The geometric series a + ar + ar^2 + ar^3 + d | vedclass

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(B) The sum of the infinite geometric series is given by $S = \frac{a}{1-r} = 7 \quad \dots(1)$The terms with odd powers of $r$ are $ar, ar^3, ar^5, \

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$\frac{5}{2}$

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(B) The sum of the infinite geometric series is given by $S = \frac{a}{1-r} = 7 \quad \dots(1)$The terms with odd powers of $r$ are $ar, ar^3, ar^5, \dots$,which form a geometric series with first term $ar$ and common ratio $r^2$.The sum of this series is $\frac{ar}{1-r^2} = 3 \quad \dots(2)$Dividing $(2)$ by $(1)$:$\frac{ar}{1-r^2} \div \frac{a}{1-r} = \frac{3}{7}$$\frac{ar}{(1-r)(1+r)} 	imes \frac{1-r}{a} = \frac{3}{7}$$\frac{r}{1+r} = \frac{3}{7}$$7r = 3 + 3r$ $\Rightarrow 4r = 3$ $\Rightarrow r = \frac{3}{4}$Substitute $r = \frac{3}{4}$ into $(1)$:$\frac{a}{1 - 3/4} = 7$ $\Rightarrow \frac{a}{1/4} = 7$ $\Rightarrow a = \frac{7}{4}$Now,calculate $(a^2 - r^2)$:$a^2 - r^2 = (\frac{7}{4})^2 - (\frac{3}{4})^2 = \frac{49}{16} - \frac{9}{16} = \frac{40}{16} = \frac{5}{2}$

The geometric series $a + ar + ar^2 + ar^3 + \dots \infty$ has a sum of $7$,and the sum of the terms involving odd powers of $r$ is $3$. Then the value of $(a^2 - r^2)$ is -

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