CBSE BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions2 Marks
Question
Show that the sequence defined by $\text{a}_\text{n}=\frac{2}{3^\text{n}},\text{n}\in\text{N}$ is a G.P.
✓
Answer
$\text{a}_\text{n}=\frac{2}{3^\text{n}},\text{n}\in\text{N}$
Put n = 1, 2, 3 ... because n is natural number
$\frac{2}{3},\frac{2}{3^2},\frac{2}{3^3},\dots$
$\frac{\text{t}_3}{\text{t}_2}=\frac{\frac{2}{3^3}}{\frac{2}{3^2}}=\frac13$
$\frac{\text{t}_2}{\text{t}_1}=\frac{\frac{2}{3^2}}{\frac{2}{3}}=\frac13$
Ratio of consecutive terms is solve
$\therefore\frac{1}{3}$ is common ratio, Hence it is G.P. $\forall\text{n}\in\text{N}.$
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