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

The sum of three numbers which are consecutive terms an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.


    Open in App
    Solution

    Let the first term of an A.P. is a and its common difference be d.

    a1+a2+a3=21

    a+(a+d)+(a+2d)=21

    3a+3d=21

    a+d=7 ..........(i)

    Now, according to the questiong :

    a,a+d1 and a+2d+1 are in G.P.

    (a+d1)2=a(a+2d+1)

    (7+aa1)2=a[a+2(7a)+1]

    (6)2=a(15a)

    36=15aa2

    a215a+36=0

    (a3)(a12)=0

    a=3,12

    Now, putting a = 2, 12 in equation (i), we get d = 5, -5, respectively.

    Thus, for a = 2 and d = 5, the numbers are 2, 7 and 12.

    And, for a = 12 and d = -5, the numbers are 12, 7 and 2.


    flag
    Suggest Corrections
    thumbs-up
    12
    Join BYJU'S Learning Program
    similar_icon
    Related Videos
    thumbnail
    lock
    Geometric Progression
    MATHEMATICS
    Watch in App
    Join BYJU'S Learning Program
    CrossIcon