Insert two numbers between 3 and 81 so that t | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Let $G_{1}$ and $G_{2}$ be two numbers between $3$ and $81$ such that the series, $3, G_{1}, G_{2}, 81,$ forms a $G.P.$

Let $a$ be the first term a

Vedclass · Accepted Answer

$9,27$

Vedclass · Answer

Let $G_{1}$ and $G_{2}$ be two numbers between $3$ and $81$ such that the series, $3, G_{1}, G_{2}, 81,$ forms a $G.P.$

Let $a$ be the first term and $r$ be the common ratio of the $G.P.$

$	herefore 81=(3)(r)^{3}$

$\Rightarrow r^{3}=27$

$	herefore r=3$ (Talking real roots only)

For $r=3$

$G_{1}=a r=(3)(3)=9$

$G_{2}=a r^{2}=(3)(3)^{2}=27$

Thus, the required two numbers are $9$ and $27$

Insert two numbers between $3$ and $81$ so that the resulting sequence is $G.P.$

Similar Questions

Let $S_1$ be the sum of areas of the squares whose sides are parallel to coordinate axes. Let $S_2$ be the sum of areas of the slanted squares as shown in the figure. Then, $\frac{S_1}{S_2}$ is equal to

The number which should be added to the numbers $2, 14, 62$ so that the resulting numbers may be in $G.P.$, is

If $2^{10}+2^{9} \cdot 3^{1}+28 \cdot 3^{2}+\ldots+2 \cdot 3^{9}+3^{10}=S -211$ then $S$ is equal to

The interior angle of a $'n$' sided convex polygon are in $G.P$.. The smallest angle is $1^o $ and common ratio is $2^o $ then number of possible values of $'n'$ is

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals