Ans. (i) 2, 7, 12… to 10 terms 

Here First term = a = 2, Common difference = d = 7 – 2 = 5 and n = 10

Applying formula,  to find sum of n terms of AP,

(ii) –37, –33, –29… to 12 terms

Here First term = a = –37, Common difference = d = –33 – (–37) = 4

And n = 12

Applying formula,  to find sum of n terms of AP,

(iii) 0.6, 1.7, 2.8… to 100 terms

Here First term = a = 0.6, Common difference = d = 1.7 – 0.6 = 1.1

And n = 100

Applying formula,  to find sum of n terms of AP,

Sn=1002[1.2+(100−1)1.1]=50(1.2+108.9)=50×110.1=5505Sn=1002[1.2+(100−1)1.1]=50(1.2+108.9)=50×110.1=5505

(iv)  to 11 terms

Here First term =a =  Common difference = d = 

Applying formula,  to find sum of n terms of AP,

Ans. (i)  

Here First term = a = 7, Common difference = d = 

And Last term = l = 84

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula , to find n^th term of arithmetic progression,

[7 + (n − 1) (3.5)] = 84

⇒ 7 + (3.5) n − 3.5 = 84

⇒ 3.5n = 84 + 3.5 – 7

⇒ 3.5n = 80.5

⇒ n = 23

Therefore, there are 23 terms in the given AP.

It means n = 23.

Applying formula, to find sum of n terms of AP,

⇒ 

(ii) 34 + 32 + 30 + … + 10

Here First term = a = 34, Common difference = d = 32 – 34 = –2

And Last term = l = 10

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula , to find n^th term of arithmetic progression,

[34 + (n − 1) (−2)] = 10

⇒ 34 – 2n + 2 = 10

⇒ −2n = −26⇒ n = 13

Therefore, there are 13 terms in the given AP.

It means n = 13.

Applying formula,  to find sum of n terms of AP,

(iii) −5 + (−8) + (−11) + … + (−230)

Here First term = a = –5, Common difference = d = –8 – (–5) = –8 + 5 = –3

And Last term = l = −230

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula , to find n^th term of arithmetic progression,

[−5 + (n − 1) (−3)] = −230

⇒ −5 − 3n + 3 = −230

⇒ −3n = −228 ⇒ n = 76

Therefore, there are 76 terms in the given AP.

It means n = 76.

Applying formula,  to find sum of n terms of AP,

Ans. (i) Given a = 5, d = 3, , find. 

Using formula , to find n^th term of arithmetic progression,

= 5 + (n − 1) (3)

⇒ 50 = 5 + 3n − 3

⇒ 48 = 3n⇒ n = 16

Applying formula,  to find sum of n terms of AP,

Therefore, n = 16 and 

(ii) Given a = 7, , find.

Using formula , to find n^th term of arithmetic progression,

= 7 + (13 − 1) (d)

⇒ 35 = 7 + 12d

⇒ 28 = 12d⇒ d = 

Applying formula,  to find sum of n terms of AP,

Therefore, d =  and 

(iii) Given , d = 3, find.

Using formula , to find n^th term of arithmetic progression,

= a + (12 − 1) 3

⇒ 37 = a + 33 ⇒ a = 4

Applying formula,  to find sum of n terms of AP,

Therefore, a = 4 and S₁₂ = 246

(iv) Given , find.

Using formula , to find n^th term of arithmetic progression,

= a + (3 − 1) (d)

⇒ 15 = a + 2d

⇒ a = 15 − 2d… (1)

Applying formula,  to find sum of n terms of AP,

⇒ 125 = 5 (2a + 9d) = 10a + 45d

Putting (1) in the above equation,

125 = 5 [2 (15 − 2d) + 9d] = 5 (30 − 4d + 9d)

⇒ 125 = 150 + 25d

⇒ 125 – 150 = 25d

⇒ −25 = 25d⇒ d = −1

Using formula , to find n^th term of arithmetic progression,

= a + (10 − 1) d

Putting value of d and equation (1) in the above equation,

= 15 − 2d + 9d = 15 + 7d

= 15 + 7 (−1) = 15 – 7 = 8

Therefore, d = −1 and  = 8

(v) Given d = 5,.

Applying formula,  to find sum of n terms of AP,

⇒ 

⇒ 150 = 18a + 360

⇒ −210 = 18a

⇒ a = 

Using formula , to find n^th term of arithmetic progression,

=  + (9 − 1) (5)

= 

Therefore, a =  and a₉ = 

(vi) Given a = 2, d = 8,.

Applying formula,  to find sum of n terms of AP,

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 2n (n − 5) + 9 (n − 5) = 0

⇒ (n − 5) (2n + 9) = 0

⇒ n = 5,−9/2

We discard negative value of n because here n cannot be in negative or fraction.

The value of n must be a positive integer.

Therefore, n = 5

Using formula ,to find n^th term of arithmetic progression,

a₅ = 2 + (5 − 1) (8) = 2 + 32 = 34

Therefore, n = 5 and 

(vii) Given a = 8, , find n and d.

Using formula , to find n^th term of arithmetic progression,

62 = 8 + (n − 1) (d) = 8 + nd – d

⇒ 62 = 8 + nd − d

⇒ nd – d = 54

⇒ nd = 54 + d… (1)

Applying formula,  to find sum of n terms of AP,

Putting equation (1) in the above equation,

⇒ ⇒ n = 6

Putting value of n in equation (1),

6d = 54 + d ⇒ d = 

Therefore, n = 6 and d = 

(viii) Given , find n and a.

Using formula , to find n^th term of arithmetic progression,

4 = a + (n − 1) (2) = a + 2n − 2

⇒ 4 = a + 2n – 2

⇒ 6 = a + 2n

⇒ a = 6 − 2n… (1)

Applying formula,  to find sum of n terms of AP,

⇒ 

Putting equation (1) in the above equation, we get

−28 = n [2 (6 − 2n) + 2n – 2]

⇒ −28 = n (12 − 4n + 2n − 2)

⇒ −28 = n (10 − 2n)

⇒ 

⇒ 

⇒ 

⇒ n (n − 7) + 2 (n − 7) = 0

⇒ (n + 2) (n − 7) = 0

⇒ n = −2, 7

Here, we cannot have negative value of n.

Therefore, we discard negative value of n which means n = 7.

Putting value of n in equation (1), we get

a = 6 − 2n = 6 – 2 (7) = 6 – 14 = −8

Therefore, n = 7 and a = −8

(ix)Given a = 3, n = 8, S = 192, find d.

Using formula,  to find sum of n terms of AP, we get

192 = [6 + (8 − 1) d] = 4 (6 + 7d)

⇒ 192 = 24 + 28d

⇒ 168 = 28d ⇒ d = 6

(x) Given l = 28, S = 144, and there are total of 9 terms. Find a.

Applying formula, , to find sum of n terms, we get

144 =  [a + 28]

⇒ 288 = 9 [a + 28]

⇒ 32 = a + 28⇒ a = 4

Ans. First term = a = 9, Common difference = d = 17 – 9 = 8, S_n = 636 

Applying formula,  to find sum of n terms of AP, we get

636 = [18 + (n − 1) (8)]

⇒ 1272 = n (18 + 8n − 8)

⇒ 

⇒ 

⇒ 

Comparing equation with general form , we get

a = 4, b = 5 and c = −636

Applying quadratic formula, and putting values of a, b and c, we get

⇒ 

⇒ 

⇒ 

We discard negative value of n here because n cannot be in negative, n can only be a positive integer.

Therefore, n = 12

Therefore, 12 terms of the given sequence make sum equal to 636.

Ans. First term = a = 5, Last term = l = 45,  

Applying formula,  to find sum of n terms of AP, we get

⇒ ⇒ n = 16

Applying formula,  to find sum of n terms of AP and putting value of n, we get

⇒ 400 = 8 (10 + 15d)

⇒ 400 = 80 + 120d

⇒ 320 = 120d

⇒ d = 

Ans. First term = a = 17, Last term = l = 350 and Common difference = d = 9 

Using formula , to find nth term of arithmetic progression, we get

350 = 17 + (n − 1) (9)

⇒ 350 = 17 + 9n − 9

⇒ 342 = 9n ⇒ n = 38

Applying formula,  to find sum of n terms of AP and putting value of n, we get

⇒ = 19 (34 + 333) = 6973

Therefore, there are 38 terms and their sum is equal to 6973.

Ans. It is given that 22nd term is equal to 149   

Using formula , to find nth term of arithmetic progression, we get

149 = a + (22 − 1) (7)

⇒ 149 = a + 147⇒ a = 2

Applying formula,  to find sum of n terms of AP and putting value of a, we get

⇒ = 11 (4 + 147)

⇒ = 1661

Therefore, sum of first 22 terms of AP is equal to 1661.

Ans. It is given that second and third term of AP are 14 and 18 respectively. 

Using formula , to find nth term of arithmetic progression, we get

14 = a + (2 − 1) d

⇒ 14 = a + d … (1)

And, 18 = a + (3 − 1) d

⇒ 18 = a + 2d … (2)

These are equations consisting of two variables.

Using equation (1), we get, a = 14 − d

Putting value of a in equation (2), we get

18 = 14 – d + 2d

⇒ d = 4

Therefore, common difference d = 4

Putting value of d in equation (1), we get

18 = a + 2 (4)

⇒ a = 10

Applying formula,  to find sum of n terms of AP, we get

Therefore, sum of first 51 terms of an AP is equal to 5610.

Ans. It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289. 

Applying formula,  to find sum of n terms of AP, we get

49

⇒ 98 = 7 (2a + 6d)

⇒ 7 = a + 3d ⇒ a = 7 − 3d … (1)

And, 

⇒ 578 = 17 (2a + 16d)

⇒ 34 = 2a + 16d

⇒ 17 = a + 8d

Putting equation (1) in the above equation, we get

17 = 7 − 3d + 8d

⇒ 10 = 5d ⇒ d = 2

Putting value of d in equation (1), we get

a = 7 − 3d = 7 – 3 (2) = 7 – 6 = 1

Again applying formula, to find sum of n terms of AP, we get

⇒ 

⇒ ⇒ S_n = n²

Therefore, sum of n terms of AP is equal to n².

Ans. (i)We need to show that  form an AP where  

Let us calculate values of  using 

= 3+4=7

a₂= 3 + 4 (2) = 3 + 8 = 11

= 3+12=15

a₄= 3 + 4 (4) = 3 + 16 = 19

So, the sequence is of the form 7, 11, 15, 19 …

Let us check difference between consecutive terms of this sequence.

11 – 7 = 4, 15 – 11 = 4, 19 – 15 = 4

Therefore, the difference between consecutive terms is constant which means terms a₁, a₂ … a_n form an AP.

We have sequence 7, 11, 15, 19 …

First term = a = 7 and Common difference = d = 4

Applying formula,  to find sum of n terms of AP, we get

Therefore, sum of first 15 terms of AP is equal to 525.

(ii)We need to show that  form an AP where 

Let us calculate values of  using 

= 9-5 = 4        a₂= 9 – 5 (2) = 9 – 10 = −1

= 9-15=-6      a₄= 9 – 5 (4) = 9 – 20 = −11

So, the sequence is of the form 4, −1, −6, −11 …

Let us check difference between consecutive terms of this sequence.

–1 – (4) = –5,–6 – (–1)

= –6 + 1 = –5,–11 – (–6)

= –11 + 6 = –5

Therefore, the difference between consecutive terms is constant which means terms  form an AP.

We have sequence 4, −1, −6, −11 …

First term = a = 4 and Common difference = d = –5

Applying formula,  to find sum of n terms of AP , we get

Therefore, sum of first 15 terms of AP is equal to –465.

Ans. It is given that the sum of n terms of an AP is equal to  

It means 

Let us calculate  using 

= 4 – 1 = 3

= 8 – 4 = 4

First term = a = = 3 … (1)

Let us find common difference now.

We can write any AP in the form of general terms like a , a + d, a + 2d …

We have calculated that sum of first two terms is equal to 4 i.e. 

Therefore, we can say that a + (a + d) = 4

Putting value of a from equation (1), we get

2a + d = 4

⇒ 2 (3) + d = 4

⇒ 6 + d = 4

⇒ d = –2

Using formula a_n = a + (n – 1) d, to find n^th term of arithmetic progression,

Second term of AP =  = a + (2 – 1) d = 3 + (2 – 1) (–2) = 3 – 2 = 1

Third term of AP =  = a + (3 – 1) d = 3 + (3 – 1) (–2) = 3 – 4 = –1

Tenth term of AP =  = a + (10 – 1) d = 3 + (10 – 1) (–2) = 3 – 18 = –15

n^th term of AP = = a + (n – 1) d = 3 + (n – 1) (–2) = 3 – 2n + 2 = 5 – 2n

Ans. The first 40 positive integers divisible by 6 are 6, 12, 18, 24 … 40 terms. 

Therefore, we want to find sum of 40 terms of sequence of the form:

6, 12, 18, 24 … 40 terms

Here, first term = a = 6 and Common difference = d = 12 – 6 = 6, n = 40

Applying formula, to find sum of n terms of AP, we get

= 

= 20 (12 + 234)

= 

Ans. The first 15 multiples of 8 are 8,16, 24, 32 … 15 terms 

First term = a = 8 and Common difference = d = 16 – 8 = 8, n = 15

Applying formula, to find sum of n terms of AP, we get

Ans. The odd numbers between 0 and 50 are 1, 3, 5, 7 … 49 

It is an arithmetic progression because the difference between consecutive terms is constant.

First term = a = 1, Common difference = 3 – 1 = 2, Last term = l = 49

We do not know how many odd numbers are present between 0 and 50.

Therefore, we need to find n first.

Using formula a_n = a + (n − 1) d, to find nth term of arithmetic progression, we get

49 = 1 + (n − 1) 2

⇒ 49 = 1 + 2n − 2

⇒ 50 = 2n ⇒ n = 25

Applying formula,  to find sum of n terms of AP, we get

Ans. Penalty for first day = Rs 200, Penalty for second day = Rs 250 

Penalty for third day = Rs 300

It is given that penalty for each succeeding day is Rs 50 more than the preceding day.

It makes it an arithmetic progression because the difference between consecutive terms is constant.

We want to know how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

So, we have an AP of the form200, 250, 300, 350 … 30 terms

First term = a = 200, Common difference = d = 50, n = 30

Applying formula, to find sum of n terms of AP, we get

⇒ 

⇒ = 15 (400 + 1450) = 27750

Therefore, penalty for 30 days is Rs. 27750.

Ans. It is given that sum of seven cash prizes is equal to Rs 700. 

And, each prize is R.s 20 less than its preceding term.

Let value of first prize = Rs. a

Let value of second prize =Rs (a − 20)

Let value of third prize = Rs (a − 40)

So, we have sequence of the form:

a, a − 20, a − 40, a – 60 …

It is an arithmetic progression because the difference between consecutive terms is constant.

First term = a, Common difference = d = (a – 20) – a = –20

n = 7 (Because there are total of seven prizes)

{given}

Applying formula, to find sum of n terms of AP, we get

⇒ 

⇒ 200 = 2a – 120

⇒ 320 = 2a ⇒ a = 160

 

Therefore, value of first prize = Rs 160

Value of second prize = 160 – 20 = Rs 140

Value of third prize = 140 – 20 = Rs 120

Value of fourth prize = 120 – 20 = Rs 100

Value of fifth prize = 100 – 20 = Rs 80

Value of sixth prize = 80 – 20 = Rs 60

Value of seventh prize = 60 – 20 = Rs 40

Ans. There are three sections of each class and it is given that the number of trees planted by any class is equal to class number. 

The number of trees planted by class I = number of sections 

The number of trees planted by class II = number of sections 

The number of trees planted by class III = number of sections 

Therefore, we have sequence of the form 3, 6, 9 … 12 terms

To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 … 12 terms.

First term = a = 3, Common difference = d= 6 – 3 = 3 and n = 12

Applying formula, to find sum of n terms of AP , we get

Ans. Length of semi–circle =  

Length of semi-circle of radii 0.5 cm = (0.5) cm

Length of semi-circle of radii 1.0 cm = (1.0) cm

Length of semi-circle of radii 1.5 cm = (1.5) cm

Therefore, we have sequence of the form:

(0.5), (1.0), (1.5) … 13 terms {There are total of thirteen semi–circles}.

To find total length of the spiral, we need to find sum of the sequence (0.5), (1.0), (1.5) … 13 terms

Total length of spiral = (0.5) + (1.0) + (1.5) … 13 terms

⇒ Total length of spiral = (0.5 + 1.0 + 1.5) … 13 terms … (1)

Sequence 0.5, 1.0, 1.5 … 13 terms is an arithmetic progression.

Let us find the sum of this sequence.

First term = a = 0.5, Common difference = 1.0 – 0.5 = 0.5 and n = 13

Applying formula, to find sum of n terms of AP, we get

Therefore, 0.5 + 1.0 + 1.5 + 2.0 … 13 terms = 45.5

Putting this in equation (1), we get

Total length of spiral= (0.5 + 1.5 + 2.0 + … 13 terms) =  (45.5) = 143 cm

Ans. The number of logs in the bottom row = 20 

The number of logs in the next row = 19

The number of logs in the next to next row = 18

Therefore, we have sequence of the form 20, 19, 18 …

First term = a = 20, Common difference = d = 19 – 20 = –1

We need to find that how many rows make total of 200 logs.

Applying formula, to find sum of n terms of AP, we get

⇒ 400 = n (40 – n + 1)

⇒ 400 = 

⇒ 

It is a quadratic equation, we can factorize to solve the equation.

⇒ − 25n − 16n + 400 = 0

⇒ n (n − 25) – 16 (n − 25) = 0

⇒ (n − 25) (n − 16)

⇒ n = 25, 16

We discard n = 25 because we cannot have more than 20 rows in the sequence. The sequence is of the form: 20, 19, 18 …

At most, we can have 20 or less number of rows.

Therefore, n = 16 which means 16 rows make total number of logs equal to 200.

We also need to find number of logs in the 16th row.

Applying formula, to find sum of n terms of AP, we get

200 = 8 (20 + l)

⇒ 200 = 160 + 8I

⇒ 40 = 8l ⇒ l = 5

Therefore, there are 5 logs in the top most row and there are total of 16 rows.

Ans. The distance of first potato from the starting point = 5 meters 

Therefore, the distance covered by competitor to pick up first potato and put it in bucket = = 10 meters

The distance of Second potato from the starting point = 5 + 3 = 8 meters {All the potatoes are 3 meters apart from each other}

Therefore, the distance covered by competitor to pick up 2nd potato and put it in bucket = = 16 meters

The distance of third potato from the starting point = 8 + 3 = 11 meters

Therefore, the distance covered by competitor to pick up 3rd potato and put it in bucket = = 22 meters

Therefore, we have a sequence of the form 10, 16, 22 … 10 terms

(There are ten terms because there are ten potatoes)

To calculate the total distance covered by the competitor, we need to find:

10 + 16 + 22 + … 10 terms

First term = a = 10, Common difference = d = 16 – 10 = 6

n = 10{There are total of 10 terms in the sequence}

Applying formula, to find sum of n terms of AP, we get

Therefore, total distance covered by competitor is equal to 370 meters.