Let b_1, b_2,......, b_n be a geometric seque | vedclass

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Question

$b_{1}+b_{2}=1 \Rightarrow b_{1}(1+r)=1 \Rightarrow b_{1}=\frac{1}{1+r}$

$\sum\limits_{k = 1}^\infty  {{b_k} = } \frac{1}{{(1 + r)(1 - r)}} =

Vedclass · Accepted Answer

$2 + \sqrt 2 $

Vedclass · Answer

$b_{1}+b_{2}=1 \Rightarrow b_{1}(1+r)=1 \Rightarrow b_{1}=\frac{1}{1+r}$

$\sum\limits_{k = 1}^\infty  {{b_k} = } \frac{1}{{(1 + r)(1 - r)}} = \frac{1}{{1 - {r^2}}} = 2$

$ \Rightarrow r = \frac{{ - \sqrt 2 }}{2}$

$b_{1}=\frac{1}{1+r}=\frac{1}{1-\frac{\sqrt{2}}{2}}=(2+\sqrt{2})$

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

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