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Byju's Answer

Standard XII

Mathematics

Least Integer Function

If r=0, pn ...

Question

If $r = 0, p_{n}$ equals

2 k / n

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(k + 1) / n

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(2 k + 1) / n

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k / n

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Solution

The correct option is A $2 k / n$
$a^{2} - 1$ is divisible by 10 if and only if last digit of a is 1 or 9.
If $r = 0$ , there are $2 k$ ways to choose $a$ .
$∴ p_{n} = 2 k / n$ .

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Similar questions

Q. If

D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

and

\sum_{k = 1}^{n} D_{k} = 56

, then

n

equals ?

If $D_{k} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $n \sum k = 1 D_{k} = 56,$ where $n \in N,$ then $n$ is equal to

Q. Let

P (n) : 2^{n} > n, \forall n \in N

and

2^{k} > k, \forall n = k

, then which of the following is true

\forall k \geq 2

Q. If

S_{k} = sin (\frac{2 π k}{n + 1}) - i cos (\frac{2 π k}{n + 1})

and

P_{k} = cos (\frac{π}{2^{k}}) + i sin (\frac{π}{2^{k}}),

Let

m = ∣ ∣ ∣ n \sum k = 1 S_{k} + \infty \prod k = 1 P_{k} ∣ ∣ ∣

, then the value of

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S_{k} = sin (\frac{2 π k}{n + 1}) - i cos (\frac{2 π k}{n + 1})

and

P_{k} = cos (\frac{π}{2^{k}}) + i sin (\frac{π}{2^{k}}),

Let

m = ∣ ∣ ∣ n \sum k = 1 S_{k} + \infty \prod k = 1 P_{k} ∣ ∣ ∣

, then the value of

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If r=0,pn equals

The correct option is A 2k/na2−1 is divisible by 10 if and only if last digit of a is 1 or 9.If r=0, there are 2k ways to choose a.∴pn=2k/n.

If $r = 0, p_{n}$ equals

The correct option is A $2 k / n$
$a^{2} - 1$ is divisible by 10 if and only if last digit of a is 1 or 9.
If $r = 0$ , there are $2 k$ ways to choose $a$ .
$∴ p_{n} = 2 k / n$ .