And difference, d = 9

We know,

a_n = a + (n – 1)d

350 = 17 + (n – 1) (9)

350 = 8 + 9n

342 = 9n

n = 38

We know sum of n terms, S_n = [a + l]

S₃₈ = [17 + 350]

= 19 * 367 = 6973

Hence, number of terms, n = 38

Sum of n terms, S_n = 6973

Given, First term, a = 2

Last term, a_n = l = 29

And, Sum of terms, S_n = [a + l]

155 = [2 + 29]

155 * 2 = n * 31

n = = = 10

We know, a_n = a + (n – 1) d

29 = 2 = (10 -1) d

29 – 2 = 9d

27 = 9d

d = 3

Hence, common difference, d = 3

Given: Sum of first ten terms that is, S₁₀ = -150 and Sum of next ten terms is -550.

To find: The A.P.

Solution: Let, first term = a

Last term = a_n

We know in an A.P,

a_n = a + (n – 1) d

a_n = a + (10 -1) d

a_n = a + 9d

We know, Sum of n terms

S_n = [a + l] ...... (1)

l and a_n means the last term.

Put the value of l in (1),

So,

S₁₀ = [a + a + 9d]

Substitute the known values,

-150 = 5 (2a + 9d)

-150 = 5 (2a + 9d)

-150= 10a + 45d

10a = -150 – 45d

a = ..... (2)

For next 10 terms, First term = a₁₁

Last term = a₂₀

For 11^th term,

a₁₁ = a + 10d

For 20^th term,

a₂₀ = a +19d

And S^'_n = [a + l] ...... (3)

Let the sum of next 10 terms to be S^'₁₀ ,


Now a will be equal to a₁₁ in this case and l will be equal to a + 19d


Put the value of a and l in the equation (3).


S’₁₀ = [a + 10d + a + 19d]

-550 = 5 [2a + 29d]

-550 = 10a + 145d

10a = -550 – 145d

a = .....(4)

From (2) and (4), we get

=

-150 – 45d = -550 – 145d

-45d + 145d = -550 + 150

100d = -400

d = -4

Put the value of d in (2), we get

a =

a=


We know A.P is of the form,

a_{1 ,} a_{2 ,}.......... ,a_n

Where a₁ is the first term of an A.P,

d is the common difference.

and

a_n = a+ ( n-1 )d

So,

a₁ = 3

a₂ = a + d = 3 + (-4) = -1

a₃ = a + 2d = 3 + 2(-4) = -5

Conclusion : The A.P. is 3, -2, -5,…

and a = 3, d = -4

Given, a = 10

S₁₄ = 1505

S₁₄ = [2a + (n – 1) d]

1505 = [2(10) + (14 – 1) d]

1505 = 7 [20 + 13d]

215 = 20 + 13d

13d = 215 – 20

13d = 195

d = 15

Now, a₂₅ = a + (n – 1) d

= 10 + (25 – 1) 15

= 10 + 24 * 15

= 10 + 360 = 370

Hence, value of 25^th term is 370

Given that

S_n = 5n² + 3n

Put n = 1

S₁= T₁= 5 + 3 = 8

Put n = 2

S₂= 5(2)² + 3 + 2 = 26

T₂= S₂ – S₁ = 26 – 8 = 18

S₃= 5(3)² + 3 + 3 = 54

T₃= S₃ – S₂ = 54 – 26

= 28

Therefore, first term, a = 8 and common difference = 18 – 8 = 10

T_m = a + (m – 1) d

168 = 8 + (m – 1) 10

168 = 8 + 10m – 10

170 = 10m

m = 17

T₂₀ = 8 + (20 – 1) 10

= 8 + 19 * 10 = 198

Given that: S_q= 63q – 3q²

Put q = 1

S₁ = T₁ = 63 – 3 = 60

Put q = 2

S₂ = 63 * 2 – 3 * (2)²

= 126 – 12 = 114

T₂ = S₂ – S₁ = 114 – 60 = 54

Put q = 3

S₃ = 63 * 3 – 3(3)²

= 189 – 27 = 162

T₃ = S₃ – S₂

= 162 – 114 = 48

Therefore, first term of this A.P. is 60 and Common difference is 54 - 60 = -6

T_p = a + (p – 1) d

-60 = 60 + (p – 1) (-6)

-120 = (p – 1) (-6)

(p – 1) = 20

p = 21

Now 11^th term of this A.P

T₁₁ = 60 + (11 – 1) (-6)

= 60 – 60 = 0

S_m = 4m² – m

Put m = 1