Arithmetic Progression And 1. Geometric Progression o Introduction : : We have seen different types of numbers since our childhood. Like set of positive even numbers, odd numbers etc. In all these sets some specific pattern is followed. In nature also we see that some pattern is followed e.g. Arrangement of petals in flower or arrangement in fruits or in tree. All these arrangements in nature look beautiful because some pattern is followed in all of them. o Sequence : A sequence is a collection of numbers arranged in a definite order according to some definite rule. Each number in the sequence is called a term of the sequence. The number in the first position is called the first term and it is denotedby t . 1 Similarly the number in the second and third positions are denoted as t 2 and t respectively. 3 In general, the number in the nth position is called nth term and is denoted by t . A sequence is usually denoted by { t } or < t > and read as sequencet . n n n n o Examples of sequences : 1. 2, 4, 6, 8, ...... Here t = 2, t = 4, t = 6, t = 8, ....... 1 2 3 4 2. 3, 6, 9, 12, ...... Here t = 3, t = 6, t = 9, t = 12, ....... 1 2 3 4 3. 5, 25, 125, 625, ...... Here t = 5, t = 25, t = 125, t = 625, ....... 1 2 3 4 4. – 4, – 2, 0, 2, ...... Here t = – 4, t = – 2, t = 0, t = 2, ....... 1 2 3 4 1 1 1 1 1 1 1 1 5. , , , , ...... Here t = , t = , t = , t = , ....... 2 6 18 54 1 2 2 6 3 18 4 54 o Sum of first n terms of a sequence : S = t 1 1 S = t + t 2 1 2 S = t + t + t 3 1 2 3 S = t + t + t + t 4 1 2 3 4 In general S = t + t + t + ....... + t . n 1 2 3 n From the above equations we observed that S = t 1 1 S – S = t 2 1 2 S – S = t 3 2 3 In general S – S = t . n n–1 n  t = S – S . n n n–1 SCHOOL SECTION 1 ALGEBRA MT EDUCARE LTD. o Types of sequence : If the number of terms in a sequence is finite then it is called a finite sequence. e.g. : 5, 9, 13, 17. If the number of terms in the sequence is not finite then it is called infinite sequence. e.g. : 7, 10, 13, 16, ...... EXERCISE - 1.1 (TEXT BOOK PAGE NO. 4) : 1. For each sequence, find the next four terms. (i) 1, 2, 4, 7, 11, ..... (1 mark) Sol. t = 1 + 0 = 1 1 t = 1 + 1 = 2 2 t = 2 + 2 = 4 3 t = 4 + 3 = 7 4 t = 7 + 4 = 11 5 t = 11 + 5 = 16 6 t = 16 + 6 = 22 7 t = 22 + 7 = 29 8 t = 29 + 8 = 37 9  The next four terms of the sequence are 16, 22, 29 and 37. (ii) 3, 9, 27, 81, ..... (1 mark) Sol. t = 31 = 3 1 t = 32 = 9 2 t = 33 = 27 3 t = 34 = 81 4 t = 35 = 243 5 t = 36 = 729 6 t = 37 = 2187 7 t = 38 = 6561 8  The next four terms of the sequence are 243, 729, 2187 and 6561. (iii) 1, 3, 7, 15, 31, .... (1 mark) Sol. t = 0 + 20 = 0 + 1 = 1 1 t = 1 + 21 = 1 + 2 = 3 2 t = 3 + 22 = 3 + 4 = 7 3 t = 7 + 23 = 7 + 8 = 15 4 t = 15 + 24 = 15 + 16 = 31 5 t = 31 + 25 = 31 + 32 = 63 6 t = 63 + 26 = 63 + 64 = 127 7 t = 127 + 27 = 127 + 128 = 255 8 t = 255 + 28 = 255 + 256 = 511 9  The next four terms of the sequence are 63, 127, 255 and 511. (iv) 192, – 96, 48, – 24, .... (1 mark) Sol. t = 192 1 192 t = = – 96 2 –2 –96 t = = 48 3 –2 48 t = = – 24 4 –2 2 SCHOOL SECTION MT EDUCARE LTD. ALGEBRA –24 t = = 12 5 –2 12 t = = – 6 6 –2 –6 t = = 3 7 –2 3 3 t = = – 8 –2 2 3  The next four terms of the sequence are 12, – 6, 3 and – . 2 (v) 2, 6, 12, 20, 30, ... (1 mark) Sol. t = 0 + 2 = 2 1 t = 2 + 4 = 6 2 t = 6 + 6 = 12 3 t = 12 + 8 = 20 4 t = 20 + 10 = 30 5 t = 30 + 12 = 42 6 t = 42 + 14 = 56 7 t = 56 + 16 = 72 8 t = 72 + 18 = 90 9  The next four terms of the sequence are 42, 56, 72 and 90. (vi) 0.1, 0.01, 0.001, 0.0001, .... (1 mark) Sol. t = 0.1 1 0.1 t = = 0.01 2 10 0.01 t = = 0.001 3 10 0.001 t = = 0.0001 4 10 0.0001 t = = 0.00001 5 10 0.00001 t = = 0.000001 6 10 0.000001 t = = 0.0000001 7 10 0.0000001 t = = 0.00000001 8 10  The next four terms of the sequence are 0.00001, 0.000001, 0.0000001 and 0.00000001. (vii) 2, 5, 8, 11, .... (1 mark) Sol. t = 2 1 t = 2 + 3 = 5 2 t = 5 + 3 = 8 3 t = 8 + 3 = 11 4 SCHOOL SECTION 3 ALGEBRA MT EDUCARE LTD. t = 11 + 3 = 14 5 t = 14 + 3 = 17 6 t = 17 + 3 = 20 7 t = 20 + 3 = 23 8  The next four terms of the sequence are 14, 17, 20 and 23. (viii) – 25, – 23, – 21, – 19, ..... (1 mark) Sol. t = – 25 1 t = – 25 + 2= – 23 2 t = – 23 + 2= – 21 3 t = – 21 + 2= – 19 4 t = – 19 + 2= – 17 5 t = – 17 + 2= – 15 6 t = – 15 + 2= – 13 7 t = – 13 + 2= – 11 8  The next four terms of the sequence are – 17, – 15, – 13 and – 11. (ix) 2, 4, 8, 16, ..... (1 mark) Sol. t = 21 = 2 1 t = 22 = 4 2 t = 23 = 8 3 t = 24 = 16 4 t = 25 = 32 5 t = 26 = 64 6 t = 27 = 128 7 t = 28 = 256 8  The next four terms of the sequence are 32, 64, 128 and 256. 1 1 1 1 (x) , , , , .... (2 mark) 2 6 18 54 1 Sol. t = 1 2 1 1 1 t =  = 2 2 3 6 1 1 1 t =  = 3 6 3 18 1 1 1 t =  = 4 18 3 54 1 1 1 t =  = 5 54 3 162 1 1 1 t =  = 6 162 3 486 1 1 1 t =  = 7 486 3 1458 1 1 1 t =  = 8 1458 3 4374 1 1 1 1  The next four terms of the sequence are , , and . 162 486 1458 4374 4 SCHOOL SECTION MT EDUCARE LTD. ALGEBRA EXERCISE - 1.1 (TEXT BOOK PAGE NO. 4) : 2. Find the first five terms of the following sequences, whose ‘nth’ terms are given : (i) t = 4n – 3 (1 mark) n Sol. t = 4n – 3 n  t = 4 (1) – 3 = 4 – 3 = 1 1  t = 4 (2) – 3 = 8 – 3 = 5 2  t = 4 (3) – 3 = 12 – 3 = 9 3  t = 4 (4) – 3 = 16 – 3 = 13 4  t = 4 (5) – 3 = 20 – 3 = 17 5  The first five terms of the sequence are 1, 5, 9, 13 and 17. (ii) t = 2n – 5 (1 mark) n Sol. t = 2n – 5 n  t = 2 (1) – 5 = 2 – 5 = – 3 1  t = 2 (2) – 5 = 4 – 5 = – 1 2  t = 2 (3) – 5 = 6 – 5 = 1 3  t = 2 (4) – 5 = 8 – 5 = 3 4  t = 2 (5) – 5 = 10 – 5 = 5 5  The first five terms of the sequence are – 3, – 1, 1, 3 and 5. (iii) t = n + 2 (1 mark) n Sol. t = n + 2 n  t = 1 + 2 = 3 1  t = 2 + 2 = 4 2  t = 3 + 2 = 5 3  t = 4 + 2 = 6 4  t = 5 + 2 = 7 5  The first five terms of the sequence are 3, 4, 5, 6 and 7. (iv) t = n2 – 2n (1 mark) n Sol. t = n2 – 2n n  t = 1 – 2 (1) = 1 – 2 = – 1 1  t = 22 – 2 (2) = 4 – 4 = 0 2  t = 32 – 2 (3) = 9 – 6 = 3 3  t = 42 – 2 (4) = 16 – 8 = 8 4  t = 52 – 2 (5) = 25 – 10= 15 5  The first five terms of the sequence are – 1, 0, 3, 8 and 15. (v) t = n3 (1 mark) n Sol. t = n3 n  t = 13 = 1 1  t = 23 = 8 2  t = 33 = 27 3  t = 43 = 64 4  t = 53 = 125 5  The first five terms of the sequence are 1, 8, 27, 64 and 125. 1 (vi) t = (1 mark) n n1 1 Sol. t = n n 1 SCHOOL SECTION 5 ALGEBRA MT EDUCARE LTD. 1 1  t = = 1 11 2 1 1  t = = 2 2 1 3 1 1  t = = 3 3 1 4 1 1  t = = 4 4 1 5 1 1  t = = 5 5 1 6 1 1 1 1 1  The first five terms of the sequence are , , , and . 2 3 4 5 6 EXERCISE - 1.1 (TEXT BOOK PAGE NO. 4) : 3. Find the first three terms of the sequences for which S isgiven below : n (i) S = n2 (n + 1) (2 marks) n Sol. S = n2 (n + 1) n  S = 12 (1 + 1) = 1 (2) = 2 1  S = 22 (2 + 1) = 4 (3) = 12 2  S = 32 (3 + 1) = 9 (4) = 36 3 We know that, t = S = 2 1 1 t = S – S = 12 – 2 = 10 2 2 1 t = S – S = 36 – 12= 24 3 3 2  The first three terms of the sequence are 2, 10 and 24. n2(n1)2 (ii) S = (2 marks) n 4 n2 (n 1)2 Sol. S = n 4 12 (11)2 1(2)2 1 4  S = = = = 1 1 4 4 4 22 (2 1)2 4(3)2  S = = = 9 2 4 4 32 (3 1)2 9 42  S = = = 9 × 4 = 36 3 4 4 We know that, t = S = 1 1 1 t = S – S = 9 – 1 = 8 2 2 1 t = S – S = 36 – 9 = 27 3 3 2  The first three terms of the sequence are 1, 8 and 27. n(n1) (2n1) (iii) (2 marks) 6 n(n 1)(2n 1) Sol. S = n 6 1(11)[2(1)1] 1 2 3 6  S = = = = 1 1 6 6 6 6 SCHOOL SECTION MT EDUCARE LTD. ALGEBRA 2(2 1)[2(2)1] 2 35 30  S = = = = 5 2 6 6 6 3(3 1)[2(3)1] 3 4 7  S = = = 14 3 6 6 We know that, t = S = 1 1 1 t = S – S = 5 – 1 = 4 2 2 1 t = S – S = 14 – 5= 9 3 3 2  The first three terms of the sequence are 1, 4 and 9. o Progression : Whenever we write some numbers one after the other, keeping a fix relation between two consecutive terms then we say the numbers are in progression. e.g. 1, 5, 10, 19, .......... o Arithmetic Progression : Arithmetic Progression is a sequence in which the difference between two consecutive terms is a constant. e.g. 4, 8, 12, 16, ........... Here, the difference between any two consecutive terms is 4 which is a constant. Therefore, the sequence is an A.P. e.g. 2, 4, 8, 16, ........... Here, the difference between any two consecutive terms is not a constant. Therefore, the sequence is not an A.P. EXERCISE - 1.2 (TEXT BOOK PAGE NO. 8) : 1. Which of the following lists of numbers are Arithmetic Progressions? Justify. (i) 1, 3, 6, 10, ..... (1 mark) Sol. t = 1, t = 3, t = 6, t = 10 1 2 3 4 t – t = 3 – 1 = 2 2 1 t – t = 6 – 3 = 3 3 2 t – t = 10 – 6 = 4 4 3 The difference between two consecutive terms is not constant.   The sequence is not an A.P. (ii) 3, 5, 7, 9, 11, ..... (1 mark) Sol. t = 3, t = 5, t = 7, t = 9, t = 11 1 2 3 4 5 t – t = 5 – 3 = 2 2 1 t – t = 7 – 5 = 2 3 2 t – t = 9 – 7 = 2 4 3 t – t = 11 – 9 = 2 5 4 The difference between two consecutive terms is 2 which is constant.   The sequence is an A.P. (iii) 1, 4, 7, 10, .... (1 mark) Sol. t = 1, t = 4, t = 7, t = 10 1 2 3 4 t – t = 4 – 1 = 3 2 1 t – t = 7 – 4 = 3 3 2 t – t = 10 – 7 = 3 4 3 The difference between two consecutive terms 3 which is constant.   The sequence is an A.P. SCHOOL SECTION 7 ALGEBRA MT EDUCARE LTD. (iv) 3, 6, 12, 24, .... (1 mark) Sol. t = 3, t = 6, t = 12, t = 24 1 2 3 4 t – t = 6 – 3 = 3 2 1 t – t = 12 – 6 = 6 3 2 t – t = 24 – 12 = 12 4 3 The difference between two consecutive terms is not constant.   The sequence is not an A.P. (v) 22, 26, 28, 31, ... (1 mark) Sol. t = 22, t = 26, t = 28, t = 31 1 2 3 4 t – t = 26 – 22 = 4 2 1 t – t = 28 – 26 = 2 3 2 t – t = 21 – 28 = 3 4 3 The difference between two consecutive terms is not constant.   The sequence is not an A.P. (vi) 0.5, 2, 3.5, 5, ... (1 mark) Sol. t = 0.5, t = 2, t = 3.5, t = 5 1 2 3 4 t – t = 2 – 0.5 = 1.5 2 1 t – t = 3.5 – 2 = 1.5 3 2 t – t = 5 – 3.5 = 1.5 4 3 The difference between two consecutive terms is 1.5 which is constant.   The sequence is an A.P. (vii) 4, 3, 2, 1, .... (1 mark) Sol. t = 4, t = 3, t = 2, t = 1, 1 2 3 4 t – t = 3 – 4 = – 1 2 1 t – t = 2 – 3 = – 1 3 2 t – t = 1 – 2 = – 1 4 3 The difference between two consecutive terms is –1 which isconstant.   The sequence is an A.P. (viii) – 10, – 13, – 16, – 19, ..... (1 mark) Sol. t = – 10, t = – 13, t = – 16, t = – 19 1 2 3 4 t – t = – 13 – (– 10) = – 13 + 10 = – 3 2 1 t – t = – 16 – (– 13) = – 16 + 13 = – 3 3 2 t – t = – 19 – (– 16) = – 19 + 16 = – 3 4 3 The difference between two consecutive terms is – 3 which is constant.   The sequence is an A.P. EXERCISE - 1.2 (TEXT BOOK PAGE NO. 8) : 2. Write the first five terms of the following Arithmetic Progressions where, the common difference ‘d’ and the first term ‘a’ are given : (i) a = 2, d = 2.5 (1 mark) Sol. a = 2, d = 2.5 Here, t = a = 2 1 t = t + d = 2 + 2.5 = 4.5 2 1 t = t + d = 4.5 + 2.5 = 7 3 2 t = t + d = 7 + 2.5 = 9.5 4 3 t = t + d = 9.5 + 2.5 = 12 5 4  The first five terms of the A.P. are 2, 4.5, 7, 9.5 and 12. 8 SCHOOL SECTION MT EDUCARE LTD. ALGEBRA (ii) a = 10, d = – 3 (1 mark) Sol. a = 10, d = – 3 Here, t = a = 10 1 t = t + d = 10 + (– 3) = 10 – 3 = 7 2 1 t = t + d = 7 + (– 3) = 7 – 3 = 4 3 2 t = t + d = 4 + (– 3) = 4 – 3 = 1 4 3 t = t + d = 1 + (– 3) = 1 – 3 = – 2 5 4  The first five terms of the A.P. are 10, 7, 4, 1 and – 2. (iii) a = 4, d = 0 (1 mark) Sol. a = 4, d = 0 Here, t = a = 4 1 t = t + d = 4 + 0 = 4 2 1 t = t + d = 4 + 0 = 4 3 2 t = t + d = 4 + 0 = 4 4 3 t = t + d = 4 + 0 = 4 5 4  The first five terms of the A.P. are 4, 4, 4, 4 and 4. (iv) a = 5, d = 2 (1 mark) Sol. a = 5, d = 2 Here, t = a = 5 1 t = t + d = 5 + 2 = 7 2 1 t = t + d = 7 + 2 = 9 3 2 t = t + d = 9 + 2 = 11 4 3 t = t + d = 11 + 2 = 13 5 4  The first five terms of the A.P. are 5, 7, 9, 11 and 13. (v) a = 3, d = 4 (1 mark) Sol. a = 3, d = 4 Here, t = a = 3 1 t = t + d = 3 + 4 = 7 2 1 t = t + d = 7 + 4 = 11 3 2 t = t + d = 11 + 4 = 15 4 3 t = t + d = 15 + 4 = 19 5 4  The first five terms of the A.P. are 3, 7, 11, 15 and 19. (vi) a = 6, d = 6 (1 mark) Sol. a = 6, d = 6 Here, t = a = 6 1 t = t + d = 6 + 6 = 12 2 1 t = t + d = 12 + 6 = 18 3 2 t = t + d = 18 + 6 = 24 4 3 t = t + d = 24 + 6 = 30 5 4  The first five terms of A.P. are 6, 12, 18, 24 and 30. NOTE : In an A.P. the difference between two consecutive terms is constant and it is denoted as ‘d’ and firstterm of anA.P. is denoted as ‘a’. SCHOOL SECTION 9 ALGEBRA MT EDUCARE LTD. o General term of an A.P. (t ) : n t = a = a + 0d  t = a + (1 – 1) d 1 1 t = a + d = a + 1d  t = a + (2 – 1) d 2 2 t = a + d + d = a + 2d  t = a + (3 – 1) d 3 3 t = a + d + d + d = a + 3d  t = a + (4 – 1) d 4 4 t = a + d + d + d + d = a + 4d  t = a + (5 – 1) d 5 5 In general nth term, of an A.P. t = a + (n – 1) d n EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) : 1. Find the twenty fifth term of the A. P. : 12, 16, 20, 24, ..... (2 marks) Sol. For the given A.P. 12, 16, 20, 24, ..... Here, a = t = 12 1 d = t – t = 16 – 12 = 4 2 1 We know, t = a + (n – 1) d n  t = a + (25 – 1) d 25  t = 12 + 24 (4) 25  t = 12 + 96 25  t = 108 25  The twenty fifth term of A.P. is 108. EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) : 2. Find the eighteenth term of the A. P. : 1, 7, 13, 19, ..... (2 marks) Sol. For the given A.P. 1, 7, 13, 19, ..... Here, a = t = 1 1 d = t – t = 7 – 1 = 6 2 1 We know, t = a + (n – 1) d n  t = a + (18 – 1) d 18  t = 1 + 17 (6) 18  t = 1 + 102 18  t = 103 18  Eighteenth term of A.P. is 103. PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) : 1. Find t from the following A.P. 4, 9, 14, ..... . (2 marks) 11 Sol. For the A.P. 4, 9, 14, ..... a = 4, d = 5 t = a + (n – 1) d n  t = 4 + (11 – 1) 5 11  t = 4 + 50 11  t = 54 11 EXERCISE - 1.3 (TEXT BOOK PAGE NO. 12) : 3. Find t for an Arithmetic Progression where t = 22, t = – 20. (3 marks) n 3 17 Given : For an A.P. t = 22 and t = – 20 3 17 Find : t . n Sol. t = a + (n – 1) d n t = a + (3 – 1) d 3 22 = a + 2d  a + 2d = 22 ......(i) t = a + (17 – 1) d 17 – 20 = a + 16d  a + 16d = – 20 ......(ii) 10 SCHOOL SECTION