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Misc 4
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 numbers in G.P. be a, ar, and ar2.
It is given that
Sum of these number is 56
a + ar + ar2 = 56
ar2 = 56 a ar
Also,
When 1, 7, 21 subtracting from these number respectively,
the new numbers are in AP
So, (a 1) , (ar 7) , (ar2 21) are in AP
Common difference is same
(ar 7) (a 1) = (ar2 21) (ar 7)
ar a 6 = ar2 ar 14
ar2 ar ar + a 14 + 6 = 0
ar2 2ar + a 8 = 0
From (1) putting ar2 = 56 a ar
(56 a ar) 2ar + a 8 = 0
56 a ar 2ar + a 8 = 0
-a + a ar 2ar + 56 8 = 0
0 3ar + 48 = 0
-3ar = -48
ar = ( 48)/( 3)
ar = 16
a = 16/
Putting a = (16 )/ in (1)
a + ar + ar2 = 56
a(1 + r + r2) = 56
16/ (1 + r + r2) = 56
(16 (1+ + 2))/ = 56
(16 (1+ + 2))/ = 56
16(1 + r + r2) = 56r
16 + 16r + 16r2 56r = 0
16r2 + 16r 56r + 16 = 0
8(2r2 5r + 2) = 0
(2r2 5r + 2) = 0/8
2r2 5r + 2 = 0
2r2 4r r + 2 = 0
2r(r 2) 1(r 2) = 0
(2r 1)(r 2) = 0
2r 1 = 0 or r 2 = 0
2r = 1 or r = 2
r = 1/2 or r = 2
Now, finding numbers
Thus the numbers are 8, 16, 32 for r = 1/2 & a =32
& 32, 16, 8 for r = 2 & a = 8

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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