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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the $20^{th}$ term of the series $2\times 4+4\times 6+6\times 8+..........$

$\begin{array}{1 1}1848 \\ 1680 \\ 1520 \\ 440 \end{array} $

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Given series is $(2\times 4)+(4\times 6)+(6\times 8)+..........$
Each bracket consists of product of two terms.
The first term in each bracket forms the sequence $2,4,6.......$
This is an A.P. with $1^{st}$ term =$a=2$ and common difference = $d=2$
$\therefore$The $20^{th}$ term of this sequence is $2+(20-1)2=40$
Similarly the$2^{nd}$ term of each bracket forms the sequence $4,6,8.....$
This sequence is also an A.P. with $a=4$ and $d=2$
$\therefore$ The $20^{th}$ term of this sequence is $4+(20-1)2=42$
$\Rightarrow\:$ The $20^{th}$ term of the given series is the product of $40$ and $42$.
$i.e.,\:\:t_{20}=40\times 42=1680$
answered Apr 2, 2014 by rvidyagovindarajan_1
 

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