    Question

# If a,b,c are in GP and a,p,q are in AP such that 2a,b+p,c+q are in GP then the common difference of AP is

A
2a
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(2+1)(ab)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2(a+b)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(21)(ba)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct options are B (√2+1)(a−b) D (√2−1)(b−a)Let common difference of the A.P is d.Given: b2=ac .......(1)2p=a+q .......(2)(b+p)2=(c+q)2a .......(3)(3)-(1) we get, ⇒p(2b+p)=ac+2aq⇒b2−(2p)b+2aq−p2=0 from (1)⇒b2−(a+q)+8aq−(a+q)24=0∴b=a+q±√a2+q2+2aq+a2+q2+2aq−8aq2⇒2b=a+q±√2a2+2q2−4aq=a+q±√2(q−a)=a(1∓√2)+q(1±√2)=a(1∓√2)+(a+2d)(1±√2) ⇒2b=2a+2d(1±√2)⇒d=b−a1±√2or d=(√2−1)(b−a),(√2+1)(a−b)Hence, Options B and D are correct.  Suggest Corrections  0      Similar questions
Join BYJU'S Learning Program
Select...  Related Videos   Arithmetic Progression
MATHEMATICS
Watch in App  Explore more
Join BYJU'S Learning Program
Select...