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

cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.3

obtaining the  sum to a particular number of  terms of an AP when the nth term of the arithmetic progression is given, 

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.

 

 an = 3+4n

replace n with (n-1) 

a(n-1) = 3 +4(n-1) = 3 +4n-4 =4n-1

an  - a(n-1) = [3+4n] - [4n-1] = 4= constant

therefore a1 , a2 , . . ., an , . . . form an AP with difference d = 4

put n = 1 

first term a = a1 =3+4(1) = 7

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

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

 with  n= 15

 

S(15) = [15/2][2(7) + (15-1)(4) ] = [15/2][14+56]

= [15/2][70]=525

(ii) an = 9 – 5n

replace n with (n-1)

a(n-1) = 9 -5 (n-1) = 9 -5n+5 = 14-5n

 an  - a(n-1) = [9-5n]-[14-5n]=(-5)= constant

therefore a1 , a2 , . . ., an , . . . form an AP with difference d =(-5)

 

put n=1

a1=9-5(1) = 4

 

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

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

 

with  n= 15

S(15) =  [15/2] [2(4)+(15-1)(-5)]=[15/2][8+14(-5)]

=[15/2][8-70]=[15/2][-62]=15*[-31]=(-465)

 

obtaining the  sum to a particular number of  terms of an AP when the two terms of the arithmetic progression are given, 

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

given that second and third terms are 14 and 18 respectively which means

t2=14 

t3=18

using the formula for nth term of an arithmetic progression AP

tn = a+(n-1)d

with n = 2 and n=3

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

t3 =a+(3-1)d=a+2d

using the given condition

a+ d =14

a+2d=18

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

-d =(-4)

d=4

use 

a+ d =14

a+4=14

a=14-4

a=10 

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

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

 with  n=51

 

S(51)  = [51/2][2(10) + (51-1)(4)]

=[51/2][20+50*4]=[51/2][20+200]=[51/2][220]=5610

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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

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