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Question
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
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Solution
Let the first term of the geometric progression = a,
Common ratio = r
∴ a4 = ar3 = x ....(1)
a10 = ar9 = y ....(2)
a16 = ar15 = z ....(3)
Dividing (2) by (1), we obtain
`y/x = (ar^9)/(ar^3) = y/x = r^6`
Dividing (3) by (2), we obtain
`z/y = (ar^15)/(ar^3) = z/y = r^6`
∴ `y/x = z/y`
Thus, x, y, z are in G.P.
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