Geometric Progressions
Last updated at August 14, 2026 by Teachoo
Transcript
Exercise Question 1 (Page 188) (i) Check whether the following sequences are geometric progressions and find their nth terms. (i) 2, 10, 50, 250,โฆ Given sequence 2, 10, 50, 250,โฆ If its GP, then common ratio is same So, we need to show (๐^๐๐ ๐๐๐๐)/(๐^๐๐ ๐๐๐๐)=" " (๐^๐๐ ๐๐๐๐)/(๐^๐๐ ๐๐๐๐) Now, ๐๐/๐=๐๐/๐๐ 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 = ๐๐^(๐โ1) Putting values = ๐ ร (๐)^(๐โ๐) 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,โฆ Given sequence 4, 8/3, 16/9, 32/27,โฆ If its GP, then common ratio is same So, we need to show (๐^๐๐ ๐๐๐๐)/(๐^๐๐ ๐๐๐๐)=" " (๐^๐๐ ๐๐๐๐)/(๐^๐๐ ๐๐๐๐) Now, (๐/๐)/๐=(๐๐/๐)/(๐/๐) 8/(3 ร 4)=16/9 ร 3/8 ๐/๐=๐/๐ Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 4 Common ratio = r = ๐/๐ Now, nth term = ๐_๐ = ๐๐^(๐โ1) Putting values = ๐ ร (๐/๐)^(๐โ๐) Exercise Question 1 (Page 188) (iii) Check whether the following sequences are geometric progressions and find their nth terms. (iii) 3,(โ3)/2, 3/4,(โ3)/8,โฆ Given sequence 3,(โ3)/2, 3/4,(โ3)/8,โฆ If its GP, then common ratio is same So, we need to show (๐^๐๐ ๐๐๐๐)/(๐^๐๐ ๐๐๐๐)=" " (๐^๐๐ ๐๐๐๐)/(๐^๐๐ ๐๐๐๐) Now, ((โ๐)/๐)/๐=(๐/๐)/((โ๐)/๐) (โ3)/(2 ร 3)=3/4 ร 2/(โ3) (โ๐)/๐=(โ๐)/๐ Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 3 Common ratio = r = (โ๐)/๐ Now, nth term = ๐_๐ = ๐๐^(๐โ1) Putting values = ๐ ร ((โ๐)/๐)^(๐โ๐)