Let mathrm{a} and mathrm{b} be be two distinc | vedclass

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

Question

$ 1^{	ext {st }} G P  \Rightarrow t_1=a, t_3=b=a r^2 \Rightarrow r^2=\frac{b}{a} $

$ t_{11}  =a r^{10}=a\left(r^2ight)^5=a \cdot\left(

Vedclass · Accepted Answer

$21$

Vedclass · Answer

$ 1^{	ext {st }} G P  \Rightarrow t_1=a, t_3=b=a r^2 \Rightarrow r^2=\frac{b}{a} $

$ t_{11}  =a r^{10}=a\left(r^2ight)^5=a \cdot\left(\frac{b}{a}ight)^5 $

$2^{	ext {nd }} 	ext { G.P. }  \Rightarrow T_1=a, T_5=a r^4=b $

$\Rightarrow r^4  =\left(\frac{b}{a}ight) \Rightarrow r=\left(\frac{b}{a}ight)^{1 / 4} $

$ T_p  =a r^{p-1}=a\left(\frac{b}{a}ight)^{\frac{p-1}{4}} $

$ t_{11}  =T_p \Rightarrow a\left(\frac{b}{a}ight)^5=a\left(\frac{b}{a}ight)^{\frac{p-1}{4}} $

$ \Rightarrow 5  =\frac{p-1}{4} \Rightarrow p=21$

Let $\mathrm{a}$ and $\mathrm{b}$ be be two distinct positive real numbers. Let $11^{\text {th }}$ term of a $GP$, whose first term is $a$ and third term is $b$, is equal to $p^{\text {th }}$ term of another $GP$, whose first term is $a$ and fifth term is $b$. Then $\mathrm{p}$ is equal to

Similar Questions

The terms of a $G.P.$ are positive. If each term is equal to the sum of two terms that follow it, then the common ratio is

If ${a^2} + a{b^2} + 16{c^2} = 2(3ab + 6bc + 4ac)$, where $a,b,c$ are non-zero numbers. Then $a,b,c$ are in

If three successive terms of a$G.P.$ with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[\mathrm{r}]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to :

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

The sum of a $G.P.$ with common ratio $3$ is $364$, and last term is $243$, then the number of terms is