Mathematics - Sequence & Series Question with Solution | TestHub

Option A: Sum of $n$ A.M.s between $a$ and $b$ is $n\cdot \frac{a+b}{2}$. Here $10\cdot \frac{1}{2}=5$. True.

Option B: Product of $n$ G.M.s between $a$ and $b$ is $(ab{)}^{n/2}$. Here $(4\tan 15^{\circ }\cot 15^{\circ }{)}^5=4^5=1024$. True.

Option C: A.M. $\ge$ G.M. gives $27abc\le (a+b+c{)}^3$, so the given condition forces equality, $a=b=c$. Then $12a=12$, $a=b=c=1$ and the sum is 3, not 10. False.

Option D: True.