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Question
Answer the following:
Find five numbers in G.P. such that their product is 243 and sum of second and fourth number is 10.
Solution
Let the five numbers in G.P. be `"a"/"r"^2, "a"/"r","a", "ar", "ar"^2`
Since their product is 243,
`"a"/"r"^2."a"/r."a"."ar"."ar"^2`= 243
∴ a5 = 35
∴ a = 3
Also, the sum of second and fourth is 10
∴ `"a"/"r" + "ar"` = 10
∴ `3/"r" + 3"r"` = 10r ...[∵ a = 3]
∴ 3 + 3r2 = 10r
∴ 3r2 − 10r + 3 = 0
∴ (r – 3)(3r – 1) = 0
∴ r – 3 = 0 or 3r – 1= 0
∴ r = 3 or r = `1/3`
Taking r = `3, "a"/"r"^2 = 3/9 = 1/3, "a"/"r" = 3/3` = 1, ar = 3 × 3 = 9,
ar2 = 3(3)2 = 27 and the five numbers are `1/3`, 1, 3, 9, 27
Taking r = `1/3, "a"/"r"^2 = 3/((1/9)) = 27, "a"/"r" = 3/((1/3))` = 9,
ar = `3(1/3)` = 1, ar2 = `3(1/3)^2 = 1/3`
and the five numbers are 27, 9, 3, 1, `1/3`
Hence, the required numbers in G.P. are `1/3`, 1, 3, 9, 27 or 27, 9, 3, 1, `1/3`.
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