Ans.3×5,7×5,9×5 i.e.15,35 and 45.
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Call the numbers of the AP a, a+d and a+2d. So (a+2d)/a=9/3=3, so a+2d=3a, a=d. So the middle term of the AP is a+d=2a or 2d. The AP is a, 2a, 3a. But the middle term, 2a, is 5 less than the middle of the required series, so (2a+5)/a=7/3; 6a+15=7a; a=15. 2a+5=35 and the three numbers are 15, 35, 45.

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