Solve the followings Questions.

1. Find the sum of the following AP’s.

(i) 2, 7, 12… to 10 terms

(ii) –37, –33, –29… to 12 terms

(iii) 0.6, 1.7, 2.8… to 100 terms

(iv)1/15 ,1/12,1/10—to 11 terms

Answer:

(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,

(iv) to 11 terms

Here First tern =a = 1/15Common difference = 

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

2. Find the sums given below
(i) 7 + 21/2+ 14 + ……………… +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Answer:

(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 nth 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 nth 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,

3. In an AP
(i) Given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(v) Given d = 5, S₉ = 75, find a and a₉.
(vi) Given a = 2, d = 8, S_n = 90, find n and an.
(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) Given an = 4, d = 2, S_n = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Answer:

(i) Given a = 5, d = 3, a_n = 50,find n and S_n.

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

⇒ 50 = 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 S_n  = 440

(ii) Given a = 7, a₁₃ = 35, find d and S₁₃..

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

a_n = a + (n-1)d

a₁₃= 7 + (13 − 1) (d)

⇒ 35 = 7 + 12d

⇒ 28 = 12d⇒ d = 28/12 = 7/3

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

Therefore, d = 7/3 and S₁₃ = 273

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂..

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

a_n = a + (n-1)d

a₁₂ = 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 a₃ = 15, S₁₀ = 125, find d and a₁₀..

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

a_n = a + (n-1)d

a₃ = 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 a_n = a + (n-1)d, to find nth term of arithmetic progression,

a_n = a + (n-1)d

a₁₀= a + (10 − 1) d

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

a₁₀= 15 − 2d + 9d = 15 + 7d

= 15 + 7 (−1) = 15 – 7 = 8

Therefore, d = −1 and a₁₀= 8

(v) Given d = 5,S₉ = 75, find a and a₉..

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

⇒ 150 = 18a + 360

⇒ −210 = 18a

⇒ a = 

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

a₉ = + (9 − 1) (5)

a₉ =

Therefore, a = and a9 = 

(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.

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 a_n = a + (n-1)d ,to find nth term of arithmetic progression,

a₅ = 2 + (5 − 1) (8) = 2 + 32 = 34

Therefore, n = 5 and a5 = 34

(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.

Using formula a_n = a + (n-1)d, to find nth 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,

⇒ n = 6

Putting value of n in equation (1),

6d = 54 + d ⇒ d = 

Therefore, n = 6 and d =

(viii) Given Given a_n = 4, d = 2, S_n = − 14, find n and a

Using formula a_n = a + (n-1)d , 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,

⇒ (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, a_n = a + (n-1)d 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

4. How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?

Answer:

First term = a = 9, Common difference = d = 17 – 9 = 8, Sn = 636

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

636 =[18 + (n − 1) (8)]

⇒ 1272 = n (18 + 8n − 8)

⇒

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.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer.

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

We get,

45 = 5 + (16 – 1)d

45 = 5 + (15)d

45 – 5 = 15d

6. The first and the last terms of an AP are 17 and 350 respectively. If, the common difference is 9, how many terms are there and what is their sum?

Answer:

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.

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Answer:

It is given that 22nd term is equal to 149

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

a_n = a + (n-1)d

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)

⇒ 11(151)= 1661

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

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer:

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

Using formula a_n = a + (n-1)d, 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.

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer:

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

⇒ 98 = 7 (2a + 6d)

⇒ 7 = a + 3d ⇒ a = 7 − 3d … (1)

And, s₁₇= 289

⇒ 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

10. Show that a₁, a₂ … , a_n , … form an AP where an is defined as below
(i) a_n = 3 + 4_n
(ii) a_n = 9 − 5_n
Also find the sum of the first 15 terms in each case.

Answer:

(i)

is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

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

(ii) 

is same every time. Therefore, this is an AP with common difference as -5 and first term as 4.

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

11. If the sum of the first n terms of an AP is 4_n − n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly find the 3^rd, the10^th and the nth terms.

Answer:

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

It means 

So here, we can find the first term by substituting n = 1 ,

=4-1=3

Similarly, the sum of first two terms can be given by,

= 8 – 4

= 4

Now, as we know,

So,

 = 4 – 3

= 1

Now, using the same method we have to find the third, tenth and nth term of A.P.

So, for the third term,

=

= (12 – 9) – (8 – 4)

= 3 – 4

= – 1

Also, for the tenth term,

=

= (40 – 100) – (36 – 81)

= – 60 + 45

= 15

So, for the nth term,

= 

=

=

= 5 – 2n

Therefore,=

12. Find the sum of the first 40 positive integers divisible by 6.

Answer:

The first positive integers that are 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

a = 6

d = 6

S₄₀ = ?

=

= 20 [12 + (39) (6 )]

= 20 (12 + 234)

= 20 x 246

= 4920

13. Find the sum of the first 15 multiples of 8.

Answer:

The first 15 multiples of 8 are 8,16, 24, 32 … 15 times

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

I  = a + (n-1) d

  = 8 + (15 – 1) 8

 = 120

Reruired Sum =

 =

Hence, the required sum is 960.

14. Find the sum of the odd numbers between 0 and 50.

Answer:

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 an = 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

=

=

= 625

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Answer:

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)

 = 15 x 1850

 = 27750

Therefore, penalty for 30 days is Rs. 27750.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is Rs 20 less than its preceding term, find the value of each of the prizes.

Answer:

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)

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

=

⇒

=

= 700 = 7 (a – 60)

On further specification, we get

 ⇒ 700/7 = a – 60

⇒ 100 + 60 = a

⇒  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

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g, a section of Class I will plant 1 tree, a section of class II will plant two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Answer:

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

=

= 6 (2+11)

= 6 x 13

= 78

Therefore, number of trees planted by 1 section of classes = 78

number of trees planted by 3 section of classes = 3 x 78 = 234

Therefore, 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … What is the total length of such a spiral made up of thirteen consecutive semicircles.

Answer:

Length of Semi-perimeter of circle = πr

Length of semi-circle of radii 0.5 cm = π(0.5) cm = π/2

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

Length of semi-circle of radii 1.5 cm = π(1.5) cm = 3π/2

Therefore, we have sequence of the form:

π(0.5), π(1.0), π(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 π(0.5), π(1.0), π(1.5) … 13 terms

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

=

=

= (13/2) (7π)

= (91π / 2)

= ((91 x 22) / 2) x (7)

= (13 x 11)

= 143

Therefore, the lenght of such spiral of 13 consecutive semi – circles will be 143 cm.

19. 200 logs are stacked in the following manner:20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer:

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 = n (41 – n)

⇒ 400 = 41n – n²

⇒ n² – 41n + 400 = 0

⇒ n² – 16n -25n + 400 = 0

⇒ n (n – 16) – 25 (n – 16) = 0

⇒ (n – 16) (n – 25) = 0

Either (n – 16) = 0 or n – 25 = 0

⇒ n = 16 or n = 25

⇒ a_n  = a + (n – 1)d

⇒ a_{16 = 20 + (16 – 1) (- 1)}

⇒ a_{16 = 20 – 15}

⇒ a_{16 = 5}

Similarly,

⇒ a_{25 = 20 + (25 – 1) (- 1)}

⇒ a_{25 = 20 – 24}

 _{= – 4}

_{Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which is not possible.}

Therefore,  200 logs can be placed in 16 rows and the number of logs in 16th row is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 meters from the first potato, and the other potatoes are placed 3 meters apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Answer:

The distance of potatoes are as follows :

5, 8, 11, 14, …..

It can be observed that these distances are in A.P.

  a = 5

 d =  8 – 5 = 3

From the formula, we get

= 

= 5 [10 + 9 x 3]

= 5 (10 + 27)

= 5 x 37

= 185

 As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, the total distance that the competitor will be run = 2 x 185 = 370m.

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2