Question

Assertion :Let the positive numbers $a,b,c$ be in A.P., then $\frac{1}{bc},\frac{1}{ac},\frac{1}{ab}$ are also in A.P. Reason: If each term of an A.P. is divided by $abc$, then the resulting sequence is also in A.P.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

if $a,b,c$ are in AP, then

$b-a=c-b$

$a+c=2b$

For $\frac{1}{bc},\frac{1}{ac},\frac{1}{ab}$

$\frac{1}{ac}-\frac{1}{bc}=\frac{b-a}{abc}$

$\frac{1}{ab}-\frac{1}{ac}=\frac{c-b}{abc}$

From above two equations,

$\frac{1}{bc},\frac{1}{ac},\frac{1}{ab}$ are also in AP.

And their common difference is divided by $abc$ since numbers are also divided by $abc$.

So, if $a,b,c$ are in AP and you divide the numbers by their product $abc$ they will remain in AP.