NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.Misc.Q.10
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.
Let the three numbers in G.P. be a, ar, and ar2. From the given condition,
a + ar + ar2 = 56
a (1 + r + r2) = 56 … (1)
a – 1, ar – 7, ar2 – 21 forms an A.P.
So, (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)
ar – a – 6 = ar2 – ar – 14
ar2 – 2ar + a = 8
ar2 – ar – ar + a = 8
a (r2 + 1 – 2r) = 8
a (r – 1)2 = 8 ……………………………… (2)
From equation (1) and (2), we get
7(r2 – 2r + 1) = 1 + r + r2
7r2 – 14 r + 7 – 1 – r – r2 = 0
6r2 – 15r + 6 = 0
6r2 – 12r – 3r + 6 = 0
6r (r – 2) – 3 (r – 2) = 0
(6r – 3) (r – 2) = 0
r = 2,
When r = 2, a = 8
When r = , a = 32
Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When, r = , the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.