$\begin{array}{1 1} 2,10,18,26 \\ 1,7,15,19 \\ 1,8, 15,21 \\ 2,8,14,20 \end{array}$

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Answer : (d) 2,8,14,20

Explanation : Let the 4 terms in AP = $a-3d\;,a-d\;,a+d\;,a+3d$

$sum=44$

$a+3d+a+d+a-d+a-3d=44$

$4a=44\quad\;a=11$

$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=5/14$

$14\;(a^2-9d^2)=5(a^2-d^2)$

$a=11\quad,1694-126d^2=605-5d^2$

$121d^2=1089$

$d^2=\frac{1089}{121}=9\quad\;d=\pm3$

$terms\; are\;11-9\;,11-3\;,11+3\;,11+9$

$=2,8,14,20.$

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