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Question

Consider the ten numbers $$ar,\,a{r^2},\,ar3,\,..........a{r^{10}}$$.
If their sum is $$18$$ and the sum of their reciprocal is $$6$$ then the product of these ten numbers, is 


A
35
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B
38
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C
310
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D
315
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Solution

The correct option is B $$3^5$$

$$ar,ar^2,ar^3,............ar^{10}$$ is a GP with first term as $$ar$$ and common ration as $$r$$

$$=>sum_1=ar+ar^2+ar^3=.......ar+{10}=18$$

$$=>sum_2=\dfrac{1}{ar}+\dfrac{1}{ar^2}+...........+\dfrac{1}{ar{10}}=6$$

multiply $$sum_2$$ by $$a^2r{11}$$

i.e.,

$$=>ar+ar^2+ar^3+..........+ar^{10}=6\times a^2r^{11}$$

$$=>sum_1=6a^2r^{10}$$

$$=>18=6a^2r^{10}$$

$$=>a^2r^{11}=3$$

Now product of the ratio terms of the GP

$$=>a^{10}r^{1+2+3+......+10}$$

$$=>a^{10}r^{10\times 11}$$

$$=>a^{10}r^{55}={(a^2r^{11})}^5$$

$$=>3^5$$


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