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Implicit Differentiation

If a,a1,a2,...

Question

If $a, a_{1}, a_{2}, a_{3}, a_{4}, . . . . . ., a_{2 n}, b$ are in AP and $a, g_{1}, g_{2}, g_{3}, g_{4}, . . . . . . . ., g_{2 n}, b$ are in GP and
$h$ is the HM of $a$ and $b$ , then $\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{2} + a_{2 n - 1}}{g_{2} g_{2 n - 1}} + . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}}$

A

\frac{2 n}{h}

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B

2 n h

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C

n h

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D

\frac{n}{h}

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Solution

The correct option is A $\frac{2 n}{h}$
The denominators $g_{1} g_{2 n}, g_{2} g_{2 n - 1}, . . . ., g_{n} g_{n + 1}$ are equal to $a b$ as it is the property of G.P
Also all the numerators $a_{1} + a_{2 n}, a_{2} + a_{2 n - 1}, . . ., a_{n} + a_{n + 1}$ are all equal to $a + b$ as it is the property of A.P
So the entire expression is equal to $n \times \frac{(a + b)}{a b}$ i.e, $\frac{2 n}{h}$

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Similar questions

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

a, g_{1}, g_{2}, g_{3}, . . . ., g_{2 n}, b

are in GP and h is the harmonic mean of a and b then

\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{2} + a_{2 n - 1}}{g_{2} g_{2 n - 1}} + . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}}

a, a_{1}, a_{2}, . . . . . a_{2 n}, b

a, g_{1}, g_{2}, . . . . . g_{2 n}

a, h, b

are in H.P then:

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

a, a_{1}, a_{2}, . . . . ., a_{2 n}, b

are in arithmetic progression and

a, g_{1}, g_{2}, . . . . g_{2 n}, b

are in geometric progression and

h

is the harmonic mean of

a

b

\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{2} + a_{2 n - 1}}{g_{2} g_{2 n - 1}} + . . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}}

a_{r} > 0, r ϵ N

a_{1}

a_{2}

a_{3}

a_{2 n}

\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; \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}}} =

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