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

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 $42$.

$i.e.,\:\:t_{20}=40\times 42=1680$

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