Check whether the following sequences are geometric progressions - Geometric Progressions

part 2 - Exercise Question 1 (Page 188) - Geometric Progressions - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress - Maths Class 9
part 3 - Exercise Question 1 (Page 188) - Geometric Progressions - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress - Maths Class 9 part 4 - Exercise Question 1 (Page 188) - Geometric Progressions - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress - Maths Class 9 part 5 - Exercise Question 1 (Page 188) - Geometric Progressions - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress - Maths Class 9 part 6 - Exercise Question 1 (Page 188) - Geometric Progressions - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress - Maths Class 9 part 7 - Exercise Question 1 (Page 188) - Geometric Progressions - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress - Maths Class 9 part 8 - Exercise Question 1 (Page 188) - Geometric Progressions - Chapter 8 - Predicting What Comes Next: Exploring Sequences & Progress - Maths Class 9

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Exercise Question 1 (Page 188) (i) Check whether the following sequences are geometric progressions and find their nth terms. (i) 2, 10, 50, 250,โ€ฆ Given sequence 2, 10, 50, 250,โ€ฆ If its GP, then common ratio is same So, we need to show (๐Ÿ^๐’๐’… ๐’•๐’†๐’“๐’Ž)/(๐Ÿ^๐’”๐’• ๐’•๐’†๐’“๐’Ž)=" " (๐Ÿ‘^๐’“๐’… ๐’•๐’†๐’“๐’Ž)/(๐Ÿ^๐’๐’… ๐’•๐’†๐’“๐’Ž) Now, ๐Ÿ๐ŸŽ/๐Ÿ=๐Ÿ“๐ŸŽ/๐Ÿ๐ŸŽ 5 = 5 Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 2 Common ratio = r = 5 Now, nth term = ๐’‚_๐’ = ๐‘Ž๐‘Ÿ^(๐‘›โˆ’1) Putting values = ๐Ÿ ร— (๐Ÿ“)^(๐’โˆ’๐Ÿ) Exercise Question 1 (Page 188) (ii) Check whether the following sequences are geometric progressions and find their nth terms. (ii) 4, 8/3, 16/9, 32/27,โ€ฆ Given sequence 4, 8/3, 16/9, 32/27,โ€ฆ If its GP, then common ratio is same So, we need to show (๐Ÿ^๐’๐’… ๐’•๐’†๐’“๐’Ž)/(๐Ÿ^๐’”๐’• ๐’•๐’†๐’“๐’Ž)=" " (๐Ÿ‘^๐’“๐’… ๐’•๐’†๐’“๐’Ž)/(๐Ÿ^๐’๐’… ๐’•๐’†๐’“๐’Ž) Now, (๐Ÿ–/๐Ÿ‘)/๐Ÿ’=(๐Ÿ๐Ÿ”/๐Ÿ—)/(๐Ÿ–/๐Ÿ‘) 8/(3 ร— 4)=16/9 ร— 3/8 ๐Ÿ/๐Ÿ‘=๐Ÿ/๐Ÿ‘ Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 4 Common ratio = r = ๐Ÿ/๐Ÿ‘ Now, nth term = ๐’‚_๐’ = ๐‘Ž๐‘Ÿ^(๐‘›โˆ’1) Putting values = ๐Ÿ’ ร— (๐Ÿ/๐Ÿ‘)^(๐’โˆ’๐Ÿ) Exercise Question 1 (Page 188) (iii) Check whether the following sequences are geometric progressions and find their nth terms. (iii) 3,(โˆ’3)/2, 3/4,(โˆ’3)/8,โ€ฆ Given sequence 3,(โˆ’3)/2, 3/4,(โˆ’3)/8,โ€ฆ If its GP, then common ratio is same So, we need to show (๐Ÿ^๐’๐’… ๐’•๐’†๐’“๐’Ž)/(๐Ÿ^๐’”๐’• ๐’•๐’†๐’“๐’Ž)=" " (๐Ÿ‘^๐’“๐’… ๐’•๐’†๐’“๐’Ž)/(๐Ÿ^๐’๐’… ๐’•๐’†๐’“๐’Ž) Now, ((โˆ’๐Ÿ‘)/๐Ÿ)/๐Ÿ‘=(๐Ÿ‘/๐Ÿ’)/((โˆ’๐Ÿ‘)/๐Ÿ) (โˆ’3)/(2 ร— 3)=3/4 ร— 2/(โˆ’3) (โˆ’๐Ÿ)/๐Ÿ=(โˆ’๐Ÿ)/๐Ÿ Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 3 Common ratio = r = (โˆ’๐Ÿ)/๐Ÿ Now, nth term = ๐’‚_๐’ = ๐‘Ž๐‘Ÿ^(๐‘›โˆ’1) Putting values = ๐Ÿ‘ ร— ((โˆ’๐Ÿ)/๐Ÿ)^(๐’โˆ’๐Ÿ)

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