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 sum $11^{2} + 12^{2} + \cdots + 20^{2} + 21^{2} = ?$

The sum of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ up to $n$ terms will be:

If the harmonic mean between $a$ and $b$ is $H$,then $\frac{H + a}{H - a} + \frac{H + b}{H - b} = $

The odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:

If $a_1, a_2, a_3, \dots, a_n$ are in $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 114$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module