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Definition of Relations

Let a1, a2,...

Question

Let $a_{1}, a_{2}, a_{3}, . . . ., a_{10}$ be in G.P. with $a_{1} > 0$ for $i = 1, 2, . . .10$ and $S$ be the set of pairs $(r, k), r k \in N$ (the set of natural numbers) for which
$∣ ∣ ∣ ∣ ∣ \begin{matrix} {log}_{e} a_{1}^{r} a_{2}^{k} & {log}_{e} a_{2}^{r} a_{3}^{k} & {log}_{e} a_{3}^{r} a_{4}^{k} {log}_{e} a_{4}^{r} a_{5}^{k} & {log}_{e} a_{5}^{r} a_{6}^{k} & {log}_{e} a_{6}^{r} a_{7}^{k} {log}_{e} a_{7}^{r} a_{8}^{k} & {log}_{e} a_{8}^{r} a_{9}^{k} & {log}_{e} a_{9}^{r} a_{10}^{k} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
Then the number of elements in $S$ , is :

A

Infinitely many

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B

4

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C

10

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D

2

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Solution

The correct option is A Infinitely many

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Similar questions

a_{1}, a_{2}, a_{3}, \dots, a_{10}

be in G.P. with

a_{i} > 0

i = 1, 2, \dots, 10

S

be the set of pairs

(r, k), r, k \in N

(the set of natural numbers) for which

∣ ∣ ∣ ∣ ∣ \begin{matrix} {log}_{e} a_{1}^{r} a_{2}^{k} & {log}_{e} a_{2}^{r} a_{3}^{k} & {log}_{e} a_{3}^{r} a_{4}^{k} {log}_{e} a_{4}^{r} a_{5}^{k} & {log}_{e} a_{5}^{r} a_{6}^{k} & {log}_{e} a_{6}^{r} a_{7}^{k} {log}_{e} a_{7}^{r} a_{8}^{k} & {log}_{e} a_{8}^{r} a_{9}^{k} & {log}_{e} a_{9}^{r} a_{10}^{k} \end{matrix} ∣ ∣ ∣ ∣ ∣

Then the number of elements in

S

A

be the set of all permutations

a_{1}, a_{2}, . . . ., a_{6}

1, 2, . . ., 6

a_{1}, a_{2}, . . . . a_{k}

is not a permutation of

1, 2, . . ., k

k, 1 \leq k \leq 5

. Then the number of elements in

A

Q. For any real number

r

A_{r} = {e^{i π r n} : n is a natural number}

be a set of complex numbers. Then

a_{1}, a_{2}, a_{3}, \dots

a_{1} = a

and common ratio

r,

a

r

positive integers, then the number of ordered pairs

(a, r)

12 \sum k = 1 {log}_{8} a_{k} = 2010

Given that a_i $>$ 0 and i belongs to a set of natural numbers. If $a_{1}, a_{2}, a_{3} . . . . . a_{2 n}$ are in AP, then find the value of $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n} - 1}{{\sqrt{a}}_{2} + \sqrt{a_{3}}} . . . . . . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$

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