Question

If a₁, a₂, a₃ ..........a₁₂ are in A.P. and

D₁= $\left| \begin{matrix} a_{1}a_{5} & a_{1} & a_{2} \\ a_{2}a_{6} & a_{2} & a_{3} \\ a_{3}a_{7} & a_{3} & a_{4} \end{matrix} \right|$, D₂ = $\left| \begin{matrix} a_{2}a_{10} & a_{2} & a_{3} \\ a_{3}a_{11} & a_{3} & a_{4} \\ a_{4}a_{12} & a_{4} & a_{5} \end{matrix} \right|$ Then D₁ : D₂ =

• A. 1 : 2
• B. 1 : 1
• C. 2 : 1
• D. None

Answer 1

Answer: (B)

1 : 1

Answer 2

D₁ = $\left| \begin{matrix} a_{1}(a_{1} + 4d) & a_{1} & d \\ a_{2}(a_{2} + 4d) & a_{2} & d \\ a_{3}(a_{3} + 4d) & a_{3} & d \end{matrix} \right|$

= $\left| \begin{matrix} a_{1}^{2} & a_{1} & d \\ a_{2}^{2} & a_{2} & d \\ a_{3}^{2} & a_{3} & d \end{matrix} \right|$ + 4d $\left| \begin{matrix} a_{1} & a_{1} & d \\ a_{2} & a_{2} & d \\ a_{3} & a_{3} & d \end{matrix} \right|$

= $\left| \begin{matrix} a_{1}^{2} & a_{1} & d \\ a_{2}^{2}–a_{1}^{2} & a_{2}–a_{1} & 0 \\ a_{3}^{2}–a_{2}^{2} & a_{3}–a_{2} & 0 \end{matrix} \right|$ + 0 $\left\lbrack \begin{aligned} & R_{2} \rightarrow R_{2}–R_{1} \\ & R_{3} \rightarrow R_{3}–R_{2} \end{aligned} \right\rbrack$

= d² $\left| \begin{matrix} a_{1}^{2} & a_{1} & d \\ a_{2} + a_{1} & 1 & 0 \\ a_{3} + a_{2} & 1 & 0 \end{matrix} \right|$ [Qa₂ –a₁ = a₃ – a₂ = d]

= – d³ (a₃ – a₁) = – 2d⁴

Similarly D₂ = – 2d⁴ \ D₁ : D₂ = 1 : 1