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Given that $r,s$ and $t$ are consecutive odd numbers and $r<s<t.$

We can take $r=x,s = x+2$ and $t = x+4$

Take option one by one and verified.

A.$rs = t$

$\implies x(x+2) = x+ 4$

$\implies x^{2} + 2x = x+ 4\:\text{(False)}$

B.$r+t = 2t-s$

$\implies x + x + 4 = 2(x+ 4) –(x+2)$

$\implies 2x + 4 = 2x+ 8 –x-2$

$\implies 2x + 4 = x + 6\:\text{(False)}$

C.$r+s = t-2$

$\implies x + x + 2 = x+ 4 – 2$

$\implies 2x + 2 = x + 2\:\text{(False)}$

D.$r + t = 2s$

$\implies x+ x+ 4 = 2(x+2)$

$\implies 2x + 4 = 2x + 4\:\textbf{(True)}$

$\textbf{Shortcut Method:}$ We can take $r = 1,s = 3$ and $t = 5$

So, the correct answer is $(D).$

We can take $r=x,s = x+2$ and $t = x+4$

Take option one by one and verified.

A.$rs = t$

$\implies x(x+2) = x+ 4$

$\implies x^{2} + 2x = x+ 4\:\text{(False)}$

B.$r+t = 2t-s$

$\implies x + x + 4 = 2(x+ 4) –(x+2)$

$\implies 2x + 4 = 2x+ 8 –x-2$

$\implies 2x + 4 = x + 6\:\text{(False)}$

C.$r+s = t-2$

$\implies x + x + 2 = x+ 4 – 2$

$\implies 2x + 2 = x + 2\:\text{(False)}$

D.$r + t = 2s$

$\implies x+ x+ 4 = 2(x+2)$

$\implies 2x + 4 = 2x + 4\:\textbf{(True)}$

$\textbf{Shortcut Method:}$ We can take $r = 1,s = 3$ and $t = 5$

So, the correct answer is $(D).$