Question
Insert 5 geometric means between 16 and $\frac{1}{4}.$
Let G1, G2, G3, G4, G5 be 5 geometric means between a = 16 and
$\text{b}=\frac{1}{4}$Then, 16, G1, G2, G3, G4, G5
$\frac{1}{4}$ is a G.P. with a = 16, $\text{b}=\frac{1}{4}$$\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{\text{n}+1}}$
$=\Bigg(\frac{\frac{1}{4}}{16}\Bigg)^{\frac{1}{5+1}}=\Big(\frac{1}{26}\Big)^{\frac{1}{6}}=\Big(\frac{1}{2}\Big)$
$\therefore\text{G}_1=\text{ar}=16\times\Big(\frac12\Big)=8$
$\text{G}_2=\text{ar}^2=16\times\frac{1}{4}=4$
$\text{G}_3=\text{ar}^3=16\times\frac{1}{8}=2$
$\text{G}_4=\text{ar}^4=16\times\frac{1}{16}=1$
$\text{G}_5=\text{ar}^5=16\times\frac{1}{{2}^5}=\frac{1}{2}$
Hence,
$8,4,2,1,\frac{1}{2}$ are 5 geometric means between 16 and $\frac{1}{4}.$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Age | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |
| Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
[Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]