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# If 1 , 3 , 8 are first three terms of an arithmetic - geometric progression (with +ve common difference ) , the sum of next three terms is :

$(a)\;180\qquad(b)\;160\qquad(c)\;140\qquad(d)\;120$

Explanation : Given , a=1
$(a+d)\;r=3$
$(a+2d)\;r^2=8$
$(1+d)\;r=3$
$(1+2d)\;r^2=8------(1)$
$(1+d)^2\;r^2=9$
$(1+2d+d^2)\;r^2=9----(2)$
Subtracting (1) from (2)
$d^2\;r^2=1$
$d\;r=1$
$d=\large\frac{1}{r}$
$(1+d)\;r=3$
$(1+\large\frac{1}{r})\;r=3$
$r+1=3$
$r=2\quad\;d=\frac{1}{2}$
Next three terms ,
$(1) \quad\; (1+3d)\;r^3 =(1+\large\frac{3}{2})\;8=20$
$(2)\quad\;(1+4d)\;r^4=(1+\large\frac{4}{2})\;16=48$
$(3)\quad\;(1+5d)\;r^5=(1+\large\frac{5}{2})\;32=112$
Sum of next three terms = $\;20+48+112=180\;.$