Question

Assertion :STATEMENT- 1 : $3,6,12$ are in $G.P$., then $9,12,18$ are in $H.P$ Reason: STATEMENT- 2 : If middle term is added in three consecutive terms of a $G.P$, resultant will be in $H.P$

• A. Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation for Statement -1
• B. Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement -1 is True, Statement -2 is False
• D. Statement -1 is False, Statement -2 is True

Answer 1

The correct option is A Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation for Statement -1

If $a,b,c$ are in $G.P$.,
then $a+b,b+b,c+b$ are in $H.P$.
$\to \left(2b\right)=\frac{2\ast (a+b)(b+c)}{(a+b)(c+b)}$
$\to b(a+2b+c)=(a+b)(b+c)$
$\to b(a+c)+2b^2=ab+ac+b^2+bc$
$\to b^2=ac$ ( because $a,b,c$ are in $GP$)
Hence, option 'A' is correct.