If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S 1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.3

obtaining the first second and nth terms using the expression of sum to n terms of an AP

 

11.If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

 

S(n) =  4n –(n^2)

 

put n = 1

S(1) = 4(1) - (1^2) = 4 -1 =3

note that S(1) = the first term 

so t1 = S(1) =3

 

put n=2

S(2) =4(2)-(2^2)=8-4=4

t2 =S(2) - S(1)

second term t2=4-3 =1

 

 

put n=3

S(2) =4(3)-(3^2)=12-9=3

t3 =S(3) - S(2)

third term t3=4-3 =3-4 = (-1)

for 10th term we need S(10) and  S(9)

put n =10 and n=9

S(10)=4(10)-(10^2)=40-100=(-60)

S(9)=4(9)-(9^2)=36-81=(-45)

10th term t(10) =S(10) - S(9) =(-60) - (-45) = (-60)+45=(-15)

for nth term we need S(n) and  S(n-1)

 S(n) =  4n –(n^2)  which is given

S(n-1) = 4(n-1) -(n-1)^2 = 4(n-1)-[(n^2)-2n+1] =4n-4-(n^2)+2n-1 

S(n-1) = -(n^2) +6n-5

nth term tn =   S(n) - S(n-1) = [4n –(n^2) ] -[-(n^2) +6n-5]

tn =  4n –(n^2) +(n^2) -6n +5

tn =5-2n

 

note that since we know t1=3 and t2=1 

we get d=t2-t1=1-3=(-2)

so that we can also use tn=a+(n-1)d 

tn=3+(n-1)(-2)=3-2n+2=5-2n

 

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.

S(7)=49

S(17)=289

using the formula for sum of n terms of an arithmetic progression ( AP )

Sn = (n/2)*[ 2a + (n-1)d ]

 

put n=7 andn=17

S(7) =(7/2)[2a+(7-1)d] = (7/2)[2a+6d]=(7/2)(2)[a+3d] =7[a+3d] 

given S(7)=49

7[a+3d] =49

a+3d =49/7

a+3d=7---------------(1)

n=17
S(17) =(17/2)[2a+(17-1)d] = (17/2)[2a+16d]=(17/2)(2)[a+8d] =17[a+8d]

given
S(17)=289

 17[a+8d]=289

a+8d=289/17

a+8d=17---------------(2)

a+3d=7---------------(1)

a+8d=17---------------(2)  

------------------------------------subtracting

(-5)d=(-10)

d=(-10)/(-5)

d=2

 

use (1)

a+3d=7

a+3(2)=7

a=1

 

using the formula for sum of n terms of an arithmetic progression ( AP )

Sn = (n/2)*[ 2a + (n-1)d ]

S(n) = (n/2)[2( 1) +(n-1)(2)]

S(n) =(n/2) [2+2n-2]

S(n)=(n/2)[2n]

S(n)=n^2

 

 

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m 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?

 solution

 

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?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre 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.

solution 

 

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 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

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 prize, find the value of each of the prizes. 

solution

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? 

solution

 

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

 solution

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

solution

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

solution

11.If the sum of the first n terms of an AP is 4n –(n^2) , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

solution

 

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.

  solution

 

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Find the sum of the first 15 multiples of 8.

 

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.3

 

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

 the first 15 multiples of 8 are terms in an arithmetic progression (AP)

a= t1 =1*8 =8

t2 = 2*8 =16 .....

last term L =15*8=120

d = t2-t1 =8

using the formula for the sum of n terms of an AP with n=15

Sn = (n/2) [a +L]

S(15) = (15/2) [8 +120]

=(15/2)*128

=15*64

=960 

 

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

the first 40 positive multiples of 8 are terms in an arithmetic progression (AP)

with a = t1 =6

t2 = 12

....

last term L=t(40)=40*6 =240

n=40

 

using the formula for the sum of n terms of an AP with n=40

Sn = (n/2) [a +L]

S(40) = (40/2) [6+240]

=20*246=4920

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m 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?

 solution

 

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?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre 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.

solution 

 

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 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

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 prize, find the value of each of the prizes. 

solution

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? 

solution

 

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

 solution

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

solution

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

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

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?

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.3

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?

Using the concept of Arithmetic progressions (AP)

t1 =200, t2 = 250 , t3 =300 ...

with first term    a = 200

d =t2 - t1 = 250-200 = 50

n =30 

using the formula for sum of n terms of an AP

Sn = (n/2)*[ 2a + (n-1)d ]

total penalty = 

S(30) = (30/2) *[ 2(200) + (30-1)(50) ]

=(15)*[400 +29*50]

=15*[400+1450]

=15*1850

=Rs. 27,750 /- 

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

selecting the odd numbers between 0 and 50 and adding 

1 + 3 +5 + 7 +11 +. . . +49

Using the concept of Arithmetic progressions (AP)

t1 =1, t2 = 3 , t3 =5 ...  last term L=49

with first term    a = 1

d =t2 - t1 = 3-1=2

n =?

use the concept of tn = a + (n-1)d

or n = [(L-a)/d] +1

n=[(49-1)/2] +1

n=[48/2] +1

n =24+1

n=25

using the formula for sum of n terms of an AP

Sn = (n/2) [a +L]

S(25) = (25/2) [ 1 +49]

= (25/2)(50)

=25 *25 

=625 

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m 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?

 solution

 

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?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre 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.

solution 

 

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 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

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 prize, find the value of each of the prizes. 

solution

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? 

solution

 

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

 solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

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 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.3

 

 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 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

 

since there are 3 sections for each class 

 

Total number of trees

= 3*[ 1 + 2 + . . . + 12]  

using the formula for sum of n terms of an AP inside the  [ ... ]

 with n=12  first term a =1 last term L = 12

Sn = (n/2) [a +L]

Total number of trees

= 3*[ 1 + 2 + . . . + 12]  

= 3* [ (12/2) [1 +12]  ]

=3*[(6)*13]

=234 trees

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 prize, find the value of each of the prizes.

 A sum of Rs.700 is to be used means Sn= 700

given that, there are  seven cash prizes so n =7

each prize is Rs.20 less than its preceding prize

means that d =(-20) where the minus sign is used for " less "

if the first term a stands for the first prize.

using the formula for sum of n terms of an AP

Sn = (n/2)*[ 2a + (n-1)d ]

 

700 = (7/2) * [ 2a + (7-1)(-20) ]

700 = (7/2) * [ 2a -120]

2*700  = 7 [2a - 120]

1400 = 14a -840

1400+840 =14a

2240 = 14a 

a = 2240/14

a = 160

 

first prize = a = Rs.160

second prize =a+d = 160 + (-20) = Rs.140

third prize = a+2d = 160 + 2*(-20) = 160-40=Rs.120

fourth prize = a+3d =  160 + 3*(-20) = Rs.100

fifth prize = a+4d =  160 + 4*(-20) = Rs.80

sixth prize = a+5d = 160 + 5*(-20) = Rs.60

seventh prize =a+6d =160 + 6*(-20) = Rs.40

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m 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?

 solution

 

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?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre 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.

solution 

 

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 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

solution 

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 prize, find the value of each of the prizes. 

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

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?

cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.3

 

 

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?

 

using arithmetic progressions AP

 t1 = 20

t2 = 19

t3=18

a =20 

d=t2 - t1 = 19 - 20 = (-1)

total number of logs is 200

so S(n) = 200 , n = ?

Sn = (n/2)*[ 2a + (n-1)d ]

 

200 = (n/2)*[ 2(20) + (n-1)(-1) ]

200 =  (n/2)*[ 40 - n +1 ]

200= (n/2)*[ 41 - n  ]

200*2 = n[41-n]

400 = 41n -(n^2)

 

(n^2) - 41n +400 =0

 

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

n = 16, n=25

if n = 16

tn = a+(n-1)d

t(16) = 20 + (16-1)(-1) = 20+(15)(-1) = 20-15 = 5 logs in the top row

if n = 25

tn = a+(n-1)d

t(25) = 20 + (25-1)(-1) = 20+(24)(-1) = 20-24= (-4) logs which is not possible.

 

So number of rows is 16, and number of logs in the top row is 5.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre 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.

 

length of arc of semi circle   = pi*r

Total length of spiral = pi*0.5 +pi*1.0+pi*1.5 + .... (13 terms)

=pi [0.5 + 1.0 +1.5 + ...  (13 terms) ]

using formula for sum of n terms of an AP

Sn = (n/2)*[ 2a + (n-1)d ]

with n =13, a=0.5, d=1.0-0.5 = 0.5

 

Total length of spiral =pi { (13/2) [ 2(0.5)+(13-1)(0.5) ] }

=pi  { (13/2) [1.0 +12*0.5 ] }

=pi { (13/2) [1.0 +6.0 ] }

=pi {(13/2)[7]}

=(22/7) {(13/2)[7]}

=11*13

=143 cm

 

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ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m 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?

 solution

 

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?

solution 

 

 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre 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.

solution 

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A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m. Calculate the total volume of concrete required to build the terrace.

 

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.4 optional exercise

 

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 

thinking of each step as a cuboid with length =50m, breadth =(1/2)m

and  height (1/4)m for the first step with increase of (1/4)m for each further step

volume for  first step

V1 = length * breadth * height =50*(1/2)*(1/4)

 

for second step 

 length =50m, breadth =(1/2)m, height =(1/4)+(1/4) =(1/2)m

volume for  second step

V2 = length * breadth * height =50*(1/2)*(1/2)

 

for third step 

 length =50m, breadth =(1/2)m, height =(1/4)+(1/4)+(1/4) =(3/4)m

volume for  second step

V2 = length * breadth * height =50*(1/2)*(3/4)

 

and so on

 

so total volume

= [50*(1/2)*(1/4)]+[50*(1/2)*(1/2)]+[50*(1/2)*(3/4)] +... (15 terms)

=50*(1/2) { (1/4) +(1/2) +(3/4) + ....(15 terms) }

=25 *{ (1/4) +(1/2) +(3/4) + ....(15 terms) } ,

using sum of n=15 terms of an AP with first term a=(1/4),  d =(1/4) for { ...}

Sn = (n/2)*[ 2a + (n-1)d ]

total volume = 25 *{ (15/2) *[2(1/4) +(15-1)(1/4) ] }

=25* { (15/2) [ (1/2)+ (14/4)  ] }

=25* {15/2 [(1/2) +(7/2)}

=25* {(15/2)*(4)}

=25*15*2=750 cubic metres

 

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m 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?

the distance covered to bring the first potato to the bucket = 5+5=10m

 

After that ,the distance covered for bringing the second potato to the bucket

=5 +3 + 3 +5 =16m

 

After that ,the distance covered for bringing the third potato to the bucket

=5 +3 + 3 +3+ 3 +5 =22m 

and so on for the 10 potatoes

 

so total distance = 10 +16 +22 + ...  (10 terms )

 

using sum of n=10 terms of an AP with first term a=10 , d=t2-t1=16-10=6

Sn = (n/2)*[ 2a + (n-1)d ]

 

total distance = 10 +16 +22 + ...  (10 terms )

= (10/2)*[2(10)+(10-1)(6)]

=5*74 =370m

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of (1/4) m and a tread of (1/2)m.   Calculate the total volume of concrete required to build the terrace.

 solution

 

chapter 5 arithmetic progressions, exercise 5.3

 exercise 5.3

 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato,and the other potatoes are placed 3 m 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?

 solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.4 optional exercise

 

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

top and last rungs are [ 2 and(1/2) ]m apart means

top and last rungs are [5/2]m apart

 

changing to cm.

top and last rungs are [5/2]*100 =250 cm apart

 

now consecutive rungs are 25cm apart 

so number of rungs =[250/25] +1 = 10+1 =11 rungs

The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top means 

 

counting upwards

 the length of the rungs form an AP  with 

n=11terms

a=45cm 

last term (11th term ) t(11) =25cm

 

To find the total length of wood we find the sum of lengths S(11)

S(n) = (n/2) [a + L ] , L refers to the last term , L = t(11) =25cm

 

S(11) = (11/2) [ 45 + 25 ]

=(11/2)[70] = 11*35

=385 cm

 

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

a = 1

The house numbers form an AP with

a=1 and 

kth term =t(k) =k for k =1 to 49 

using S(n) = (n/2) [a + L ] , where 

where L refers to the last term 

 

Sx = (x/2)[1+x]

 

S(49) = (49/2) [1+49] = (49/2)[50]=49*25=1225

 

S(x-1) = [(x-1)/2][1+(x-1)]= [(x-1)/2][x]

 

by given conditions

 S(x-1) = S(49) - S(x)

[(x-1)/2][x] = 1225 - (x/2)[1+x]

 

multiply each term with 2

(x-1)x = 2450 -x(1+x)

(x^2) - x = 2450 -x - (x^2) 

2(x^2)  -2450 = 0

 dividing by 2

(x^2)  - 1225 = 0

 

(x^2) = 1225

taking square root

x = 35 , because x cannot be negative

=================================================

ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

Which term of the AP : 121, 117, 113, . . ., is its first negative term? 

solution

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

solution

 3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

solution

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means