If the 4^{th}, 7^{th} and 10^{th} terms of a | vedclass

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

Question

(C) Let the first term of the $G.P.$ be $A$ and the common ratio be $r$.We know that the $n^{th}$ term of a $G.P.$ is given by $T_n = A r^{n-1}$.Given

Vedclass · Accepted Answer

$b^2 = ac$

Vedclass · Answer

(C) Let the first term of the $G.P.$ be $A$ and the common ratio be $r$.We know that the $n^{th}$ term of a $G.P.$ is given by $T_n = A r^{n-1}$.Given terms are:$T_4 = a = A r^3$$T_7 = b = A r^6$$T_{10} = c = A r^9$Now,consider the product $ac$:$ac = (A r^3) 	imes (A r^9) = A^2 r^{12}$Also,consider $b^2$:$b^2 = (A r^6)^2 = A^2 r^{12}$Since $b^2 = A^2 r^{12}$ and $ac = A^2 r^{12}$,it follows that $b^2 = ac$.Therefore,$a, b, c$ are in $G.P.$

If the $4^{th}, 7^{th}$ and $10^{th}$ terms of a $G.P.$ are $a, b, c$ respectively,then the relation between $a, b, c$ is

Similar Questions

The sum of the series $\frac{3}{1! + 2! + 3!} + \frac{4}{2! + 3! + 4!} + \frac{5}{3! + 4! + 5!} + \dots + \frac{2008}{2006! + 2007! + 2008!}$ is equal to

If $\frac{3 + 5 + 7 + \dots \text{ to } n \text{ terms}}{5 + 8 + 11 + \dots \text{ to } 10 \text{ terms}} = 7$,then the value of $n$ is

The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is

If $a, b, c, d$ are in $H.P.$,then

Find the sum of all odd numbers of four digits which are divisible by $9.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module