Arithmetic Progressions Video Lecture | Mathematics (Maths) Class 10

0:0:0: imagine a person climbing a staircase
0:0:2: the first step he takes he moves One
0:0:5: Step Above in the Second Step he moves
0:0:8: two steps above in the third step he
0:0:10: moves three steps above how many steps
0:0:14: above will he move in the fourth step
0:0:17: right he will move four steps above and
0:0:21: how did you know that well you saw the
0:0:24: pattern in this problem and predicted
0:0:26: the fourth step now if we write down
0:0:30: these numbers it will be 1 2 4 7
0:0:33: 11 this is called a sequence a group of
0:0:37: numbers with a particular
0:0:39: pattern now let's make it simple what if
0:0:43: we made the height of each step equal
0:0:46: that means on First Step he would move
0:0:49: up by two steps on the Second Step he
0:0:51: would also move up by two steps and in
0:0:54: the third step two he would move up by
0:0:56: two steps now tell me how many steps
0:1:0: above will he move in the fourth step
0:1:3: correct he will move up by two steps
0:1:6: only here the height or difference of
0:1:9: each step is the same we can write these
0:1:13: groups of numbers as 1 3 5 7 9 this is
0:1:18: called an arithmetic progression or
0:1:21: AP now tell me whether the following
0:1:23: group of numbers is a sequence or AP or
0:1:27: both 2 4 6 8 10 correct it's an AP
0:1:34: because the difference between each
0:1:36: consecutive number is the same that is
0:1:38: two now this is also a sequence because
0:1:42: it has a fixed pattern and anything that
0:1:45: has a pattern is a
0:1:47: sequence now let's move on to the next
0:1:50: example 1 2 4 7 is this a sequence or an
0:1:56: AP yes this is a sequence as you can see
0:1:59: by the pattern but it's not an AP
0:2:2: because the difference between the
0:2:4: numbers is not
0:2:6: fixed so remember this an AP is always a
0:2:10: sequence but a sequence may or may not
0:2:12: be an
0:2:14: AP now come back to the example of our
0:2:17: staircase the AP can be written as 1 3 5
0:2:22: 7
0:2:23: 9 now here one is the first step we
0:2:26: denote this by a or A1 a 2 denotes the
0:2:30: second term which is three here and
0:2:33: similarly A3 denotes the third term
0:2:36: which is five here now as we saw the
0:2:39: height of each step is 2 which is A2
0:2:41: minus A1 is equal to 2 this is called
0:2:45: the common difference of the AP we
0:2:48: denote this by D so in a general
0:2:52: arithmetic progression a n denotes the
0:2:54: nth term of an AP and D denotes the
0:2:57: common difference let's see an example 1
0:3:2: 4 7 10 tell me what is A1 A2 and A3 here
0:3:8: correct A1 is 1 A2 is 4 and A3 is 7 now
0:3:14: tell me what is a common difference of
0:3:16: this AP D is = to A2 - A1 is = 4 - 1
0:3:22: isal 3 now let's visualize this A1 is 1
0:3:27: and A2 is 4 the common difference
0:3:30: between A1 and A2 is 3 that is A2 - A1
0:3:35: is equal to 3 so if we add D to A1 we
0:3:39: will get
0:3:41: A2 right 1 + 3 = 4 correct so A2 is
0:3:49: equal to A1 + D now similarly if we add
0:3:52: D to A2 we will get
0:3:54: A3 right let's write it A3 is equal to
0:3:59: A2 + + D which is A3 is equal to A1 + D
0:4:3: + D therefore A3 is equal to A1 + 2 D
0:4:9: easy right so tell me what will be
0:4:13: A4 correct it will be a 1 +
0:4:17: 3D so we can write this AP as a 1 a 1 +
0:4:22: 1 D A 1 + to D A 1 + 3D and so on this
0:4:27: is how we represent an AP
0:4:30: in this way if anyone asks us what is
0:4:32: the value of the fourth term we can just
0:4:35: put the value of A1 and D in a 1 + 3D
0:4:39: which gives us 1 + 3 * 3 which equal
0:4:43: 10 now an AP is of two types it can be
0:4:46: finite or infinite a finite AP is an AP
0:4:51: in which the number of terms is
0:4:53: finite it ends after the last term an
0:4:57: infinite AP is an AP that is never
0:4:59: ending and so it has unlimited terms for
0:5:3: example the days in the month are a
0:5:5: finite
0:5:6: AP that is 1 2 3 4 5 6 7 till
0:5:12: 31 the last term is 31 and the common
0:5:15: difference is one the group of odd
0:5:18: numbers is an infinite AP we can write
0:5:21: them as 1 3 5 7 9 it is never
0:5:27: ending now let's do some Qui questions
0:5:29: to understand this topic better question
0:5:33: one for the following AP write the first
0:5:36: term and the common
0:5:37: difference - 5 -1 3
0:5:44: 7 so your first term is min -5 and the
0:5:48: difference is A2 - A1 which is -1 - -5
0:5:53: which becomes -1 + 5 which is 4 let's
0:5:57: verify this now A3 - A2 is = to 3 - -1
0:6:2: that is 3 + 1 which equal to 4 so A3 -
0:6:7: A2 is equal to A2 -
0:6:10: A1 so D is equal to 4 question two write
0:6:15: first four terms of the AP when the
0:6:17: first term a and the common difference D
0:6:20: are given a is equal to 4 d is equal to
0:6:23: minus 3 we learned how to write an AP in
0:6:27: terms of a and d right let's apply that
0:6:30: here first term is a that is four second
0:6:34: term is a + D third term is a plus 2D
0:6:38: and similarly fourth term is a +
0:6:40: 3D let's put the values and the second
0:6:43: term becomes 4 + - 3 that is 1 third
0:6:48: term becomes 4 + 2 into - 3 which is 4 -
0:6:52: 6 which equal to
0:6:54: -2 fourth term becomes 4 + 3 into - 3
0:6:59: which is is 4 - 9 which equal to - 5 so
0:7:4: the AP becomes 4 1 - 2 - 5 question
0:7:9: three in the given situation does the
0:7:12: list of numbers involved make an
0:7:14: arithmetic progression and why the
0:7:17: amount of air present in a cylinder when
0:7:19: a vacuum pump removes 1/4 of the air
0:7:22: remaining in the cylinder at a time
0:7:25: solution in these type of questions we
0:7:28: just need to write the first three terms
0:7:30: according to the
0:7:31: situation let's write first term here so
0:7:35: in the starting the cylinder will be
0:7:37: completely filled with air let's
0:7:40: consider total air is equal to X so the
0:7:42: cylinder is filled with x amount of air
0:7:46: so our first term is X now the pump
0:7:48: removes 1/4 of this air that means the
0:7:51: air left in the cylinder is now x - 1X
0:7:55: 4X we can simplify it and write it like
0:7:58: 3 by
0:8:0: 4X so our second term becomes 3x
0:8:4: 4X now the pump again removes 1x4 of air
0:8:9: left that means the air left now is 3x
0:8:12: 4x - 1X 4 into 3x 4X which is 9x 16x so
0:8:18: our third term becomes 9x 16 x now we
0:8:22: have our three first terms let's write
0:8:26: them down together x 3X 4 x 9 by 16 x
0:8:32: now remember what is a property of
0:8:34: arithmetic
0:8:36: progression yes its difference between
0:8:39: two consecutive terms is same we denote
0:8:42: it by D let's find out difference
0:8:45: between A2 and A1 here 3x 4x - x is =
0:8:50: -1x 4 now A3 - A2 is = 9x 16 x - 3x
0:8:58: 4X which is equal to - 3x
0:9:2: 16x now as we can see difference between
0:9:6: A2 and A1 is not equal to difference
0:9:8: between A3 and A2 so this is not an AP
0:9:14: with this we have come to the end of
0:9:16: this video remember math is not just
0:9:19: about numbers it's about understanding
0:9:22: the patterns and sequences around us we
0:9:26: hope that you have understood the
0:9:27: concept of arithmetic program
0:9:29: questions now here's a secret on how you
0:9:32: can study effectively with the urv app
0:9:36: you can learn with chapter notes watch
0:9:38: video lectures and solve ncer based McQ
0:9:42: tests of this chapter on
0:9:44: urv and that's not all you also get
0:9:47: amazing courses for class 10 and much
0:9:49: more for your class 10
0:9:51: preparation thank you