Sequences & SerieseasyFree

a,b,c in AP and b-a,c-b,a in GP: Find a:b:c | JEE

JEE Maths question with a full step-by-step solution.

Question
If a,b,ca, b, c are three distinct numbers such that a,b,ca, b, c are in A.P. and b−a, c−b, ab-a,\ c-b,\ a are in G.P., then a:b:ca:b:c is
A2:3:42:3:4
B3:4:53:4:5
C1:3:51:3:5
D1:2:31:2:3correct
Solution
Step 1: A.P. gives 2b=a+c2b=a+c and equal differences b−a=c−b.b-a=c-b. Step 2: G.P. of b−a, c−b, ab-a,\ c-b,\ a gives (c−b)2=a(b−a)(c-b)^2=a(b-a). Replace c−bc-b by b−ab-a:
(b−a)2=a(b−a) ⇒ b−a=a.(b-a)^2=a(b-a)\ \Rightarrow\ b-a=a.
 Step 3: So b=2a, c=3a⇒a:b:c=1:2:3.b=2a,\ c=3a\Rightarrow a:b:c=1:2:3. Correct answer: (4)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · medium
Let α,β\alpha, \beta be the roots of x2−x+p=0x^2 - x + p = 0 and γ,δ\gamma, \delta be the roots of x2−4x+q=0x^2 - 4x + q = 0, where p,q∈Zp, q \in \mathbb{Z}. If α,β,γ,δ\alpha, \beta, \gamma, \delta are in G.P., then ∣p+q∣|p + q| equals:
Sequences & Series · easy
a,b,c,da, b, c, d are in increasing G.P. If the A.M. between aa and bb is 66 and the A.M. between cc and dd is 5454, then the A.M. of aa and dd is
Sequences & Series · medium
If the sum of the first 10 terms of the series 11+14⋅4+21+24⋅4+31+34⋅4+⋯\dfrac{1}{1+1^4\cdot4}+\dfrac{2}{1+2^4\cdot4}+\dfrac{3}{1+3^4\cdot4}+\cdots is mn\dfrac{m}{n} with gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to:
Sequences & Series · medium
Let a1,a2,…,a4001a_1, a_2, \ldots, a_{4001} be in A.P. such that 1a1a2+1a2a3+⋯+1a4000a4001=10\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\cdots+\frac{1}{a_{4000} a_{4001}}=10 and a2+a4000=50a_2+a_{4000}=50. Then ∣a1−a4001∣|a_1-a_{4001}| equals

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath