### Discussion :: Letter and Symbol Series - Type 1 (Q.No.2)

In these series, you will be looking at both the letter pattern and the number pattern. Fill the blank in the middle of the series or end of the series.

Priya said: (Nov 30, 2010) | |

Hi guys, I didnt understand the above logic. |

Bala said: (Dec 1, 2010) | |

i didn't get the logic |

Gurram said: (Dec 8, 2010) | |

I didn't understand. |

Nagendra said: (Dec 25, 2010) | |

please explain clearly |

Rashmi said: (Jan 26, 2011) | |

Hi guys. Can anyone explain above logic clearly? |

Narender Mgu said: (Jan 31, 2011) | |

Hi friends, can any one explain the above logic. |

Manish Gupta said: (Jan 31, 2011) | |

I didn't realize it fully.... |

Anusha said: (Feb 1, 2011) | |

In the question, there are only three alphabets in sequence B,C,D....The Numbers are also in sequence where the 3rd sequence/number is missing hence BC3D is the answer. Replace the question with the given answer: B2CD, (BC3D), BCD4, B5CD, BC6D Now the numbers are in sequence 2,3,4,5,6... Hope you got it now guys! |

Ajay said: (Mar 2, 2011) | |

Concentrate on the no's dude first it comes to b then c then d and again started from b, c and so on. |

Diya M said: (Apr 7, 2011) | |

The letters are the same for each case only the no.s 2,3,4....... are moving in cyclic order i.e. from b to c & from c to d and and when one cycle is finished we again go back to the previous one i.e. to B. |

Anand Upadhyay Sagar said: (Apr 8, 2011) | |

watch anusha comments she is giving right answer |

Deepak said: (May 16, 2011) | |

The letter are same for each case only no are change in every order for eg. first b2cd then one no increase it make bc3d |

Murali said: (Jul 22, 2011) | |

Just keep concentrate on the numbers only. Which are given in series. Their is missing a number i.e '3', in the series 2, 3, 4, 5, 6. So the answer is 'B'. |

Ritesh Sinha said: (Sep 19, 2011) | |

Hi, In the above quest they are asking for sequential series ok...just concentrate the series in first term b2cd = b+b+c+d(4 letter) in third term i.e bcd4 = b+c+d+d+d+d(total 6 letter)similarly 4th term contain 7 letter and last term contain 8 letter therefore according to series there must be contain 5 letter in 2nd term. now see the option i.e how many option u have that contain 5 letter only two option are there i.e option A & opt B so we go for option B only because no one term contain two word simultaneously more than one times except option B i.e in option A(it contain 5 letter but here two word i.e B,B and C,C simultaneously)thats y we go for option B. this is the right ans. |

Sravanthi said: (Jan 13, 2012) | |

It is very simple thing letter is same its didn't change but number is in sequence the sequence is 2 3 4 5 6 and in the options b and d contains 3, but in d option it also had 2 so it is not correct option then b is the correct. |

Narendra said: (Apr 30, 2012) | |

Tell me what is logic in this? |

Megha Singhal said: (Jun 17, 2012) | |

In the above question they are asking for sequential series ok. Just concentrate the series in first term B_{2}CD = B+B+C+D (4 letter) in third term i.e., BCD_{4} = B+C+D+D+D+D (total 6 letter) similarly 4th term contain 7 letter and last term contain 8 letter therefore according to series there must be contain 5 letter in 2nd term. Now see the option i.e., how many option you have that contain 5 letter only two option are there i.e., option A & opt B so we go for option B only because no one term contain two word simultaneously more than one times except option B i.e., in option A (it contain 5 letter but here two word i.e., B, B and C, C simultaneously) thats why we go for option B. This is the right answer. |

Chandana said: (Dec 3, 2012) | |

It is very simple see the question don't concentrate on letter please concentrate on numbers. |

Sameera said: (Aug 28, 2013) | |

By looking at the numbers I got the answer its BC3D. |

Dinesh said: (Nov 16, 2013) | |

Just see the base of 1st letter is 2, next time the base of second letter is 3, next time the base of 3rd letter is 4, Similarly for next series... b2c3d4, b5c6d7, |

Swap said: (Feb 18, 2014) | |

Because all numbers are moving in circular path. First number 2 on B (i.e B2Cd) , then it get incremented by one i.e. 3 moves to next alphabet (i.e C) form BC3D (Ans). Then again number get incremented by one i.e 4 moves to next alphabet (i.e D) form BCD4, Now Here Alphabet are Three only (B, C, D) which are in circular form (means After D here B will come) , again numbers moves to 1st alphabet i.e 'B' Hence next incremented number get moved to B (i.e B5CD) by this way After BC6D number will come BCD7 (In Advanced). |

Santhosh Kumar P said: (May 3, 2015) | |

Very nice and I understanding this logic its good thanks. |

Dee said: (Jul 11, 2015) | |

Thanks @Anusha. |

Sunil Brahmajosyula said: (Aug 18, 2015) | |

Base values are in the order of 4, ---, 6, 7, 8 find the base value of 5 from the options. |

Surya Teja.M said: (Nov 6, 2015) | |

B2CD, _BC3D_, BCD4 = in B2CD B has number 2, in BCD4 D has a number 4 C has no number so answer is BC3D. The series of the number is 2, 3, 4. B2 has 2 in first set of words, C has missing number so we have entered 3 for C in second set of words i.e., C3, D4 in third set of words. |

Rubini said: (Jan 24, 2016) | |

Hai friends I got the answer. |

Harsha said: (Mar 13, 2016) | |

I can't understand this logic. |

Harsha said: (Mar 13, 2016) | |

@Rubini tell me about the your answer. |

Williamkery said: (May 3, 2016) | |

First number B2CD = B + B + C + D = 4 "B2". First series start with 2. Second number BC3D = B + C + C + C +D = 5 "C3". Second series Sequence 3. Third Numbr BCD4 = B + C + D + D + D + D = 6 "D4" Third Series sequence 4. Fourth Number B5CD = B + B + B + B + B + C + D =7 "B5" Alternator continue with number 5. Fifth number BC6D =B +C + C + C + C + C + C + D =8 "C6" This the sequence with 6. |

Gomathi said: (Oct 27, 2016) | |

On this letter series the number will take b to c, and c to d, and again d to b like triangle wise. So first, b series they had been mentioned the substring of 2. So, what will happen, next to b, c series will get the next process, after that d, next c. |

Pandu said: (Dec 16, 2016) | |

How will come? please explain. |

Prakash said: (Jan 13, 2017) | |

Someone please explain me to get the answer. |

Krishna said: (Jun 2, 2017) | |

B2CD, -------, BCD4, B5CD, BC6D. SOLVE, 2+1=3 after one word BC3D. 3+1=4 after one word BCD4. 4+1=5 after one word (repeat from first word because D is end) B5CD. 5+1=6 after one word BC6D. |

Nithin said: (Aug 16, 2017) | |

I didn't get logic, please explain me. |

Swati said: (Nov 25, 2017) | |

Please explain the last term. |

Anjali said: (Dec 28, 2017) | |

Steps. 1. Give number to all alphabets as 1, 2, 3, 4 (a-1, b-2, c-3). 2. You will notice that number series 234 is repeated. 3. Observe that in the first set, the first alphabet is repeated 2 times, in the second set should repeat. Times, in third series third alphabet is repeated four time. The circle is repeated with increasing number of repeating alphabet. So at the blank space second alphabet should repeat 3 times. |

Sushri said: (Jun 10, 2018) | |

Please, explain it clearly that how they came in order? |

Rupa said: (Sep 8, 2018) | |

I can understand now thanks for your explanation. |

Veeresh said: (Jan 23, 2019) | |

B2CD, (BC3D), BCD4, B5CD, BC6D. Series is 2,3,4,5,6. 1st B-2. 2nd-C-3. 3rd-D-4. . . . Cycle get repeats with an increase in numbers of the Conservative letters. 4th B-5. 6th-C-6. 8th -D-7. Thanks . |

#### Post your comments here:

Name *:

Email : (optional)

» Your comments will be displayed only after manual approval.