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  • Three numbers are in an increasing geometric progression with common ratio . If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference

Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r^{2}$, then $r^{2}-d is equal to :
Option: 1 7-\sqrt{3}
Option: 2 7+3 \sqrt{3}
Option: 3 7-7 \sqrt{3}
Option: 4 7+\sqrt{3}

Answers (1)

best_answer

T_{4}= 3r^{2}
\Rightarrow T_{3}= 3r,T_{2}= 3,T_{1}= \frac{3}{r}
Given \; \; \; 2T_{2}= T_{1}+T_{3}
\Rightarrow 2\times 6= 3r+\frac{3}{r}
\Rightarrow r^{2}-4r+1= 0
\Rightarrow r= \frac{4\pm \sqrt{16-4}}{2}= 2\pm \sqrt{3}
But since it is an increasing  A.P,
r= 2+\sqrt{3}
so T_{1}= \frac{3}{2+\sqrt{3}}= 3\left ( 2-\sqrt{3} \right )= 6-3\sqrt{3}
2T_{2}= 6
so\: d= 6-\left ( 6-3\sqrt{3} \right )= 3\sqrt{3}
r^{2}= \left ( 2+\sqrt{3} \right )^{2}= 7+4\sqrt{3}
r^{2}-d= \ 7+4\sqrt{3}-3\sqrt{3}= 7+\sqrt{3}
option (4)

Posted by

Kuldeep Maurya

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