A) 36
B) 24
C) 28
D) 32
Correct Answer: C
Solution :
Let 3 consecutive terms are \[\frac{a}{r},\,\,a,\,\,ar\]of a G.P. \[\therefore \]\[{{a}^{3}}=512\Rightarrow a=8\] Now,\[\frac{8}{r}+4,12,8r\]are in A.P. \[\therefore \]\[24=\frac{8}{r}+4+8r\Rightarrow 24r=8+4r+8{{r}^{2}}\] \[\Rightarrow \]\[2{{r}^{2}}-5r+2=0\Rightarrow (r-2)(2r-1)=0\] \[\Rightarrow \]\[r=2\]or\[\frac{1}{2}\] When r = 2, terms are 4, 8, 16 When \[r=\frac{1}{2},\]terms are 16, 8, 4 So, required sum \[=16+8+4=28\]You need to login to perform this action.
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