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Solved Examples based on GP

Solved examples :


The 7th term of a G.P. is 8 times the 4th term. Find the G.P. when its 5th term is 48.


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 ……


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?


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).


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.


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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