The sum of the first four terms of a geometri | vedclass

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

Question

(D) Let the first four terms be $a, ar, ar^2, ar^3$.The sum of the first four terms is $S_4 = a(1+r+r^2+r^3) = a\frac{r^4-1}{r-1} = \frac{65}{12} \qua

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the first four terms be $a, ar, ar^2, ar^3$.The sum of the first four terms is $S_4 = a(1+r+r^2+r^3) = a\frac{r^4-1}{r-1} = \frac{65}{12} \quad (1)$.The sum of their reciprocals is $\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} = \frac{r^3+r^2+r+1}{ar^3} = \frac{1}{a} \frac{r^4-1}{r^3(r-1)} = \frac{65}{18} \quad (2)$.Dividing $(1)$ by $(2)$,we get $\frac{a\frac{r^4-1}{r-1}}{\frac{1}{a}\frac{r^4-1}{r^3(r-1)}} = a^2 r^3 = \frac{65/12}{65/18} = \frac{18}{12} = \frac{3}{2}$.Given the product of the first three terms is $a \cdot ar \cdot ar^2 = a^3 r^3 = 1$,which implies $ar = 1$,so $a = \frac{1}{r}$.Substituting $a = \frac{1}{r}$ into $a^2 r^3 = \frac{3}{2}$,we get $(\frac{1}{r})^2 r^3 = r = \frac{3}{2}$.Thus,$a = \frac{1}{3/2} = \frac{2}{3}$.The third term $\alpha = ar^2 = \frac{2}{3} 	imes (\frac{3}{2})^2 = \frac{2}{3} 	imes \frac{9}{4} = \frac{3}{2}$.Therefore,$2\alpha = 2 	imes \frac{3}{2} = 3$.

The sum of the first four terms of a geometric progression $(G.P.)$ is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$. If the product of the first three terms of the $G.P.$ is $1$,and the third term is $\alpha$,then $2\alpha$ is ....... .

Similar Questions

If the sum of infinite terms of a $G.P.$ is $3$ and the sum of the squares of its terms is $3$,then its first term and common ratio are:

$A$ $G.P.$ 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

The sum of the first ten terms of a geometric progression is $S_1$ and the sum of the next ten terms ($11^{th}$ to $20^{th}$) is $S_2$. What is the common ratio?

If the $4^{th}$,$7^{th}$,and $10^{th}$ terms of a Geometric Progression $(GP)$ are $a, b,$ and $c$ respectively,then:

If $\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx}$ and $x \neq 0$,then show that $a, b, c,$ and $d$ are in $G.P.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module