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 involving odd powers of $r$ are $ar, ar^3, a

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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 involving odd powers of $r$ are $ar, ar^3, ar^5, \dots$. This is a new geometric series with first term $A = ar$ and common ratio $R = r^2$.The sum of this series is given as $3$,so $\frac{ar}{1-r^2} = 3 \quad \dots(2)$.From $(1)$,we have $a = 7(1-r)$. Substituting this into $(2)$:$\frac{7(1-r)r}{(1-r)(1+r)} = 3$$\frac{7r}{1+r} = 3 \Rightarrow 7r = 3 + 3r \Rightarrow 4r = 3 \Rightarrow r = \frac{3}{4}$.Substituting $r = \frac{3}{4}$ into $(1)$:$a = 7(1 - \frac{3}{4}) = 7(\frac{1}{4}) = \frac{7}{4}$.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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