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Maximum of Second Term over Sum of First Three Terms of a GP | JEE

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Question
If the common ratio of a geometric progression is positive, then the maximum value of the ratio of the second term to the sum of the first three terms equals
A22
B13\dfrac13correct
C33
D12\dfrac12
Solution
Step 1: Choose a symmetric way of writing the three terms. Taking the middle term as aa and the common ratio as rr, the first three terms are
ar,a,ar.\frac{a}{r}, \quad a, \quad ar .
 This choice puts the second term - the one we want - on top by itself. Step 2: Form the required ratio.
second termsum of first three=aar+a+ar=aa(1r+1+r)=1r+1r+1.\frac{\text{second term}}{\text{sum of first three}} = \frac{a}{\dfrac{a}{r} + a + ar} = \frac{a}{a\left(\dfrac1r + 1 + r\right)} = \frac{1}{r + \dfrac1r + 1} .
 Step 3: Bound the denominator. Since r>0r > 0, apply AM ≥\ge GM to rr and 1r\dfrac1r:
r+1r2≥r⋅1r=1  ⟹  r+1r≥2.\frac{r + \dfrac1r}{2} \ge \sqrt{r \cdot \frac1r} = 1 \implies r + \frac1r \ge 2 .
 Step 4: Hence
r+1r+1≥3  ⟹  1r+1r+1≤13.r + \frac1r + 1 \ge 3 \implies \frac{1}{r + \dfrac1r + 1} \le \frac13 .
 Step 5: Check the bound is attained. Equality in AM ≥\ge GM needs r=1rr = \dfrac1r, i.e. r=1r = 1 (positive root). Then the three terms are a,a,aa, a, a and the ratio is a3a=13\dfrac{a}{3a} = \dfrac13. Answer: (2).
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