1
Question
If a, b, c, all distinct are such that a,b, and c are in AP and b-a, c-b, and a are in GP then a:b:c =?
If a, b, c, all distinct are such that a,b, and c are in AP and b-a, c-b, and a are in GP then a:b:c =?
Open in App
Solution
The correct option is C 1:2:3
Going from answer options is the best choice.
**Observe that only options (a) and (c) are in an AP.
By Assumption technique; taking a=2, b=3 and c=4
b-a=1, c-b=1 and a=2, which does not form a GP. Hence this option can never be true
Taking a=1, b=2 and c=3.
b-a=1, c-b=1 and a=1 is in a GP. Hence, answer option (c) is the right one
Going from answer options is the best choice.
**Observe that only options (a) and (c) are in an AP.
By Assumption technique; taking a=2, b=3 and c=4
b-a=1, c-b=1 and a=2, which does not form a GP. Hence this option can never be true
Taking a=1, b=2 and c=3.
b-a=1, c-b=1 and a=1 is in a GP. Hence, answer option (c) is the right one
Suggest Corrections
1
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
