A G.P. consists of an even number of terms. I | vedclass

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

Question

(C) Let the $G.P.$ have $2n$ terms,with the first term $a$ and common ratio $r$.The sum of all $2n$ terms is $S_{2n} = a\frac{(r^{2n} - 1)}{(r - 1)}$.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the $G.P.$ have $2n$ terms,with the first term $a$ and common ratio $r$.The sum of all $2n$ terms is $S_{2n} = a\frac{(r^{2n} - 1)}{(r - 1)}$.The terms at odd places are $a, ar^2, ar^4, \dots, ar^{2n-2}$. This is a $G.P.$ with $n$ terms,first term $a$,and common ratio $r^2$.The sum of terms at odd places is $S_{odd} = a\frac{((r^2)^n - 1)}{(r^2 - 1)} = a\frac{(r^{2n} - 1)}{(r^2 - 1)}$.According to the problem,$S_{2n} = 5 	imes S_{odd}$.Substituting the formulas: $a\frac{(r^{2n} - 1)}{(r - 1)} = 5 	imes a\frac{(r^{2n} - 1)}{(r^2 - 1)}$.Since $r 
eq 1$ and $r^{2n} - 1 
eq 0$,we can divide both sides by $a\frac{(r^{2n} - 1)}{(r - 1)}$:$1 = \frac{5}{r + 1}$.$r + 1 = 5$.$r = 4$.

$A$ $G.P.$ consists of an even number of terms. If the sum of all the terms is $5$ times the sum of the terms occupying odd places,then the common ratio will be equal to

Similar Questions

If the variance of the terms in an increasing $A.P.$,$b_{1}, b_{2}, b_{3}, \ldots, b_{11}$ is $90$,then the common difference of this $A.P.$ is

Let $a_1, a_2, a_3$ be any positive real numbers,then which of the following statements is not true?

The average of $n$ numbers is $a$. The first number is increased by $2$,the second one is increased by $4$,the third one is increased by $8$,and so on. The average of the new numbers is:

$11^3 + 12^3 + .... + 20^3$

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$,such that $B_n > A_n$ for all $n \geq p$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module