Progressions

Given that price for 1^st km is Rs. 20 so that is first term ‘a’ of the series.

And also we have been given that for every km after 1^st km is Rs. 8

So, for 2 km it will be 20 + 8 = 28

Similarly, 3 km = 28 + 8 + 8 = 36

In the same way it adds on for another km

If we put taxi fares in series for each km. It will look like this,

20, 28, 36, 44, 52 ……

On seeing the above series we can conclude that it is in A.P.

∵ They have same common difference between two consequent numbers (d = 8).

Let V be the initial volume of air that is present in the cylinder.

They have said that a vacuum removes of the air.

Let initial volume (v) = 1 litre

After first removal of air, the remaining will be V₁

V₁ = 1- ⇒

Similarly,

In second attempt it removes of the air remaining.

That means they remove = of air.

So, V₂ = - ⇒

After third attempt V₃ = of air will be remaining.

So, the series will be look like

1, ,

The above series is NOT looking like A.P.

∵ Their common difference is not same.

Given that cost for digging 1^st meter is Rs. 150 so that is first term ‘a’ of the series.

And also we have been given that for every meter after 1^st meter digging is Rs. 50

So, for 2 meter it will be 150 + 50 = 200

Similarly, 3 meter = 150 + 50 + 50 = 250

In the same way it adds on for every another meter digging

If we put cost of digging in series for each meter. It will look like this,

150, 200, 250, 300, ……

On seeing the above series we can conclude that it is in A.P.

∵ They have same common difference between two consequent numbers (d = 50).

Given that initially 10000 is in the account, so it is first term ‘a’.

Every year 8% compound interest will be paid for that.

That means they will pay 8% of 10000 for the first year.

⇒ will be paid.

After first year 800 will be added to account.

So the amount will be become 10800.

Since it is compound interest every year principal amount will be updated.

So, they will pay 8% of 10800 in the second year.

⇒ will be paid.

After first year 864 will be added to account.

So the amount will be become 11664.

So, the series will be 10000, 10800, 11664 …..

Since the common difference is not same.

The series is not in A.P

Given that first term (a) = 10

Common difference (d) = 10

The first four terms of the A.P is given by

a, a+d, a+2d, a+3d

⇒ 10, 10+10, 10+2×10, 10+3×10

⇒ 10, 20, 30, 40

Given that first term (a) = -2

Common difference (d) = 0

Series will be same as First Term. Since common difference (d) in the series is 0.

The first four terms in the series is -2,-2,-2,-2.

Given that first term (a) = 4

Common difference (d) = -3

The first four terms of the A.P is given by

a, a+d, a+2d, a+3d

⇒ 4, 4+ (-3), 4+ 2(-3), 4+ 3(-3)

⇒ 4, 1, -2, -5 .

Given that first term (a) = -1

Common difference (d) =

The first four terms of the A.P is given by

a, a+d, a+2d, a+3d

⇒ -1, -1+, -1+2×, -1+3×

⇒ -1, , 0,

Given that first term (a) = -1.25

Common difference (d) = -0.25

The first four terms of the A.P is given by

a, a+d, a+2d, a+3d

⇒ -1.25, -1.25+ (-0.25), -1.25+2 × (-0.25), -1.25+3 ×(-0.25)

⇒ -1.25, -1.5, -1.75, -2.

In the above series first term (a) is 3

Common difference (d) is given by 2^nd term - 1^st term

⇒ d = 1 -3

⇒ d = -2

In the above series first term (a) is -5

Common difference (d) is given by 2^nd term - 1^st term

⇒ d = -1 – (-5)

⇒ d = 4

In the above series first term (a) is

Common difference (d) is given by 2^nd term - 1^st term

⇒ d = –

⇒ d =

In the above series first term (a) is 0.6

Common difference (d) is given by 2^nd term - 1^st term

⇒ d = 1.7 -0.6

⇒ d = 1.1

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = 4 – 2 = 2

D₂ = Third term - Second term = 8 – 4 = 4

Since common difference is not equal the above series is not in AP

For a series to be in AP, the common difference (d) should be

Equal.

d₁ = second term – first term = – 2 = 0.5

d₂ = Third term - Second term = 3 – = 0.5

Since common difference is same the above series is in AP.

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by = a + (5-1)d = a + 4d = 2 + 4(0.5) = 4

6^th term is a + (6-1)d = a + 5d = 2 + 5(0.5) = 4.5

7^th term is a + (7-1)d = a + 6d = 2 + 6(0.5) = 5.

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = -3.2 – (-1.2) = -2

D₂ = Third term - Second term = -5.2 – (-3.2) = -2

Since common difference is equal the above series is in AP

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by

= a + (5-1)d = a + 4d = -1.2 + 4(-2) = -9.2

6^th term is a + (6-1)d = a + 5d = -1.2 + 5(-2) = -11.2

7^th term is a + (7-1)d = a + 6d = -1.2 + 6(-2) = -13.2

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = -6 – (-10) = 4

D₂ = Third term - Second term = -2 – (6) = 4

Since common difference is equal the above series is in AP

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by

= a + (5-1)d = a + 4d = -10 + 4(4) = 6

6^th term is a + (6-1)d = a + 5d = -10 + 5(4) = 10

7^th term is a + (7-1)d = a + 6d = -10 + 6(4) = 14

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = 3 + √ 2 - 3 = √ 2

D₂ = Third term - Second term = 3 + 2√ 2 - 3 + √ 2 = √2

Since common difference is equal the above series is in AP

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by

= a + (5-1)d = a + 4d = 3 + 4(√2) = 3 + 4√2

6^th term is a + (6-1)d = a + 5d = -10 + 5(√2) = 3 + 5√2

7^th term is a + (7-1)d = a + 6d = -10 + 6(√2) = 3 + 6√2.

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = 0.22 -0.2 = 0.02

D₂ = Third term - Second term = 0.222 – 0.22 = 0.002

Since common difference is not equal the above series is not in AP

For a series to be in AP, the common difference (d) should be

Equal.

d₁ = second term – first term = -4 – 0 = -4

d₂ = Third term - Second term = -8 – (-4) = -4

Since common difference is same the above series is in AP.

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by = a + (5-1)d = a + 4d = 0 + 4(-4) = -16

6^th term is a + (6-1)d = a + 5d = 0 + 5(-4) = -20

7^th term is a + (7-1)d = a + 6d = 0 + 6(-4) = -24.

For a series to be in AP, the common difference (d) should be

Equal.

d₁ = second term – first term = – = 0

d₂ = Third term - Second term = – = 0

Since common difference is same the above series is in AP.

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by = a + (5-1)d = a + 4d = + 4(0) =

6^th term is a + (6-1)d = a + 5d = + 5(0) =

7^th term is a + (7-1)d = a + 6d = + 6(0) = .

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = 3 - 1 = 2

D₂ = Third term - Second term = 9 – 3 = 7

Since common difference is not equal the above series is not in AP

For a series to be in AP, the common difference (d) should be

Equal.

d₁ = second term – first term = 2a – a = a

d₂ = Third term - Second term = 3a – 2a = a

Since common difference is same the above series is in AP.

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by = a + (5-1)d = a + 4d = a + 4(a) = 5a

6^th term is a + (6-1)d = a + 5d = a + 5(a) = 6a

7^th term is a + (7-1)d = a + 6d = a + 6(a) = 7a.

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = a² - a = a² - a

D₂ = Third term - Second term = a³ - a² = a³ - a²

Since common difference is not equal the above series is not in AP

He above series can be re written as

√2, √8, √18, √32

⇒ √2, √(2² × 2), √(3² × 2), √(4² × 2)

√2, 2√2, 3√2, 4√2,…..

For a series to be in AP, the common difference (d) should be

Equal.

d₁ = second term – first term = 2√2 – √2 = √2

d₂ = Third term - Second term = 3√2 – 2√2 =

Since common difference is same the above series is in AP.

The next three terms will be the 5^th, 6^th, 7^th.

5^th term will be given by

a + (5-1)d = a + 4d = √2+ 4(√2) = 5√2 = √50

6^th term is a + (6-1)d = a + 5d = √2 + 5(√2) = 6√2 = √72

7^th term is a + (7-1)d = a + 6d = √2 + 6(√2) = 7√2 = √98.

For a series to be in AP, the common difference (d) should be

Equal.

D₁ = second term – first term = √3 - √6 = √3 (1 - √2)

D₂ = Third term - Second term = √6 - √9 = √3 (√2 - 3)

Since common difference is not equal the above series is not in AP

I. Given first term (a) = 7

Common difference (d) = 3

n = 8 ,

a_n = ?

we know that a_n = a + (n-1)d

So, we have to find a₈

a₈ = a + (8-1)d = 7 + 7 × 3 = 28

II.

Given first term (a) = -18

Common difference (d) = ?

n = 10, a_n = 0

a₁₀ = a + (10-1)d

0 = -18 + 9×d

d = 2

III. Given

Common difference (d) = -3

n = 18, a_n = -5

first term (a) = ?

a₈ = a + (18-1)d

-5 = a + 17×(-3)

a = -46

IV. Given first term (a) = -18.9

Common difference (d) = 2.5

n = ?

a_n = 3.6

we know that a_n = a + (n-1)d

3.6 = -18.9 + (n-1)2.5

22.5 = (n-1)2.5

9 = n-1

n = 10

V. Given first term (a) = 3.5

Common difference (d) = 0

n = 105 ,

a_n = ?

we know that a_n = a + (n-1)d

a_n = 3.5 + (105-1)0

a_n = 3.5

Given

First term (a) = 10

Common difference (d) = 2^nd – 1^st terms

d = 7 – 10 = -3

we need to find the 30^th term

we know that a_n = a + (n-1)d

so n = 30

a₃₀ = 10 + (30-1) (-3)

a₃₀ = -77

Given

First term (a) = -3

Common difference (d) = 2^nd – 1^st terms

d = -3 – = 2.5

we need to find the 11^th term

we know that a_n = a + (n-1)d

so here, n = 11

a₁₁ = -3 + (11-1) (2.5)

a₁₁ = 22

Given that a₁ = 2 & a₃ = 26

That means a₃ = a₁ + (3-1)d

⇒ 26 = 2 + 2d

⇒ 24 = 2d

⇒ d = 12

∴ a₂ = a₁ + (2-1)d

⇒ a₂ = 2 + 12 = 14

Given that a₂ = 13 & a₄ = 3

a₂ = a₁ + (2-1)d

13 = a₁ + d ⇒ 1

a₄ = a₁ + (4-1)d

3 = a₁ + 3d ⇒ 2

By subtracting the both equations

13 – 3 = (a₁ + d ) - (a₁ + 3d)

10 = -2d

d = -5

By substituting in 1 equations we get a₁

13 = a₁ + (-5)

18 = a₁

a₃ = a₁ + (3-1)d

= 18 + 2 × (-5)

= 8.

Given that a₁ = 5 & a₄ = 9.5

a₄ = a₁ + (4-1)d

9.5 = a₁ + 3d

Substituting a₁ in above equation

9.5 = 5 + 3d

4.5 = 3d

d = 1.5

now, a₂ = a₁ + (2-1)d

= 5 + 1.5

= 6.5

a₃ = a₁ + (3-1)d

= 5 + 2× 1.5

= 8

Given that a₁ = 4

And a₆ = 6

a₆ = a₁ + (6-1)d

6 = 4 + 5d

D = 0.4

To find a₂:

a₂ = a₁ + (2-1)d

= 4 + 0.4

= 4.4

To find a₃:

a₃ = a₁ + (3-1)d

= 4 + 2×0.4

= 4.8

To find a₄:

a₄ = a₁ + (4-1)d

= 4 + 3×0.4

= 5.2

To find a₅:

a₅ = a₁ + (5-1)d

= 4 + 4×0.4

= 5.6

Given that a₂ = 38 & a₆ = -22

a₂ = a₁ + (2-1)d

38 = a₁ + d ⇒ 1

a₆ = a₁ + (6-1)d

-22 = a₁ + 5d ⇒ 2

By subtracting the both equations

38 – (-22) = (a₁ + d ) - (a₁ + 5d)

60 = -4d

d = -15

By substituting in 1 equations we get a₁

38 = a₁ + (-15)

53 = a₁

To find a₃:

a₃ = a₁ + (3-1)d

= 53 + 2×(-15)

= 23

To find a₄:

a₄ = a₁ + (4-1)d

= 53 + 3×(-15)

= 8

To find a₅:

a₅ = a₁ + (5-1)d

= 53 + 4×(-15)

= -7

Given first term (a) = 3

Common difference (d) = n₂-n₁ = 8-3 = 5

a_n = 78

⇒ a_n = a + (n-1)d

78 = 3 +(n-1) 5

75 = (n-1) 5

(n-1) = 15

n = 16

Therefore 78 is 16^th term of the series.

Given first term (a) = 7

Common difference (d) = n₂-n₁ = 13 - 7 = 6

a_n = 205

⇒ a_n = a + (n-1)d

205 = 7 +(n-1) 6

198 = (n-1) 6

(n-1) = 33

n = 34

Therefore 205 is 34^th term of the series.

Given first term (a) = 18

Common difference (d) = n₂-n₁ = 15 - 18 = 2.5

a_n = -47

⇒ a_n = a + (n-1)d

-47 = 18 +(n-1) 2.5

65 = (n-1) 2.5

(n-1) = 26

n = 27

Therefore -47 is 27^th term of the series

Given first term (a) = 18

Common difference (d) = n₂-n₁ = 8 – 11 = -3

Let us consider -150 is a term in above series.

Then, a_n = -150

a_n = a + (n-1)d

-150 = 11 + (n-1)(-3)

-161 = (n-1)(-3)

n-1 =

n =

Since n is not an integer the given number is not in the series.

Given

11^thterm = 38

⇒ a + (11-1)d = 38 → 1

16^th term = 73

⇒ a + (16-1)d = 73 → 2

By subtracting the both equations we will get ‘d’

(a +10d) – (a+15d) = 38 -73

-5d = -35

d = 7

By substituting “d” in equation 1

a +10d = 38

a + 10×7 = 38

a = -32

a₃₁ = a + (31-1) d

= -32 + 30×7

= 178

Given

a_n = 0, n = ?

3^rd term = 4

⇒ a + (3-1)d = 4 → 1

9^th term = 73

⇒ a + (9-1)d = -8 → 2

By subtracting the both equations we will get ‘d’

(a +2d) – (a+8d) = 4 – (-8)

-6d = 12

d = -2

By substituting “d” in equation 1

a +2d = 4

a + (-2)2 = 4

a = 8

a_n = a+(n-1)d

0 = 8+(n-1)(-2)

-8 = (n-1)(-2)

4 = n-1

n = 5

∴ 0 is the 5^th term in the series.

Sum of the first n term is given by (s_n) = [2a + (n-1) d]

The sum of first 7 terms S₇ = 49 → 7²

S₇ = [2×a+(7-1)d]

49 = [2a+6d]

49 = 7[a+3d]

7 = a + 3d ⇒ 1

The sum of first 17 terms S₁₇ = 289 → 17²

S₁₇ = [2×a+(17-1)d]

289 = [2a+16d]

289 = 17[a+8d]

17 = a + 8d ⇒ 2

By subtracting both the equations

17 – 7 = a + 8d – (a+3d)

10 = 5d

d =2

∴ common difference (d) = 2

Substituting‘d’ in equation 1

7 = a + 3d

7 = a + 3(2)

a = 1

∴ Sum of the first n term is given by (s_n) = [2a + (n-1) d]

= [2×1 + (n-1)2]

= n[ 1+ (n-1)]

= n²

Given that the 2 AP series has same common difference (d)

Let x, x+d, x+2d,…. X+(n-1)d &

y, y+d, y+2d,…..y+(n-1)d are the two A.P series

given that the difference between their 100th terms is 100

that is,

x+(100-1)d – (y+(100-1)d) = 100

x-y = 100

The difference between their 1000^th term is

= x+(1000-1)d –(y+(1000-1)d)

= x-y

We know that value from previous equation

= 100

∴ The difference between their 1000th terms is 100

The series of three-digit numbers divisible 7 is,

105, 112, 119,…………. 994

So here first term (a) is 105

Since it is divisible by 7 the common difference will be 7

Let a_n = 994

a_n = a + (n-1) d

994 = 105 + (n-1) 7

889 = (n-1)7

127 = n-1

n = 128

The multiples of 4 lie between 10 & 250 is

12, 16, 20,.....248

So, first term (a) = 12

Since it is multiples if 4 the common difference will be 4

a_n = 248

a_n = a + (n-1) d

248 = 12 + (n-1) 4

236 = 4(n-1)

59 = n-1

n = 60.

We have to find the value of n, so that the nth term in the both series should be equal

63, 65, 67, ….

In the above series

First term (a₁) = 63,

Common difference (d₁) = 2

3, 10, 17, ….

First term (a₂) = 3,

Common difference (d₂) = 7

Since value of n^th term is equal we equate a_n of both the series.

a_n1 = a_n2

a₁ + (n-1)d₁ = a₂ + (n-1)d₂

63 + (n-1)2 = 3 + (n-1)7

61 +2n = 7n – 4

5n = 65

n = 13.

Given

a₃ = 16; a₇ = a₅ + 12

a + (7-1) d = a + (5-1) d +12

2d = 12

d = 12

Common difference (d) = 12

Substituting value of d in a₃

a₃ = 16

a + (3-1) d = 16

a + 2 (6) = 16

a = 4

∴ The series will be a, a+d, a+2d, a+3d,….

4, 10, 16, 22,…

Given:

First term (a) = 3

Common difference (d) = n₂ – n₁ = 8 - 3 = 5

a_n = a + (n-1)d

253 = 3 + (n - 1) × 5

253 - 3 = 5(n - 1)

(n - 1) = 250/5 = 50

n = 51

Given,

a₄ + a₈ = 24;

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

2a + 10d = 24 ⇒ a+5d = 12 → 1

We have been also given that

a₆ + a₁₀ = 44

(a + 5d) + (a + 9d) = 44

2a + 14d = 44 ⇒ a +7d = 22 → 2

By subtracting two equations

(a + 5d) - (a + 7d) = 12 – 22

-2d = -10

d = 5

Substitute d in any one of the equation.

a+5d = 12

a + 5×5 = 12

a = 2

The first 3 terms of the series is given by a, a+d, a+2d

⇒ 2, 2+5, 2+2×5

⇒ 2, 7, 12

Initial salary of Subba Rao is Rs. 5000

That is first term (a) = 5000

He receives Rs.200 every year as increment

So, common difference (d) = 200

The year when he reaches Rs. 7000

So, a_n = 7000

a_n = a +(n-1)d

7000 = 5000 + (n-1) 200

2000 = 200(n-1)

10 = n-1

n = 11

So, it takes him 11 years to reach Rs. 7000

In the above series

First term (a) = 2

Common different (d) = n₂ – n₁ = 7 – 2 = 5

No. Of terms (n) = 10

The sum of the series is given by S_n = [ 2a + (n-1) d]

= [ 2×2 + (10-1)5]

= 245

In the above series

First term (a) = -37

Common different (d) = n₂ – n₁ = -33 – (-37) = 4

No. Of terms (n) = 12

The sum of the series is given by S_n = [2a + (n-1) d]

= [2×(-37) + (12-1)4]

= -180

In the above series

First term (a) = 0.6

Common different (d) = n₂ – n₁ = 1.7 - 0.6 = 1.1

No. Of terms (n) = 100

The sum of the series is given by S_n = [2a + (n-1) d]

= [2×(0.6) + (100-1)1.1]

= 5505

In the above series

First term (a) =

Common different (d) = n₂ – n₁ = - =

No. Of terms (n) = 11

The sum of the series is given by S_n = [2a + (n-1) d]

= [2×() + (11-1))]

= [ + ]

=

We have given a formulae to find sum of the series when first and last numbers are given.

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

Where n is the last term in the series

d is the common difference

a is the first term.

a_n = 84, a = 7

d = n₂ – n₁ = 10 - 7 = 3.5

a_n = a + (n-1)d

84 = 7 + (n-1)3.5

77 = 3.5(n-1)

22 = n-1

n = 23

The sum of the series is S_n = [2a + (n-1) d]

= [ 2×7 + (23-1)3.5]

= [ 14 + (22× 3.5)

The sum of the series given by

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

Where n is the last term in the series

d is the common difference

a is the first term.

a_n = 10, a = 34

d = n₂ – n₁ = 32 - 34 = 2

a_n = a + (n-1)d

10 = 34 + (n-1)(-2)

-24 = -2(n-1)

12 = n-1

n = 13

The sum of the series is S_n = [2a + (n-1) d]

= [ 2×34 + (13-1)(-2)]

= [ 44 ] ⇒ 286

The sum of the series given by

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

Where n is the last term in the series

d is the common difference

a is the first term.

a_n = -230, a = -5

d = n₂ – n₁ = (-8) – (-5) = -3

n^th term a_n = a + (n-1)d

-230 = (-5) + (n-1)(-3)

-225 = -3(n-1)

75 = n-1

n = 76

The sum of the series is S_n = [2a + (n-1) d]

= [ 2×(-5) + (76-1)(-3)]

= 38 [ -235 ]

⇒ 8930

d is the common difference = 3

a is the first term = 5

a_n = 50

a_n = a + (n-1)d

50 = 5 + (n-1)(3)

45 = 3(n-1)

15 = n-1

n = 16

The sum of the series is S_n = [2a + (n-1) d]

= [ 2×5 + (16-1)(3)]

= 8 [ 55 ]

⇒ 440

d is the common difference

a is the first term = 7

a₁₃ = 35

a₁₃ = a + (13-1)d

35 = 7 + (12)(d)

28 = 12 d

d =

The sum of the series is S_n = [2a + (n-1) d]

S₁₃ = [2×7 + (13-1) ]

= [42]

= 273

d is the common difference = 3

a is the first term

a₁₂ = 37

a₁₂ = a + (12-1)d

37 = a + (11)3

a = 4

The sum of the series is S_n = [2a + (n-1) d]

S₁₂ = [2×4 + (12-1)×3]

= 6 [41]

= 246

d is the common difference of the series

a is the first term of the series

a₃ = 15(Given in the Question)

∴S₁₀ =10/2 [2×a + (10-1)×d]

Solving (i) and (ii) we get:-

2a + 9d= 25
a + 2d= 15 (Multiplying the Equation by 2 and changing the sign)

⇒ 2a + 9d = 25
2a + 4d = 30
- - -
--------------------
⇒ 5d = (-5)
---------------------

∴ d= (-1)

After putting value of d in any of the above equation we get a=17

∴ a ₁₀ = 17 + (10-1)(-1)
= 8

First term (a) = 2

Common difference (d) = 8

S_n = 90

The sum of the series is S_n = [2a + (n-1) d]

90 = [2×2 + (n-1)8]

90 = 4n² – 2n

2n² – n - 45 = 0

By solving the above quadratic equation we will get

2n² -10n + 9n – 45 = 0

2n(n-5) + 9(n-5) = 0

n = 5

a_n = a + (n-1)d

a₅ = 2 + (5-1)8 = 34

Given, series in A.P.

n^th term in the AP a_n = 4,

Common difference (d) = 2

Sum of the series is s_n = -14

a_n = a +(n-1)d

4 = a + (n-1)2

6 = a + 2n

a = 6 – 2n ⇒ 1

Sum of the series (s_n) = [2a + (n-1) d]

-14 = [ 2×(6-2n) + (n-1)2]

-14 = n[5-n]

n² -5n -14 = 0

n² -7n + 2n -14 = 0

n(n-7) + 2(n-7) = 0

(n-7)(n+2) = 0

n-7 = 0 ⇒ n = 7 (or) n+2 = 0 ⇒ n = -2

since n should be a positive integer

we take n = 7

substitute in equation 1

a = 6 – 2n

a = 6 – 2(7)

a = -8

∴ first term is -8

Given

Last term (l) = 28

Sum of the series s_n = 144

The number of terms in the series (n) = 9

We have another formula to find sum of the series, when common difference is not given

Sum of the series s_n = [a +l]

Where a is the first term

144 = [a + 28]

32 = a + 28

4 = a

∴ first term (a) = 4.

Given first term (a) = 17, last term (a_n) = 350

Common difference (d) = 9

a_n = a + (n-1) d

350 = 17 + (n-1) 9

333 = 9(n-1)

37 = n-1

n = 38.

Sum of the series (s_n) = [2a + (n-1) d]

= ( 2×17 +(38-1)9)

= 19 × 367

= 6973

Second term (a₂) = 14; Third term (a₃) = 18

a₂ = a+(2-1)d

14 = a+d ⇒ 1

a₃ = a+(3-1)d

18 = a+2d ⇒ 2

By subtracting both the equations

14 -18 = a+d –(a+2d)

-4 = -d

d = 4

By substituting d in the equation 1

a + d = 14

a + 4 = 14

a = 10

The sum of first 51 terms is given by (s_n) = [2a + (n-1) d]

S₅₁ = [2×10 + (51-1) 4]

= 5610

Given a_n = 3+4n

So, a₁ = 3+4(1) = 7

a₂ = 3+4(2) = 11

a₃ = 3+4(3) = 15

as of now a₁, a₂, a₃ have the same common difference (d) = 4

so given one is A.P

The sum of first 15 terms is given by (s_n) = [2a + (n-1) d]

= [ 2×7 +(15-1) 4]

= 525

Given a_n = 9-5n

So, a₁ = 9 – 5(1) = 4

a₂ = 9 -5(2) = -1

a₃ = 9 – 5(3) = -6

as of now a₁, a₂, a₃ have the same common difference

so given one is A.P

The sum of first 15 terms is given by (s_n) = [2a + (n-1) d]

= [ 2×4 +(15-1) (-5)]

= -465

The sum of first n terms is given by 4n –n²

So the first term is

a₁ = 4(1) – 1² = 3

the sum of first two term S₂ = 4(2) – 2² = 4

The difference between S₂ & S₁ gives us 2^nd term

a₂ = S₂ - S₁ = 4-3 = 1

the sum of first 3 term S₃ = 4(3) – 3² = 3

The difference between S₂ & S₃ gives us 3^rd term

a₃ = S₃ – S₂ = 3-4 = -1

From this we can find common difference = a₂ – a₁ = 1-3 = -2

So the 10^th term is = a + (10-1)d

= 3 + 9×(-2) = -15

n^th term is = a + (n-1)d

= 3 + (n-1)×(-2) = 5 -2n

The first 40 positive integers divisible by 6 are,

6, 12,…..240

Here we have n = 40 terms

First term (a) = 6;

Common difference is 6 since they are multiple of 6

The sum of 6 terms is given by (s_n) = [2a + (n-1) d]

S₆ = [2×6 + (40-1)6]

= 4920

Let p be the first prize awarded to students

Since there are 7 cash prizes, n = 7

So a₁ = x

Then the preceding prizes will be Rs. 20 lesser

So the series will be

p, p-20, p-40,…. p-120.

Since preceding prizes will be Rs. 20 lesser the common difference is 20

The sum of all prizes are given by (s₇) = 700

[2×(p)+(7-1)(-20)] = 700

(2p - 120) = 200

2p = 320

P = 160

So, the each prize cost will be

160, 140, 120, 100, 80, 60, 40

Given that each section of each class will plant a trees.

e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections.

So the trees planted by each class will be like this

1 × 3, 2 × 3, 3 × 3,…. 12 × 3

⇒ 3, 6, 9,…..36

In the above series

First term (a) = 3

Common difference (d) = a₂ – a₁ = 6-3 = 3

The sum of 12 terms is given by (s_n) = [2a + (n-1) d]

S₁₂ = [2×3 + (12-1) 3]

= 6( 6+33) = 234

So, totally 234 trees will be planted by students in the school.

Firstly, we have find circumference of each semi circle to find the total length

The circumference for semicircles is given by πr

where r is radius of each semi circle

the radii of circles is given by 0.5, 1.0,….6.5

l₁ = πr₁ = = 1.57cm

l₁ = πr₂ = = 3.14cm

Likewise we calculate all the other circumferences

1.57, 3.14,…… 20.41

The first term (a) = 1.57

Common difference (d) = a₂ – a₁ = 3.14 -1.57 = 1.57

The sum of 13 terms is given by (s_n) = [2a + (n-1) d]

S₁₃ = [2×1.57 + (13-1) 1.57]

= 6.5×21.98

= 142.87 = 143 (app.)

So, the total length of the spiral made is 143 cm

Let us consider the row number as the term in the series & the no. Of logs present in the each row as the values

20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on.

Considering bottom row as first term so (a) = 20

Second term a₂ = 19

Common difference (d) = a₂ – a = 19-20 = -1

Given that 200 logs are stacked

So, s_n = 200

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

Here n is the no. Of rows are used to pile the log.

S_n = [2×20+ (n-1)(-1)]

200 = (41-n)

400 = 41n – n²

n² -41n+400 = 0

n² – 25n -16n + 400 = 0

n(n-25) -16(n-25) = 0

(n-16)(n-25) = 0

∴ n = 16 or 25

Let us take n = 16, because 25 is higher value and that is not feasible solution for the problem

a₁₆ = a+(16-1)d = 20+15×(-1) = 5

There will be 5 logs in top row.

The distance between the first ball and the bucket is 5m

So the competitor runs 10m (forward + backward)

The second ball the placed 3m ahead of first ball

For second ball he will 16.

Likewise, the distance covered by each ball is

3^rd = 22, 4^th = 28,….. 10^th = 64

Common difference (d) = 3^rd - 2^nd = 22-16 = 6

Here there are 10 balls to be picked so n = 10

Last term a₁₀ = 64

The sum of all the distance is given by S_n = (2a +(n-1)d)

=

= 5×(74) = 370

So the total distance covered by the competitor to pick all the is 370

This is not likely be to in G.P

Initially the salary of sharmila is 5,00,000

They have said that she will get 10% every year

For second year her salary will be increased by 10% of 5,00,000.

So her second year salary will be 5,50,000

For third year her salary will be increased by 10% of 5,00,000.

So her second year salary will be 6,00,000

You can see the common ratios will be same r =

= =

r = = =

This is not in GP

Since they have given common difference -2

They have said that successive steps need 2 bricks less than the previous.

It has been already given that the perimeter of any figure is drawn by using the midpoint of the same figure in GP.

If the perimeter of that triangle is “a” then the perimeter of the triangles formed by using midpoint is , ….,

Perimeter of a triangle is = s₁ + s₂ + s₃ = 24 + 24 + 24 = 72

In the same way the perimeter of given triangle is 72

The perimeter of the triangles formed by using midpoint will be 36, 18, 9

We can see that their common ratios is same

R₁ =

R₂ =

First term (a) = 4; common ratio (r) = 3

The three terms are given by

⇒ a, ar, ar².

= 4, 4×3, 4×3²

= 4, 12, 24

First term (a) = 4; common ratio (r) = 3

The three terms are given by

⇒ a, ar, ar².

= √5, √5×, √5×( ² = √5,

First term (a) = 81; common ratio (r) =

The three terms are given by

⇒ a, ar, ar².

= 81, 81×, 81ײ = 81, -27, 9

First term (a) = ; common ratio (r) = 2

The three terms are given by

⇒ a, ar, ar².

= , ×2, ײ

=

Here first term is (a) = 4

r₁ =

r₂ =

Common ratios are equal so the given series is in GP.

a₄ = a r³ = 4×2³ = 4×8 = 32

a₅ = a r⁴ = 4×2⁴ = 4×16 = 64

a₆ = a r⁵ = 4×2⁵ = 4×32 = 128

Here first term is (a) =

r₁ =

r₂ =

Common ratios are equal so the given series is in GP.

Here first term is (a) = 5

The common ratios are not equal so this is not in GP

Here first term is (a) = -2

r₁ =

r₂ =

Common ratios are equal so the given series is in GP.

a₄ = a r³ = -2×3³ = -2 × 27 = -54

a₅ = a r⁴ = -2×3⁴ = -2 × 81 = -162

a₆ = a r⁶ = -2×3⁵ = -2 × 243 = -486

Here first term is (a) =

r₁ =

r₂ =

The common ratios are not equal so this is not in GP

Here first term is (a) = 3

r₁ =

r₂ =

Common ratios are equal so the given series is in GP.

a₄ = a r³ = 3×(-3)³ = 3 × -27 = -81

a₅ = a r⁴ = 3×(-3)⁴ = 3 × 81 = 243

a₆ = a r⁵ = 3×(-3)⁵ = 3 × -243 = -729

Here first term is (a) = x

r₁ =

r₂ =

Common ratios are equal so the given series is in GP.

a₄ = a r³ = x(³ =

a₅ = a r⁴ = x(⁴ =

a₆ = a r⁵ = x(⁵ =

First term is (a) =

r₁ = = -2√2

r₂ = -2√2

Common ratios are equal so the given series is in GP.

a₄ = a r³ = (-2√2)³ = -16

a₅ = a r⁴ = (-2√2)⁴ = 32√2

a₆ = a r⁵ = (-2√2)⁵ = -128

Here first term is (a) = 0.4

r₁ =

r₂ =

Common ratios are equal so the given series is in GP.

a₄ = a r³ = 0.4×(0.1)³ = 0.0004

a₅ = a r⁴ = 0.4×(0.1)⁴ = 0.00004

a₆ = a r⁵ = 0.4×(0.1)⁵ = 0.000004

In a GP common ratio are equal

x, x+2, x+6…..

r =

(x+2)² = x(x+6)

X² + 4 + 4x = x² +6x

2x = 4

X = 2

First term (a) = 3

Common ratio (r) = =

n^th term in the GP (a_n) = ar^n-1

a_n = 3^n-1

First term (a) = 2

Common ratio (r) = =

n^th term in the GP (a_n) = ar^n-1

a_n = 2(^n-1

First term (a) = -1

Common ratio (r) = =

n^th term in the GP (a_n) = ar^n-1

a_n = 3(^n-1 = 3ⁿ

First term (a) = 5

Common ratio (r) = =

n^th term in the GP (a_n) = ar^n-1

a_n = (5) (^n-1 = (^n-1

First term (a) = 5

Common ratio (r) = =

n^th term in the GP (a_n) = ar^n-1

a_n = 5(5^n-1 = 5ⁿ

10^th term in the GP (a₁₀) = ar^10-1

a₁₀ = 5×5⁹ = 5¹⁰

First term (a) = 9

Common ratio (r) =

7^th term in the GP (a₇) = ar^7-1 = ar⁶

=

First term (a) = -12

Common ratio (r) =

6^th term in the GP (a₆) = ar^6-1 = ar⁵

First term (a) = 2

Common ratio (r)

n^th term in the GP (a_n) = ar^n-1

512 = 2×4^n-1

256 = 4^n-1

4⁴ = 4^n-1

Equating powers on both sides

n-1 = 4 = 5

First term (a) = √3

Common ratio (r)

n^th term in the GP (a_n) = ar^n-1

729 = √3×(√3)^n-1

3⁶ = √3ⁿ

Equating powers on both sides

= 6 ⇒ n = 12.

First term (a) =

Common ratio (r)

n^th term in the GP (a_n) = ar^n-1

= ^n-1 = )ⁿ

()⁷ = )ⁿ

Equating powers on both sides

n = 7

Given,

common ratio (r) is 2.

8^th term is 192

⇒ ar^8-1 = 192

a×2⁷ = 192

a = =

12^th term of a GP = ar^12-1 = ar¹¹

= ×2¹¹ = 3072.

Given,

4^th term of the series =

ar³ =

a = ⇒ 1

7^th term of a series =

ar⁶ = ⇒ 2

Substituting ‘a’ in 1

× r⁶ =

r³ =

r³ = (³

equating power and base we will get

r =

The geometric series is a, ar, ar², ar³

,²,…..

, , , ,……

Given series 162, 54, 18 ….

First term (a₁) = 162

Common ratio (r₁) = = =

, …..

First term (a₂) =

Common ratio (r₂) = = = 3

n^th term of both the series is equal.

a₁ r₁^n-1 = a₂r₂^n-1

162×()^n-1 = × 3^n-1

81² = 3^2n-2

3⁸ = 3^2n-2

Comparing the powers on both sides

8 = 2n-2

10 = 2n

n = 5