MCQ 11 Mark
Assertion $(A):$ If the numbers $\frac{-2}{7}, K, \frac{-7}{2}$ are in $GP,$ then $k = \pm 1$.
Reason $(R):$ If $a_1, a_2, a_3$ are in $GP,$ then $\frac{a_2}{a_1}=\frac{a_3}{a_2}$.
Reason $(R):$ If $a_1, a_2, a_3$ are in $GP,$ then $\frac{a_2}{a_1}=\frac{a_3}{a_2}$.
- ✓Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A.$
- C$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Answer
View full question & answer→Correct option: A.
Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
Assertion: If $\frac{-2}{7}, K, \frac{-7}{2}$ are in $G.P.$
Then, $\frac{a_2}{a_1}=\frac{a_3}{a_2}$
$\left[\because\right.$ common ratio $( r )=\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\ldots]$
$\therefore \frac{k}{\frac{-2}{7}}=\frac{\frac{-7}{2}}{k}$
$\Rightarrow \frac{7}{-2} k=\frac{-7}{2} \times \frac{1}{k}$
$\Rightarrow 7 k \times 2 k =-7 \times(-2)$
$\Rightarrow 14 k ^2=14$
$\Rightarrow k ^2=1$
$\Rightarrow k = \pm 1$
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.
Then, $\frac{a_2}{a_1}=\frac{a_3}{a_2}$
$\left[\because\right.$ common ratio $( r )=\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\ldots]$
$\therefore \frac{k}{\frac{-2}{7}}=\frac{\frac{-7}{2}}{k}$
$\Rightarrow \frac{7}{-2} k=\frac{-7}{2} \times \frac{1}{k}$
$\Rightarrow 7 k \times 2 k =-7 \times(-2)$
$\Rightarrow 14 k ^2=14$
$\Rightarrow k ^2=1$
$\Rightarrow k = \pm 1$
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.