MCQ 11 Mark
Assertion (A): If 7th, 10th and 13th term of a G.P. are $\mathrm{a}, \mathrm{b}$ and c respectively, then $\mathrm{b}_{2}=\mathrm{ac}$.
Reason (R): In a G.P., $\mathrm{a}_{\mathrm{n}}=\sqrt{a_{n-k} \times a_{n+k}}, n, k \in N$.
Reason (R): In a G.P., $\mathrm{a}_{\mathrm{n}}=\sqrt{a_{n-k} \times a_{n+k}}, n, k \in N$.
- ABoth 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→(a) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
Explanation: We know that in a G.P., terms taken at regular intervals also form a G.P.
So, $a_{n-k}, a_{n}, a_{n+k}$ are in G.P.
$\Rightarrow \mathrm{an}^{2}=\mathrm{a}_{\mathrm{n}}-\mathrm{k} \cdot \mathrm{a}_{\mathrm{n}}+\mathrm{k} \Rightarrow \mathrm{a}_{\mathrm{n}}=\sqrt{a_{n-k} \cdot a_{n+k}}$
$\therefore \mathrm{R}$ is true.
Given $\mathrm{a}_{7}=\mathrm{a}, \mathrm{a}_{10}=\mathrm{b}$ and $\mathrm{a}_{13}=\mathrm{c}$
So, $\mathrm{a}_{10}=\sqrt{a_{10-3} \times a_{10+3}}=\sqrt{a_{7} \cdot a_{13}}$
$\Rightarrow \mathrm{b}=\sqrt{a c} \Rightarrow \mathrm{~b}^{2}=\mathrm{ac}$
$\therefore \mathrm{A}$ is true and R is the correct explanation of A.
Explanation: We know that in a G.P., terms taken at regular intervals also form a G.P.
So, $a_{n-k}, a_{n}, a_{n+k}$ are in G.P.
$\Rightarrow \mathrm{an}^{2}=\mathrm{a}_{\mathrm{n}}-\mathrm{k} \cdot \mathrm{a}_{\mathrm{n}}+\mathrm{k} \Rightarrow \mathrm{a}_{\mathrm{n}}=\sqrt{a_{n-k} \cdot a_{n+k}}$
$\therefore \mathrm{R}$ is true.
Given $\mathrm{a}_{7}=\mathrm{a}, \mathrm{a}_{10}=\mathrm{b}$ and $\mathrm{a}_{13}=\mathrm{c}$
So, $\mathrm{a}_{10}=\sqrt{a_{10-3} \times a_{10+3}}=\sqrt{a_{7} \cdot a_{13}}$
$\Rightarrow \mathrm{b}=\sqrt{a c} \Rightarrow \mathrm{~b}^{2}=\mathrm{ac}$
$\therefore \mathrm{A}$ is true and R is the correct explanation of A.