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Question
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Solution
Let the three numbers of the geometric series be a, ar, ar2.
Sum of all three terms = a + ar + ar2 = 56 …......(i)
By subtracting 1, 7, 21 from these numbers the numbers
ar – 1, ar – 7, ar2 – 21 are in arithmetic progression.
∴ 2(ar – 7) = (a – 1) + (ar2 – 21)
or 2ar – 14 = ar2 + a – 22
ar2 – 2ar + a = 22 – 14 = 8 ….........(ii)
Dividing equation (i) by (ii)
= `("a"(1 + "r" + "r"^2))/("a"(1 - 2"r" + "r"^2)) = 58/8 = 7`
or 7(1 – 2r + r2) = 1 + r + r2
6r2 – 15r + 6 = 0
2r2 – 5r + 2 = 0
or (r – 2) (2r – 1) = 0 या r = 2, `1/2`
Putting r = 2 in equation (i),
a(1 + 2 + 4) = 56 or a = `56/7 = 8`
Thus there are three numbers: 8, 16, 32
Again by putting r = `1/2` in equation (i),
`"a" (1 + 1/2 + 1/4) = 56`
`"a" = (56 xx 4)/7 = 32`
∴ Three numbers 32, 16, 8
Hence, the required numbers are 8, 16, 32.
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