Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSGeometric Progressions3 Marks
Question
Insert 6 geometric means between 27 and $\frac{1}{81}.$
✓
Answer
6 Geometric means between 27 and $\frac{1}{81}.$ Let G1, G2, G3, G4, G5, G6 be 6 geometric means between a = 27 and $\text{b}=\frac{1}{81}$ Then, 27, G1, G2, G3, G4, G5, G6, $\frac{1}{81}$ is a G.P. with common ratio r given by $\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{\text{n}+1}}$ $=\Bigg(\frac{\frac{1}{81}}{27}\Bigg)^{\frac{1}{6+1}}=\Big(\frac{1}{81\times27}\Big)^{\frac{1}{7}}=\Big(\frac{1}{3^7}\Big)^{\frac{1}{7}}$ $\therefore\text{G}_1=\text{ar}=27\times\Big(\frac13\Big)=9$ $\text{G}_2=\text{ar}^2=27\times\frac{1}{9}=3$ $\text{G}_3=\text{ar}^3=27\times\frac{1}{27}=1$ $\text{G}_4=\text{ar}^4=27\times\frac{1}{27\times3}=\frac13$ $\text{G}_5=\text{ar}^5=27\times\frac{1}{{3}^5}=\frac{1}{9}$ $\text{G}_6=\text{ar}^6=27\times\frac{1}{36}=\frac{1}{27}$ Hence, $9, 3, 1, \frac{1}{3},\frac19,\frac{1}{27}$ are 6 geometric means between 27 and $\frac{1}{81}.$
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