In the following AP find the missing term (v) _ ,38, _ , _ , _ , (-22)

cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.2

 

 

in the following AP find the missing term

(v) _ ,38, _ , _ , _ , (-22)

 

a =?

t2=38

t6=(-22)

 

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d with n=2 and n=6

t2=a+(2-1)d =a+d

t6=a+5d

using given values

a+ d  =38

a+5d=(-22) 

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

  (-4)d = 60

d =60/(-4)

d=(-15)

 

substitute in a+ d  =38

 

a+ (-15) =38

a=38+15

a=53

 

missing terms are t1 ,t3,t4,t5

t1 =a =53

t3=a+2d =53+2*(-15)=53-30=23

t4=a+3d=53+3*(-15)=53-45=8

t5=a+4d=53+4*(-15)=53-60=(-7)

 

30th term of the AP: 10, 7, 4, . . . , is

a=10

d=t2 -t1 =7-10=(-3)

using the formula for the nth term of an arithmetic progression AP

tn= a+ (n-1)d with n=30

 

t(30) =a+ (30-1)d =a+29d =10 +29*(-3) = (-77)

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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 

 

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

10.  Show that a1 , a2 , . . ., an , . . . form an AP where a n is defined as below :
 an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

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

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

solution

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

 solution

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?

solution

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.

solution 

 

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

solution 

 

 3.

given a = 5, d = 3, an = 50, find n and Sn

solution

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

solution

(iii) given a(12) = 37, d = 3, find a and S(12 )

solution

(iv) given a3 = 15, S(10) = 125, find d and a(10)

 solution 

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

solution 

(vi) given a = 2, d = 8, Sn = 90, find n and an .

solution 

 vii) given a = 8, an = 62, Sn = 210, find n and d

solution

 (vii) given an = 4, d = 2, Sn = –14, find n and a.

solution 

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

solution

(x) given L= 28, S = 144, and there are total 9 terms. Find a.

solution

find the sums given below :

7 + [10 +(1/2) ] +14 + ...+84

solution

(ii) 34 + 32 + 30 + . . . + 10

 solution

(iii) –5 + (–8) + (–11) + . . . + (–230)

solution  

 Find the sum of the following APs:
 2, 7, 12, . . ., to 10 terms.

solution   

 (ii) –37, –33, –29, . . ., to 12 terms.

solution 

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms

solution  

 (iv) (1/15) +(1/12) +(1/10) + .... 11terms

solution

 

Aruna saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs.20.75, find n.

solution

exercise 5.2

19. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs.200 each year. In which year did his income reach Rs.7000?

solution  

 

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

solution  

 

17.Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

solution

 

 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

solution
 

15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . 

and 3, 10, 17, . . . equal?

solution

 14. How many multiples of 4 lie between 10 and 250?

solution

 

13.How many three-digit numbers are divisible by 7?

solution   

 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

solution   

 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

solution 

 9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

solution 

 

 8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

solution

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

solution

6. Check whether (-150) is a term of the AP : 11, 8, 5, 2 . . .

solution  

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

 

solution

5. Find the number of terms in each of the following APs :

 7,13,19, ....,205

solution

(ii) 18, [15+(1/2)],13,  . . . ,(-47)

solution  

 

4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?

solution   

 3. In the following APs, find the missing terms

2, _,26, .  .  . 

solution

(ii)  _ ,13 , _ , 3 , .  .   .

 solution

(iii) 5 , _ ,  _ , [9 +(1/2)], . . .

 solution

(iv) (-4) , _ , _ ,_ , _ , 6

solution

 (v) _ ,38, _ , _ , _ , (-22)

solution

30th term of the AP: 10, 7, 4, . . . , is

 solution

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