Question

If the sequence $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$is an A.P with common difference 3, then a/b = ________.

Answer 1

The given A.P. is: $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$

Common difference (d) = a₂ − a₁
$$\begin{gathered} \Rightarrow 3=\frac{3a-3b}{b}-\frac{a-3b}{b} \\ \Rightarrow 3=\frac{3a-3b-a+3b}{b} \\ \Rightarrow 3=\frac{2a}{b} \\ \Rightarrow 3b=2a \\ \Rightarrow 2a=3b \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2b,\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{2a}{2b}=\frac{3b}{2b} \\ \Rightarrow \frac{a}{b}=\frac{3}{2} \end{gathered}$$

Hence, if the sequence $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$is an A.P with common difference 3, then a/b = $\underline{\frac{3}{2}}$.