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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the sum of the products of the corresponding terms of the sequences $2,4,8,.......32$ and $128,32,8,......\large\frac{1}{2}$

$\begin{array}{1 1} 496 \\ 248 \\ 992 \\ 1984 \end{array} $

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1 Answer

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Given sequences are
$2,4,8,16,32$.........(i) and
$128,32,8,2,\large\frac{1}{2}$............(ii)
Both the sequences have 5 terms and are G.P.
We have to find the sum of the products of the corresponding terms
$i.e.,$
$(2\times 128)+(4\times 32)+(8\times 8)+(16\times 2)+(32\times \large\frac{1}{2})$
This can be written as
$=2^8+2^7+2^6+......2^4$
It is again a G.P. with first trerm $a=2^8$ and
common ratio $r=\large\frac{2^7}{2^8}=\frac{1}{2}$
The no. of terms of this G.P=5
$\therefore\:$ The required sum $=S_5=a\large\frac{1-r^5}{1-r}$
$=2^8.\large\frac{1-1/2^5}{1-1/2}=$$2^8.\large\frac{\large\frac{2^5-1}{2^5}}{\large\frac{1}{2}}$
$=2^4(2^5-1)=16\times 31=496$
answered Mar 1, 2014 by rvidyagovindarajan_1
 

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