If a -1- b - b 2- b 3- to - then write b in terms of a given that - b -1

A = 1 + b + b^2 + b^3 + ... ∞ a = 1/1 b [Since, S_∞ = a/ 1 r] a(1 b) = 1 a ab = 1 ab = a 1 b = a 1/a ...

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Standard XII

Mathematics

Sum of Infinite Terms of a GP

If a =1+ b + ...

Question

If $a = 1 + b + b^{2} + b^{3} + . . .$ to $\infty$ , then write b in terms of a given that $| b | < 1$ .

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Solution

$a = 1 + b + b^{2} + b^{3} + . . . \infty$

$a = \frac{1}{1 - b}$ $[Since, S_{\infty} = \frac{a}{1 - r}]$

$a (1 - b) = 1$

$a - a b = 1$

$a b = a - 1$

$b = \frac{a - 1}{a}$

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If a=1+b+b2+b3+... to ∞, then write b in terms of a given that |b|<1.

a=1+b+b2+b3+...∞ a=11−b [Since, S∞=a1−r] a(1−b)=1 a−ab=1 ab=a−1 b=a−1a

If $a = 1 + b + b^{2} + b^{3} + . . .$ to $\infty$ , then write b in terms of a given that $| b | < 1$ .

$a = 1 + b + b^{2} + b^{3} + . . . \infty$

$a = \frac{1}{1 - b}$ $[Since, S_{\infty} = \frac{a}{1 - r}]$

$a (1 - b) = 1$

$a - a b = 1$

$a b = a - 1$

$b = \frac{a - 1}{a}$