CAT 2006 — QA Question 5

${\mathrm{t}}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}+2}$

$\therefore {\mathrm{t}}_3=\frac{3}{3+2}=\frac{3}{5}$

$t_4=\frac{4}{4+2}=\frac{4}{6}$

$t_5=\frac{5}{5+2}=\frac{5}{7}$

...

$t_{51}=\frac{51}{53}$

$t_{52}=\frac{52}{54}$

$t_{53}=\frac{53}{55}$

∴ t₃ × t₄ × t₅ × ... × t₅₃ = $\frac{3}{5}$ × $\frac{4}{6}$ × $\frac{5}{7}$ × ... × $\frac{51}{53}$ × $\frac{52}{54}$ × $\frac{53}{55}$

$=\frac{(3\times 4)}{(54\times 55)}=\frac{2}{495}$

Hence, option (a).