wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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 A.P. Find the numbers.

Open in App
Solution

Let the first term of a G.P be a and its common ratio be r.
a1+a2+a3=56a+ar+ar2 = 56 a 1+r+r2 = 56 a=561+r+r2 .......i Now, according to the question:a-1, ar-7 and ar2-21 are in A.P. 2ar-7 = a-1 + ar2 - 212ar - 14 = ar2+a-22ar2-2ar+a-8 = 0a1-r2 = 8a = 81-r2 .......iiEquating (i) and (ii):81-r2=561+r+r281+r+r2 = 561+r2-2r 1+r+r2 =7 1+r2-2r1+r+r2 =7+7r2-14r6r2-15r+6=0 32r2-5r+2 = 02r2-4r-r+2=02r(r-2)-1(r-2)=0(r-2)(2r-1)=0r=2, 12When r=2, a=8. [Using (ii)]And, the required numbers are 8, 16 and 32.When r=12, a=32. [Using (ii)]And, the required numbers are 32, 16 and 8.

flag
Suggest Corrections
thumbs-up
5
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction to AP
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon