Three numbers are in the ratio 3:7:9.if 5 subtracted from the second the resulting number are in A.P find the original number
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Let the numbers be
3X, 7X and 9X
When we subtract 5 from 2nd term we get
7X - 5
Now they are in A.P. Therefore
3X + 9X = 2×(7X - 5)
12X = 14X - 10
2X = 10
X = 5
Thus the numbers are
15, 35 and 45
3X, 7X and 9X
When we subtract 5 from 2nd term we get
7X - 5
Now they are in A.P. Therefore
3X + 9X = 2×(7X - 5)
12X = 14X - 10
2X = 10
X = 5
Thus the numbers are
15, 35 and 45
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