Gujarati
8. Sequences and Series
medium

यदि $x = 1 + a + {a^2} + ....\infty ,\,(a < 1)$ $y = 1 + b + {b^2}.......\infty ,\,(b < 1)$

तब  $1 + ab + {a^2}{b^2} + ..........\infty $ का मान होगा

A

$\frac{{xy}}{{x + y - 1}}$

B

$\frac{{xy}}{{x + y + 1}}$

C

$\frac{{xy}}{{x - y - 1}}$

D

$\frac{{xy}}{{x - y + 1}}$

Solution

(a) चूंकि दी गयी श्रेणी, गुणोत्तर श्रेणी में है, तब

$x = \frac{1}{{1 – a}} $

$\Rightarrow a = \frac{{x – 1}}{x}$ एवं $y = \frac{1}{{1 – b}} $

$\Rightarrow b = \frac{{y – 1}}{y}$

$\therefore $$1 + ab + {a^2}{b^2} + ……….\infty  = \frac{1}{{1 – ab}}$

 $ = \frac{1}{{1 – \frac{{x – 1}}{x}.\frac{{y – 1}}{y}}} = \frac{{xy}}{{x + y – 1}}$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.