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Answers of maths NCERT chapter 5

Answer» 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 the fare is Rs 15 for the first km and Rs 8 for each additional km.(ii) The amount of air present in a cylinder when a vacuum pump removes one fourth\xa0of the air remaining in the cylinder at a time.(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.Ans. (i)Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23Taxi fare after 3 km = 23 + 8 = Rs 31Taxi fare after 4 km = 31 + 8 = Rs 39Therefore, the sequence is 15, 23, 31, 39...It is an arithmetic progression because difference between any two consecutive terms is equal which is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, ...)(ii)Let amount of air initially present in a cylinder =\xa0VAmount of air left after pumping out air by vacuum pump =\xa0Amount of air left when vacuum pump again pumps out air=\xa0So, the sequence we get is like\xa0Checking for difference between consecutive terms ...Difference between consecutive terms is not equal.Therefore, it is not an arithmetic progression.(iii)\xa0Cost of digging 1 meter of well = Rs 150Cost of digging 2 meters of well = 150 + 50 = Rs 200Cost of digging 3 meters of well = 200 + 50 = Rs 250Therefore, we get a sequence of the form 150, 200, 250...It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50...)Here, difference between any two consecutive terms which is also called common difference is equal to 50.(iv)Amount in bank after Ist year =\xa0… (1)Amount in bank after two years =\xa0… (2)Amount in bank after three years =\xa0… (3)Amount in bank after four years =\xa0… (4)It is not an arithmetic progression because (2) − (1) ≠ (3) − (2)(Difference between consecutive terms is not equal)Therefore, it is not an Arithmetic Progression.2. Write first four terms of the AP, when the first term a and common difference d are given as follows:(i)a = 10, d = 10(ii) a = -2, d = 0(iii) a = 4, d = -3(iv) a = -1, d =\xa0(v) a = -1.25, d = -0.25Ans. (i)\xa0First term = a = 10, d = 10Second term = a + d = 10 + 10 = 20Third term = second term + d = 20 + 10 = 30Fourth term = third term + d = 30 + 10 = 40Therefore, first four terms are: 10, 20, 30, 40(ii)\xa0First term = a = –2 , d = 0Second term = a + d = –2 + 0 = –2Third term = second term + d = –2 + 0 = –2Fourth term = third term + d = –2 + 0 = –2Therefore, first four terms are: –2, –2, –2, –2(iii)\xa0First term = a = 4, d =–3Second term = a + d = 4 – 3 = 1Third term = second term + d = 1 – 3 = –2Fourth term = third term + d = –2 – 3 = –5Therefore, first four terms are: 4, 1, –2, –5(iv)\xa0First term = a = –1, d =\xa0Second term = a + d = –1 +\xa0\xa0= −Third term = second term + d = −+\xa0= 0Fourth term = third term + d = 0 +\xa0=Therefore, first four terms are: –1, −, 0,(v)\xa0First term = a = –1.25, d = –0.25Second term = a + d = –1.25 – 0.25 = –1.50Third term = second term + d = –1.50 – 0.25 = –1.75Fourth term = third term + d= –1.75 – 0.25 = –2.00Therefore, first four terms are: –1.25, –1.50, –1.75, –2.003. For the following APs, write the first term and the common difference.(i) 3, 1, –1, –3 …(ii) –5, –1, 3, 7...(iii)\xa0(iv) 0.6, 1.7, 2.8, 3.9 ...Ans. (i)\xa03, 1, –1, –3...First term = a = 3,Common difference (d) = Second term – first term = Third term – second term and so onTherefore, Common difference (d) = 1 – 3 = –2(ii)\xa0–5, –1, 3, 7...First term = a = –5Common difference (d) = Second term – First term= Third term – Second term and so onTherefore, Common difference (d) = –1 – (–5) = –1 + 5 = 4(iii)First term = a =\xa0Common difference (d) = Second term – First term= Third term – Second term and so onTherefore, Common difference (d) =\xa0(iv)\xa00.6, 1.7, 2.8, 3.9...First term = a = 0.6Common difference (d) = Second term – First term= Third term – Second term and so onTherefore, Common difference (d) = 1.7 − 0.6 = 1.14. 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...(ii) 2,\xa0, 3,\xa0...(iii) −1.2, −3.2, −5.2, −7.2...(iv) −10, −6, −2, 2...(v)\xa0(vi) 0.2, 0.22, 0.222, 0.2222...(vii) 0, −4, −8, −12...(viii)\xa0(ix) 1, 3, 9, 27...(x)\xa0a, 2a, 3a, 4a...(xi)\xa0(xii)\xa0(xiii)\xa0(xiv)\xa0(xv)\xa0Ans. (i)\xa02, 4, 8, 16...It is not an AP because difference between consecutive terms is not equal.As4 – 2 ≠ 8 − 4(ii)2,\xa0, 3,\xa0...It is an AP because difference between consecutive terms is equal.Common difference (d) =\xa0Fifth term =\xa0\xa0Sixth term = 4 + ½ =\xa0Seventh term =\xa0Therefore, next three terms are 4,\xa0and 5.(iii)−1.2, −3.2, −5.2, −7.2...It is an AP because difference between consecutive terms is equal.\xa0−3.2 − (−1.2)= −5.2 − (−3.2)= −7.2 − (−5.2) = −2Common difference (d) = −2Fifth term = −7.2 – 2 = −9.2Sixth term = −9.2 – 2 = −11.2Seventh term = −11.2 – 2 = −13.2Therefore, next three terms are −9.2, −11.2 and −13.2(iv)\xa0−10, −6, −2, 2...It isan AP because difference between consecutive terms is equal.\xa0−6 − (−10) = −2 − (−6)= 2 − (−2) = 4Common difference (d) = 4Fifth term = 2 + 4 = 6 Sixth term = 6 + 4 = 10Seventh term = 10 + 4 = 14Therefore, next three terms are 6, 10 and 14(v)\xa0It is an AP because difference between consecutive terms is equal.\xa0Common difference (d) =\xa0Fifth term =\xa0Sixth term =\xa0Seventh term =\xa0Therefore, next three terms are\xa0(vi)\xa00.2, 0.22, 0.222, 0.2222...It is not an AP because difference between consecutive terms is not equal.0.22 − 0.2 ≠ 0.222 − 0.22(vii)\xa00, −4, −8, −12...It is an AP because difference between consecutive terms is equal.\xa0−4 – 0 = −8 − (−4)= −12 − (−8) = −4Common difference (d) = −4Fifth term = −12 – 4 =−16 Sixth term = −16 – 4 = −20Seventh term = −20 – 4 = −24Therefore, next three terms are −16, −20 and −24(viii)\xa0It is an AP because difference between consecutive terms is equal.\xa0Common difference (d) = 0Fifth term =\xa0Sixth term =\xa0Seventh term =\xa0Therefore, next three terms are\xa0(ix)\xa01, 3, 9, 27...It is not an AP because difference between consecutive terms is not equal.3 – 1 ≠ 9 − 3(x)\xa0a, 2a, 3a, 4a...It is an AP because difference between consecutive terms is equal.\xa02a\xa0–\xa0a\xa0= 3a\xa0− 2a\xa0= 4a\xa0− 3a\xa0=\xa0aCommon difference (d) = aFifth term = 4a\xa0+\xa0a\xa0= 5a\xa0Sixth term = 5a\xa0+\xa0a\xa0= 6aSeventh term = 6a\xa0+\xa0a\xa0= 7aTherefore, next three terms are 5a, 6a\xa0and 7a(xi)\xa0a,\xa0a2,\xa0a3,\xa0a4...It is not an AP because difference between consecutive terms is not equal.a2\xa0–\xa0a\xa0≠\xa0a3\xa0−\xa0a2(xii)\xa0\xa0It is an AP because difference between consecutive terms is equal.\xa0Common difference (d) =\xa0Fifth term =\xa0\xa0Sixth term =\xa0Seventh term =\xa0Therefore, next three terms are\xa0(xiii)\xa0It is not an AP because difference between consecutive terms is not equal.(xiv)\xa0It is not an AP because difference between consecutive terms is not equal.(xv)\xa0\xa01, 25, 49, 73...It is an AP because difference between consecutive terms is equal.\xa0=\xa0= 24Common difference (d) = 24Fifth term = 73 + 24 = 97 Sixth term = 97 + 24 = 121Seventh term = 121 + 24 = 145Therefore, next three terms are 97, 121 and 145Exercise 1
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