Pth term of the series 3-1n-3-2n-3-3n- will be-

The correct option is D [ 3+( 1 ) #160; ^ p .p n #160; ] =( 3 1 n #160; ) +( 3+ 2 n #160; ) +( 3 3 2 #160; ) +.... pth ter ...

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Byju's Answer

Standard IX

Mathematics

Expansion of (a ± b ± c)²

pth term of t...

Question

$p^{t h}$ term of the series $(3 - \frac{1}{n}) + (3 + \frac{2}{n}) + (3 - \frac{3}{n}) + . . . . .$ will be?

[3 + \frac{{(- 1)}^{p} . p}{n}]

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3 - \frac{p}{n}

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3 + \frac{n}{p}

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3 - \frac{n}{p}

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Solution

The correct option is D $[3 + \frac{{(- 1)}^{p} . p}{n}]$
$= (3 - \frac{1}{n}) + (3 + \frac{2}{n}) + (3 - \frac{3}{2}) + . . . .$
$p t h$ term $\to$
$= [3 + \frac{{(- 1)}^{p} . p}{n}]$

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Similar questions

p^{t h}

term of the series

(3 - \frac{1}{n}) + (3 - \frac{2}{n}) + (3 - \frac{3}{n}) . . . . . . .

will be

Q. p^th term of the series

(3 - \frac{1}{n}) + (3 - \frac{2}{n}) + (3 - \frac{3}{n}) + . . . .

will be

Q. Assertion :The

n^{t h}

term of the series

1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + . . . \infty i s \frac{n}{3^{n - 1}}

Reason: The

n^{t h}

term of an of an

A . G . P

is the product of

n^{t h}

terms of corresponding

A . P .

G . P .

Q. The sum of first

n

terms of the series

1^{3} + {3.2}^{2} + 3^{3} + {3.4}^{2} + 5^{3} + {3.6}^{2} + . . . .

, where

n

is even be

∴ S_{n} = \frac{n}{k} [n^{m} + 4 n^{2} + 10 n + p]

. Find

k + m - p

If the number of positive integral solutions of x_1 +x_2+x_3=n be denoted by P_n then
$∣ ∣ ∣ ∣ \begin{matrix} P_{n} & P_{n + 1} & P_{n + 2} P_{n + 1} & P_{n + 2} & P_{n + 3} P_{n + 2} & P_{n + 3} & P_{n + 4} \end{matrix} ∣ ∣ ∣ ∣ =$

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pth term of the series (3−1n)+(3+2n)+(3−3n)+..... will be?

The correct option is D [3+(−1)p.pn]=(3−1n)+(3+2n)+(3−32)+....pth term →=[3+(−1)p.pn]

$p^{t h}$ term of the series $(3 - \frac{1}{n}) + (3 + \frac{2}{n}) + (3 - \frac{3}{n}) + . . . . .$ will be?

The correct option is D $[3 + \frac{{(- 1)}^{p} . p}{n}]$
$= (3 - \frac{1}{n}) + (3 + \frac{2}{n}) + (3 - \frac{3}{2}) + . . . .$
$p t h$ term $\to$
$= [3 + \frac{{(- 1)}^{p} . p}{n}]$