Thursday, 22 November 2012

Arithmetic Progression

Exercise 5.1
1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when fare is Rs. 15 for the first km and Rs. 8 for each additional km.

Solution: There is a common difference d and except first term each term can be obtained by using the common difference. This is arithmetic progression.
(ii) The amount of air present in a cylinder when a vaccum pump removes ¼ of the air remaining in the cylinder at a time.

Solution: This is not an arithmetic progression. The difference will keep on decreasing as per remaining volume, so it cannot be said as a common difference.
(iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.

Solution: This is arithmetic progression. The common difference is 50
(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.

Solution: This is not an arithmetic progression, as every year the value of interest will increase, because it is a compound interest.

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10 d = 10

Solution: n1 = a + d(1-1) = 10 + 0 = 0
n2 = a + d(2-1) = 10+10=20
n3 = a + d(3-1) = 10 + 20 = 30
n4 = a + d(4-1) = 10+30 = 40

(ii) a = -2 d = 0 
 
Solution: Every term in this AP will be equal to -2 because d is 0.

(iii) a = 4, d = - 3  
   
Solution: n1 = 4
n2 = 4 – 3(1) = 1
n3 = 4 – 3(2) = -2
n4 = 4-3(3) = -5

(iv) a = -1 d = ½

Solution: n1 = -1
n2 = -1 + ½ = -1/2
n3 = -1/2 + ½ = 0
n4 = 0 + ½ = ½ 

(v) a = -1.25 d = -0.25

Solution: n1 = -1.25
n2 = -1.5
n3 = -1.75
n4 = -2

3. For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, …..

Solution: First Term a = 3
n2 – n1 = 1 – 3 = -2
n3 – n2 = -1 – 1 = -2
So, common difference = -2
(ii) -5, -1, 3, 7, ……

Solution: First Term a = -5
Common Difference = 4
(iii) 1/3, 5/3, 9/3, 13/3, …….
arithematic progression 1
(iv) 0.6, 1.7, 2.8, 3.9, …….

Solution: First term a = 0.6
n2-n1 = 1.7 – 0.6 = 1.1
n3 – n2 = 2.8 – 1.7 = 1.1
So, common difference = 1.1

4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16
Solution: This is not an AP as numbers are getting doubled
(ii) 2, 5/2, 3, 7/2, ……

Solution: This is an AP and common difference is 0.5

(iii) -1.2, -3.2 , -5.2, -7.2, …..

Solution: Common Difference is -2

(iv) -10, -6, -2, 2, …

Solution: Common Difference is 4
arithematic progression 2
(vi) 0.2, 0.22, 0.222, 0.2222, ……

Solution: This is not an AP

(vii) 0, -4, -8, -12, …..

Solution: Common Difference is -4
(viii) -1/2, -1/2, -1/2, -1/2, …..
 
Solution: Common difference is 0

(ix) 1,3, 9, 27,……

Solution: This is not an AP

(x) a, 2a, 3a, 4a, …….

Solution: Common difference is 1a

(xi) a, a2, a3, a4, ….

Solution: This is not an AP
arithematic progression 3
Solution: This is not an AP
(xiii) 12, 52, 72, 73, ……
Solution: The series can be written as follows:
1, 25, 49, 73
Common difference is 24
arithematic progression 4
(xv) 12, 32, 52, 72, ……

Solution: After rewriting we get
1, 9, 25, 49
As the difference is increasing at every term so this is not an AP.

Exercise 5.2

1. Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …….. is
(a) 97 (b) 77 (c) -77 (d) -87
Solution: a = 10, d = -3
So, n30 = a + d(30 – 1) = 10 – 3(29) = -77
11th term of the AP: -3, -1/2, 2, …….. is
(a) 28 (b) 22 (c) -38 (d) -48.5
Solution: a = -3, d = 2.5
n11 = -3 + 2.5(10) = -3 + 25 = 22
Find the missing term in following APs:
(i) 2, ….., 26
arithmetic progression 5
(ii) ……, 13, ….., 3
arithmetic progression 7
(iii) 5, ……., ……….., 9.5
Solution: We have n4 = 5 + d(3)
Or, 9.5 – 5 = 3d
Or, d = 1.5
So, the AP is 5, 6.5, 8, 9.5
(iv) -4, ……, ……., ……., ……., 6
Solution: n5 = 6 = -4 + 4d
Or, 4d = 10
Or, d = 2.5
So, AP: -4, -1.5, 1, 3.5, 6
(v) …., 38, …, ….., …….., -22
Solution: Assuming n2 to be the first term we get following equation:
-22 = 38 + 3d
Or, 3d = -22 – 38 = -60
Or, d = -20
So, AP: 58, 38, 18, -2, -22
5. Which term of the AP: 3, 8, 13, 18, ….. is 78?
Solution: Here a = 3 and d = 5
78 = 3 + 5(n-1)
Or, 5(n-1) = 75
Or, n-1 = 15
Or, n = 16
6. Find the number of terms in each of the following APs:
(i) 7, 13, 19, ….., 205
Solution: a = 7, d = 6
205 = 7 + 6(n-1)
Or, 6(n-1) = 198
Or, n-1 = 33
Or, n = 34
(ii) 18, 15.5, 13, ……., -47
Solution: a = 18, d = -2.5
-47 = 18 + d(n-1)
Or, -2.5(n-1) = -65
Or, n-1 = 26
N = 27
7. Check whether -150 is a term of the AP: 11, 8, 5, 2,…..
Solution: a = 11, d = -3
-150 = 11+ d(n-1)
Or, -3(n-1) = -161
Or, n-1 = 161/3
As the result will not be an integer so -150 cannot be a term of the given AP.
8. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution: n11 = 38 = a + 10d ……… (i)
And n16 = 73 = a + 15d ………….. (ii)
Subtracting equation (i) from (ii) we get
35 = 5d
Or, d = 7
Putting the value of d in either of equations we can find value of a
38 = a + 70
Or, a = -32
So, n31 = -32 + 7x30 = 178
9. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution: 12 = a + 2d
106 = a + 49d
So, 106-12 = 47d
Or, 94 = 47d
Or, d = 2
Hence, a = 8
And, n29 = 8 + 28x2 = 64
10. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution: -8 = a + 8d
4 = a + 2d
Or, -8 – 4 = 6d
Or, -12 = 6d
Or, d = -2
Hence, a = -8 + 16 = 8
0 = 8 + -2(n-1)
Or, 8 = 2(n-1)
Or, n-1 = 4
Or, n = 5
11. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution: n7 = a + 6d
And, n10 = a + 9d
Or, a + 9d – a – 6d = 7
Or, 3d = 7
Or, d = 7/3
12. Which term of the AP: 3. 15, 27, 39, … will be 132 more than its 54th term?
Solution: d = 12,
132/12 = 11
So, 54 + 11 = 65th term will be 132 more than the 54th term.
13. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution: The difference will be 100
14. How many three digit numbers are divisible by 7?
Solution: Smallest three digit number divisible by 7 is 105
Greatest three digit number divisible by 7 is 994
Number of terms
= {(last term – first term )/common difference }+1
= {(994-105)/7}+1
= (889/7)+1=127+1=128
15. How many multiples of 4 lie between 10 and 250?
Solution: Smallest number divisible by 4 after 10 is 12,
The greatest number below 250 which is divisible by 4 is 248
Number of terms: {(248-12)/4}+1
{236/4}+1 = 59+1 = 60
16. For what value of n, are the nth terms of two APs: 63, 65, 67,… and 3, 10, 17,… equal?
Solution: In the first AP       a = 63 and d = 2
In the second AP                 a = 3 and d = 7
As per question,
63+2(n-1) = 2+ 7(n-1)
Or, 61 = 5 (n-1)
Or, n-1 = 61/5
As the result is not an integer so there wont be a term with equal values for both APs.
17. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution: As the 7th term exceeds the 5th term by 12, so the 5th term will exceed the 3rd term by 12 as well
So, n3 = 16
n5 = 28
n7 = 40
n4 or n6 can be calculated by taking average of the preceding and next term
So, n4 = (28+16)/2 = 22
This gives the d = 6
AP: 4, 10, 16, 22, 28, 34, 40, 46, ……..
18. Find the 20th term from the last term of the AP: 3, 8, 13, ……, 253.
Solution: a = 3, d = 5
253 = 3 + 5(n-1)
Or, 5(n-1) = 250
Or, n-1 = 50
Or, n = 51
So, the 20th term from the last term = 51 – 19 = 32nd term
Now, n32 = 3 + 5x31 = 158
19. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and the 10th terms is 44. Find the first three terms of the AP.
Solution: 
a + 3d + a + 7d = 24
Or, 2a + 10d = 24
Similarly, 2a + 14d = 44
So, 44 – 24 = 4d
Or, d = 5
2a + 10x5 = 24
Or, a + 25 = 12
Or, a = -13
So, first three terms of AP: -13, -8, -3,
20. 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 reached Rs. 7000.?
Solution: 
7000 = 5000 + 200(n-1)
Or, 200(n-1) = 2000
Or, n-1 = 10
Or, n = 11
21. Ramkali 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 savings become Rs. 20.75, find n.
Solution: 
20.75 = 5 + 1.75(n-1)
Or, 1.75(n-1) = 15.75
Or, n-1 = 9
Or, n = 10

Exercise 5.3

1. Find the sum of the following APs:
(i) 2, 7, 12, ……, to 10 terms
arithmetic progression 8
(ii) -37, -33, -29, …… to 12 terms
arithmetic progression 9
(iii) 0.6, 1.7, 2.8, …… to 100 terms
arithmetic progression 10
arithmetic progression 11
2. Find the sums given below:
(i) 7+10.5+14+…..+84
arithmetic progression 12
(ii) 34+32+30+….+10
arithmetic progression 13
(iii) -5+(-8)+(-11)+…….+(-230)
arithmetic progression 14
3. In an AP:
(i) Given a = 5, d = 3, an = 50, find n and Sn.
arithmetic progression 15
(ii) Given a = 7, a13 = 35, find d and S13.
arithmetic progression 16
(iii) Given a12 = 37, d = 3, find a and S12.
arithmetic progression 17
(iv) Given a3 = 15, S10 = 125, find d and a10.
arithmetic progression 18
(v) Given d = 5, S9 = 75, find a and a9.
arithmetic progression 19
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
arithmetic progression 20
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
arithmetic progression 21
(viii) Given an = 4, d = 2, Sn = -14, find n and a.
arithmetic progression 22
(ix) Given a = 3, n = 8, S = 192, find d.
arithmetic progression 23
(x) Given nl = 28, S = 144, and there are total 9 terms. Find a.
arithmetic progression 24
4. How many terms of the AP: 9, 17, 25,…. Must be taken to give a sum of 636?
arithmetic progression 25
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.
arithmetic progression 26
6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is the sum?
arithmetic progression 27
7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
arithmetic progression 28
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
arithmetic progression 29
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.
arithmetic progression 30
10. If the sum of the first n terms of an AP is 4n – 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 3rd, the 10th and the nth term.
arithmetic progression 31
11. Find the sum of the first 40 positive integers divisible by 6.
arithmetic progression 32
12. Find the sum of the first 15 multiples of 8.
arithmetic progression 33
13. Find the sum of the odd numbers between 0 and 50.
arithmetic progression 34
14. 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?
arithmetic progression 35
15. 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.
arithmetic progression 36
16. 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?
arithmetic progression 37
17. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with cnetre A, fo radii 0.5 cm, 1 cm, 1.5 cm, 2 cm, …….. as shown in the figure. What is the total length of such spiral made up of thirteen consecutive semicircles?
arithmetic progression 38
arithmetic progression 39
18. 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?
arithmetic progression 40
19. 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 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?
arithmetic progression 41





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