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Sequences and Series question

2017 · 9 Apr · Shift 1 · Q24

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JEE Main
Mathematics
Sequences and Series
2017 · 9 Apr · Shift 1 · Q24

Sequences and Series question

2017 · 9 Apr · Shift 1 · Q24

JEE MainMathematicsSequences and SeriesMCQ+4 / −1

If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :

A
2
B
4 ${^{{1 \over 3}}}$
C
4 ${^{{2 \over 3}}}$
D
4

View written solutionFree

Correct answer: A

Use the A.P. condition

Since $a,b,c$ are in A.P., the middle term is the average of the other two: $2b=a+c.$

So $a$ and $c$ can be written as

\qquad c=b+d$$ for some real number $d$. Because all three numbers are positive, we must have $$b-d>0,\qquad b+d>0.$$ In particular, $b>0$. --- 2. **Use the product condition** Given $$abc=8,$$ substitute $a=b-d$ and $c=b+d$: $$b(b-d)(b+d)=8.$$ Using $(b-d)(b+d)=b^2-d^2$, $$b(b^2-d^2)=8.$$ So, $$b^3-bd^2=8.$$ Rearranging, $$b^3=8+bd^2.$$ Since $b>0$ and $d^2\ge 0$, we get $$b^3\ge 8.$$ Hence, $$b\ge 2.$$ --- 3. **Check whether the minimum is attainable** Equality $b=2$ occurs when $$d^2=0\Rightarrow d=0.$$ Then $$a=b-d=2,\qquad c=b+d=2.$$ Indeed, $(2,2,2)$ are in A.P. and $$abc=2\cdot 2\cdot 2=8.$$ So the minimum possible value of $b$ is actually attained. --- 4. **Evaluate the options** The minimum possible value is $$\boxed{2}.$$ So the correct option is **A**. --- 5. **Comparison with stored answer** Stored correct answer: **A** Our derived answer is also **A**, so they agree.

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