Home
Class 14
REASONING
c - abbcc - bb - cab -c - abb...

c - abbcc - bb - cab -c - abb

A

babac

B

cacbc

C

cbbca

D

abcca

Text Solution

Verified by Experts

The correct Answer is:
B
The given cyclic order series follows the pattern :
Promotional Banner

Topper's Solved these Questions

  • SERIES COMPLETION : ALPHABET

    LUCENT PUBLICATION|Exercise EXERCISE 3|8 Videos

  • SERIES COMPLETION : ALPHABET

    LUCENT PUBLICATION|Exercise EXERCISE 1(Type-III : Group of Letter) |8 Videos

  • SEQUENTIAL OUTPUT TRACING

    LUCENT PUBLICATION|Exercise Exercise -4|5 Videos

  • SERIES COMPLETION : MIXED SERIES

    LUCENT PUBLICATION|Exercise Exercise ( Type-V: Miscellaneous )|1 Videos

Similar Questions

Explore conceptually related problems

ab - abb - bba - b

abb - baa - a - bab - ab

- a - b - abaa - bab - abb

- aa bc a - b - c - - bb - c - c

Which one set of letters when sequentially placed at the gaps in the given letter series shall complete it. c-bba - cab - ac - ab - ac

c _ bba _ cab _ ac _ ab _ ac

Prove that =|a cc-a a+cc bb-c b+c a-bb-c0a-c x y z1+x+y|=0 implies that a ,b ,c are in A.P. or a ,c ,b are in G.P.

Show that the determinant |a^2+b^2+c^2b c+c a+a bb c+c a+a bb c+c a+a b a^2+b^c+c^2b c+c a+a bb c+c a+a bb c+c a+a b a^2+b^2+c^2| is always non-negative. When is the determinant zero?

If a ,b , c are nonzero real numbers such that |b cc a a b c a a bb c a bb cc a|=0,t h e n 1/a+1/(bomega)+1/(comega^2)=0 b. 1/a+1/(bomega^2)+1/(comega^)=0 c. 1/(aomega)+1/(bomega^2)+1/c=0 d. none of these