Question

Assertion (A) If $\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}$ are not in an AP, then a, b, c are not in AP.
Reason (R) If different constants are added to each term of an AP, then the resultant pattern is still an AP.
Which of the following is true?

• A. Both (A) and (R) are true and (R) correct explanation of (A)
• B. Both (A) and (R) are true but (R) is not correct explanation of (A)
• C. (A) is true and (R) is false
• D. Both (A) and (R) are false

Answer 1

The correct option is C (A) is true and (R) is false

Given, $\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}$ are not in AP.
Then, $\frac{1}{bc}+\frac{1}{ab}\ne \frac{2}{ca}$
[$\because 2b\ne a+c$, if given series is not AP]
$\Rightarrow \frac{a+c}{abc}\ne \frac{2}{ca}$
$\Rightarrow \frac{a+c}{b}\ne 2$
$\Rightarrow a+c\ne 2b$
Thus, a, b, c are not in AP.
$\therefore$ Assertion is true.
Reason is not true.
Adding different constants ( like 2, 3, 5, 8, 6...) to a given AP ( like 2, 4, 6, 8, 10...)
we get 2(+2), 4(+3), 6(+5), 8(+8), 10(+6)...
$\Rightarrow$ 4, 7, 11, 16, 16, . . . .
which is not an AP.