CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The sum of three consecutive terms in an A.P. is 6 and their product is 120. Find the three numbers.

Open in App
Solution

Let ad,a,a+d be the first three terms of an A.P.

It is given that the sum of the terms is 6 that is:

ad+a+a+d=63a=6a=63a=2......(1)

It is also given that the product of the terms is 120 that is:

(ad)(a)(a+d)=120a(a2d2)=120((x+y)(xy)=x2y2)2(22d2)=120(From(1))4d2=1202
4d2=60d2=4+60d2=64d=±64d=±8...........(2)

Now, if a=2 and d=8 then the first three terms of the A.P are:

ad=2(8)=2+8=10
a=2 and
a+d=28=6

And if a=2 and d=8 then the first three terms of the A.P are:

ad=28=6
a=2 and
a+d=2+8=10

Hence, the three terms are 6,2,10 or 10,2,6


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Arithmetic Progression
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon