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
B
38
C
310
D
315

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$$=>a^{10}r^{1+2+3+......+10}$$$$=>a^{10}r^{10\times 11}$$$$=>a^{10}r^{55}={(a^2r^{11})}^5$$$$=>3^5$$Maths

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