Question

If four positive quantities are in continued
proportion, show that the difference between the
first and last is at least three times as greater as
the difference between the other two.​

Answer 1

Answer:

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Answer 2

• Let, the four quantities be a, b, c, d.

                           $\frac{a}{b} =\frac{b}{c} =\frac{c}{d} =\frac{1}{r}$

• We know ,

              $b=ar, c=ar^{2} , d=ar^{3}$

• Therefore by using formula,

           $\frac{d-a}{c-b} =\frac{ar^{3-a} }{ar^{2}-ar }$

                  $=\frac{r^{3} -1}{r(r-1)}$

                  $=\frac{(r-1)(r^{2}+r+1) }{r(r-1)}$

                  $=\frac{r^{2}+r+1 }{r}$

                 $=\frac{r^{2}-2r+1+3r }{r}$

                  $=\frac{3r+(r-1)^{2} }{r}$

                 $=3+\frac{(r-1)^{2} }{r}$

          ∴$\frac{d-a}{c-b} \geq$ 3

Hence proved!