Solution:
Given that t7 = 8t4 ⇒ ar6 = 8ar3
⇒ r3 = 8 = 23 ⇒ r = 2.
Also t5 = 48 Þ ar4 = 48 or 16a = 48 ⇒ a = 3.
Hence the required G.P. is 3, 6, 12, 24 ……
Illustration:
Does there exists a G.P. containing 27, 8 and 12 as three of its terms? If it exists, how many such progressions are possible?
Solution:
Let 8 be the mth, 12 the nth and 27 be the tth terms of a G.P. whose first term is A and common ratio is R.
Then 8 = ARm–1, 12 = ARn–1, 27 = ARt–1
⇒ 8/12 = Rm–n = 2/3, 12/27 = Rn–t = (2/3)2, 8/27 = Rm–t =(2/3)3
⇒ 2m – 2n = n – t and 3m – 3n = m – t
⇒ 2m + t = 3n and 2m + t = 3n
⇒ 2m+t/3 = n.
There are infinite sets of values of m, n, t which satisfy this relation. For example, take m = 1, then 2+t/3 = n = k ⇒ n = k, t = 3k – 2. By giving different values to k we get integral values of n and t. Hence there are infinite numbers of G.P.’s whose terms may be 27, 8, 12 (not consecutive).
Illustration:
In a four term series if first three are in G.P. and last three are in A.P. with common different 6 and last terms is equal to the first term then find all four terms in series.
Solution:
This is very tricky question. If you read question carefully then it is clear that we have to start with A.P. because common difference is given.
Let the numbers be a + 6, a–6, a, a+6 now first three are in G.P. is (a–6)2 = a(a+6) or, a2 – 12a + 36 = a2 therefore numbers are 8, –4, 2, 8.
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