Correct Answer - 27
Let AGP be
`a,(a+d)r,(a+2d)r^(2)`,…
Comparing with given progression, we get
`a=3,(3+d)r=4,(3+2d)r^(2)=4`
`therefore4/r-3=d=(4/r^(2)-3)1/2`
`rArr3r^(2)-8r+4=0`
`rArr(3r-2)(r-2)=0`
Since we have infinite AGP, r`=2/3`.
`therefore(3+d)2/3=4`
`rArrd=3`
`therefore` Sum of infinite AGP is, `S_(oo)=a/(a-r)+(dr)/((1-r)^(2))=27`
