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Harmonic Progression

If a1 ∈ R -...

Question

If $a_{1} \in R - {0}, i = 1, 2, 3, 4$ and $x \in R$ and $(\sum_{i = 1}^{3} a_{i}^{2}) x^{2} - 2 x (\sum_{i = 1}^{3} a_{i} a_{i + 1}) + \sum_{i = 2}^{4} a_{i}^{2} \leq 0$ , then $a_{1}, a_{2}, a_{3}, a_{4}$ are in

A

A.P

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B

G.P

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C

H.P

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D

A.G.P

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Solution

The correct option is B G.P
$({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) x^{2} - 2 x (a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4}) + ({a_{2}}^{2} + {a_{3}}^{2} + {a_{4}}^{2}) \leq 0 (x^{2} {a_{1}}^{2} - 2 x a_{1} a_{2} + {a_{2}}^{2}) + ({a_{2}}^{2} x^{2} - 2 a_{2} a_{3} + {a_{3}}^{2}) + ({a_{3}}^{2} x - 2 x a_{3} a_{4} + {a_{4}}^{2}) \leq 0 {(a_{1} x - a_{2})}^{2} + {(a_{2} x - a_{3})}^{2} + {(a_{3} x - a_{4})}^{2} \leq 0$
Since each one is a square term
These sum of term cannot be less than $0$
Only possibility they can be zero
For
$a_{1} x - a_{2} = 0 \Rightarrow x = \frac{a_{2}}{a_{1}} a_{2} x - a_{3} = 0 \Rightarrow x = \frac{a_{3}}{a_{2}} a_{3} x - a_{4} = 0 \Rightarrow x = \frac{a_{4}}{a_{3}} x = \frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} = \frac{a_{4}}{a_{3}} = k$
Hence, $a_{1}, a_{2}, a_{3}, a_{4}$ are in GP

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a_{r} > 0; \forall r, n \in N

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

are in A.P, then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

a_{r} > 0, r \in N

a_{1}, a_{2}, a_{3}, . . ., a_{2 n}

are in A.P. then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

a_{1}, a_{2}, a_{3}, . . ., a_{n}

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a_{i} > 0

for all i, then show that

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

a_{1}, a_{2}, a_{3}, . . ., a_{n}

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