Let b_1, b_2, dots, b_n be a geometric sequen | vedclass

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

Question

(C) Let the first term be $b_1$ and the common ratio be $r$.Given $b_1 + b_2 = 1$,we have $b_1 + b_1 r = 1$,which implies $b_1(1 + r) = 1$,so $b_1 = \

Vedclass · Accepted Answer

$2 + \sqrt{2}$

Vedclass · Answer

(C) Let the first term be $b_1$ and the common ratio be $r$.Given $b_1 + b_2 = 1$,we have $b_1 + b_1 r = 1$,which implies $b_1(1 + r) = 1$,so $b_1 = \frac{1}{1 + r}$.The sum of an infinite geometric series is given by $S = \frac{b_1}{1 - r} = 2$.Substituting $b_1 = \frac{1}{1 + r}$ into the sum formula: $\frac{1}{(1 + r)(1 - r)} = 2$.This simplifies to $\frac{1}{1 - r^2} = 2$,which means $1 - r^2 = \frac{1}{2}$,so $r^2 = \frac{1}{2}$,giving $r = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.Since $b_2 = b_1 r  0$,which contradicts $b_2 If $r = -\frac{\sqrt{2}}{2}$,then $b_1 = \frac{1}{1 - \frac{\sqrt{2}}{2}} = \frac{2}{2 - \sqrt{2}} = \frac{2(2 + \sqrt{2})}{4 - 2} = 2 + \sqrt{2}$.Then $b_2 = b_1 r = (2 + \sqrt{2})(-\frac{\sqrt{2}}{2}) = -\sqrt{2} - 1 Thus,$b_1 = 2 + \sqrt{2}$.

Let $b_1, b_2, \dots, b_n$ be a geometric sequence such that $b_1 + b_2 = 1$ and $\sum_{k=1}^{\infty} b_k = 2$. Given that $b_2 < 0$,then the value of $b_1$ is:

Similar Questions

The $n^{th}$ term of the series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

Let $a, b, c \in R$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy$ for all $x, y \in R$,then $\sum_{n=1}^{10} f(n)$ is equal to:

The sum of three numbers in a $GP$ is $105$ and their product is $8000$. Find the numbers.

$(1^{2}+2^{2}+3^{2}+\cdots+10^{2})$ is equal to:

$A$ number is the reciprocal of the other. If the arithmetic mean of the two numbers is $\frac{13}{12}$,then the numbers are

Vedclass Test Series

Exam Paper Generator

Online Exam Module