Maharashtra 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 $G_1, G_2, G_3, G_4, G_5, G_6$ be 6 geometric means between $a =27$ and $b=\frac{1}{81}$ Then, $27, G_1, G_2, G_3, G_4, G_5, G_6, \frac{1}{81}$ is a G.P. with common ratio $r$ given by
$r=\left(\frac{b}{a}\right)^{\frac{1}{a_1}}$
$=\left(\frac{\frac{1}{81}}{27}\right)^{\frac{1}{6+1}}=\left(\frac{1}{81 \times 27}\right)^{\frac{1}{7}}=\left(\frac{1}{3^7}\right)^{\frac{1}{7}}$
$\therefore G_1=ar=27 \times\left(\frac{1}{3}\right)=9$
$G_2=ar^2=27 \times \frac{1}{9}=3$
$G_3=ar^3=27 \times \frac{1}{27}=1$
$G_4=ar^4=27 \times \frac{1}{27 \times 3}=\frac{1}{3}$
$G_5=ar^5=27 \times \frac{1}{3^5}=\frac{1}{9}$
$G_6=ar^6=27 \times \frac{1}{36}=\frac{1}{27}$
Hence, $9,3,1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}$ are $6$ geometric means between $27$ and $\frac{1}{81}$.
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