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InterviewSolution
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There are two sets A and B each of which consists of three numbers in GP whose product is 64 and R and r are the common ratios sich that `R=r+2`. If `(p)/(q)=(3)/(2)`, where p and q are sum of numbers taken two at a time respectively in the two sets. The value of `r^(R)+R^(r)` isA. 5392B. 368C. 32D. 4 |
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Answer» Correct Answer - C Let `A={(A)/(R),A,AR}` `B={(a)/(r ),a,ar}` According to the question, `(A)/(R )*A*AR=64` `implies A^(3)=64 " "implies A=4" " "……..(i)"` `(a)/(r )*a*ar=64" " implies a^(3)=64 " " implies a=4 " " "……(ii)"` and `R=r+2" " "……..(iii)"` `p=(A)/(R)*A*AR+AR*(A)/(R)` `=(A^(2))/(R)+A^(2)R+A^(2)=(16)/(R )+16R+16` `q=(a)/(r )*a+a*ar+ar*(a)/( r)` `=(a^(2))/(r )+a^(2)r+a(^2)=(16)/(r )=(16)/(r )+16r+16` Give that, `(p)/(q)=(3)/(2)` So, `((16+16R^(2)+16R)r)/((16+16r^(2)+16r)R)=(3)/(2)` `((1+R^(2)+R)r)/((1+r^(2)+r)R)=(3)/(2)` From Eq. (iii), `R=r+2` `implies ((1+r^(2)+4+4r+r+2)r)/((1+r+r^(2))(r+2))=(3)/(2)` `implies (r^(3)+5r^(2)+7r)/(r^(3)+3r^(2)+3r+2)=(3)/(2)` `implies r^(3)-r^(2)-5r+6=0` `implies (r-2)(r^(2)+r-3)=0` `impliesr=2" or " r=(-1pmsqrt(13))/(2)` So, `R=4` `r^(R)+R^(r )=(4)^(2)+(2)^(4)=16+16=32` |
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