Exercise Question 1 (Page 188)

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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,… Given sequence 2, 10, 50, 250,… If its GP, then common ratio is same So, we need to show (𝟐^𝒏𝒅 𝒕𝒆𝒓𝒎)/(𝟏^𝒔𝒕 𝒕𝒆𝒓𝒎)=" " (𝟑^𝒓𝒅 𝒕𝒆𝒓𝒎)/(𝟐^𝒏𝒅 𝒕𝒆𝒓𝒎) Now, 𝟏𝟎/𝟐=𝟓𝟎/𝟏𝟎 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 = 𝒂_𝒏 = 𝑎𝑟^(𝑛−1) Putting values = 𝟐 × (𝟓)^(𝒏−𝟏) 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,… Given sequence 4, 8/3, 16/9, 32/27,… If its GP, then common ratio is same So, we need to show (𝟐^𝒏𝒅 𝒕𝒆𝒓𝒎)/(𝟏^𝒔𝒕 𝒕𝒆𝒓𝒎)=" " (𝟑^𝒓𝒅 𝒕𝒆𝒓𝒎)/(𝟐^𝒏𝒅 𝒕𝒆𝒓𝒎) Now, (𝟖/𝟑)/𝟒=(𝟏𝟔/𝟗)/(𝟖/𝟑) 8/(3 × 4)=16/9 × 3/8 𝟐/𝟑=𝟐/𝟑 Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 4 Common ratio = r = 𝟐/𝟑 Now, nth term = 𝒂_𝒏 = 𝑎𝑟^(𝑛−1) Putting values = 𝟒 × (𝟐/𝟑)^(𝒏−𝟏) Exercise Question 1 (Page 188) (iii) Check whether the following sequences are geometric progressions and find their nth terms. (iii) 3,(−3)/2, 3/4,(−3)/8,… Given sequence 3,(−3)/2, 3/4,(−3)/8,… If its GP, then common ratio is same So, we need to show (𝟐^𝒏𝒅 𝒕𝒆𝒓𝒎)/(𝟏^𝒔𝒕 𝒕𝒆𝒓𝒎)=" " (𝟑^𝒓𝒅 𝒕𝒆𝒓𝒎)/(𝟐^𝒏𝒅 𝒕𝒆𝒓𝒎) Now, ((−𝟑)/𝟐)/𝟑=(𝟑/𝟒)/((−𝟑)/𝟐) (−3)/(2 × 3)=3/4 × 2/(−3) (−𝟏)/𝟐=(−𝟏)/𝟐 Since common ratio is same, it is a GP Now, we need to find nth term Here, First term = a = 3 Common ratio = r = (−𝟏)/𝟐 Now, nth term = 𝒂_𝒏 = 𝑎𝑟^(𝑛−1) Putting values = 𝟑 × ((−𝟏)/𝟐)^(𝒏−𝟏)