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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Given $4$ numbers, the first $3$ numbers are in $AP$ and the last $3$ numbers are in $GP$. The $1^{st}$ and $3^{rd}$ terms add upto $2$ and the $2^{nd}$ and $4^{th}$ terms add to $17$. Find the numbers.

$\begin{array}{1 1} -2,1,4,16 \\ -3,1,5,25 \\ -1,2,5,25 \\ 1,2,3,9 \end{array}$

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1 Answer

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Answer : (a) -2,1,4,16
Explanation : Let the 4 numbers be a,b,c,d
$a,b,c\;in\;AP\quad\;2b=a+c$
$b,c,d\;in\;GP\quad\;c^2=bd$
$a+c=2$
$b+d=17$
$a+c=2\quad\;2b=a+c=2$
$b=1$
$b+d=17\quad\;d=17-1=16$
$c^2=bd\quad\;c^2=1*16=16$
$c=\pm4$
$a=2-c$
$=-2\quad\;or\quad\;6$
$Series\;is\;-2,1,4,16\quad\;or\quad\;6,1,-4,16.$
answered Jan 6, 2014 by yamini.v
 

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