Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

Although the questions on progressions may not come
directly in MBA exams, the theory behind progressions is used in every place
where we need to sum up numbers. During your CAT 2010 preparations, the
formula used in progressions should become second nature to you as they will
save a lot of time. Also, the methods of summing up various types of
progressions, arithmetic, geometric or otherwise, should be very clear to you
so that you are able to instantly spot the type of series you are facing. Knowing
the basic methods of progressions also helps you simplify a lot of complex
series. So let's start with some basic progressions and their properties:
Numbers are said to be in Arithmetic Progression (A.P.) when the difference between any two consecutive numbers in the progression is constant i.e. in an Arithmetic Progression the numbers increase or decrease by a constant difference. Each of the following series forms an Arithmetical
Progression:
^{th} term of an Arithmetical
Progression is 23 and 12^{th} term is 38 find the first term and the
common difference.Answer: 7 ^{th} term = 23
= a + 6d ---- (1)12 ^{th} term = 38 = a +
11d ---- (2)Solving (1) and (2) we get a = 5 and d = 3
Let n arithmetic means m
First Method:Let the means be m _{1},
m_{2}, m_{3}... m_{10}. Therefore 4, m_{1}, m_{2},
m_{3}... m_{10}, 37 are in AP and 37 is the 12^{th} term
in the arithmetic progression. Hence, 37 = 4 + 11d --> d = 3Therefore means are 7, 10, 13 ... 34 and their sum is 205. Second Method:We know that in an AP- the sum of first term + last term = sum of second term + second last term = the sum of third term + third last term = .. and so on. Therefore, 4 + 37 = m _{1}
+ m_{10} = m_{2} + m_{9} = m_{3} + m_{8}
= m_{4} + m_{7} = m_{5} + m_{6} = 41.Hence m _{1} + m_{2}
+ m_{3}â€¦ + m_{9} + m_{10}= (m_{1} + m_{10})
+ (m_{2} + m_{9})...+ (m_{5} + m_{6})= 5 x 41 = 205.
Answer: Let the numbers be a - d, a, a + d Hence a - d + a + a + d = 30 or a = 10 The numbers are 10 - d, 10, 10 + d Therefore, (10 - d) ^{2}
+ 10^{2} + (10 + d)^{2} = 318Or d = 3, therefore the numbers are 7, 10, and 13. I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this in the CBT Club this week. You might also like: Absolute Value (Modulus) Simple and Compound Interest |

Find the sum... | |

Hi TG, I have been unable to get what is the sum of 1+4+6+5+11+6+... 200terms?? could you please help me out on this problem?? Thanks n regards |

Re: Find the sum... | |

Hi Sam, Break it into two series- 1 + 6 + 11... and 4 + 5 + 6.. Total Gadha |

Re: pls help. | |

hi ravi as per TG sir, a number can be written as sum 2 or more cosecutive nos is=no of odd idvisor-1; so here odd divisor 4-1=3
pls correct me????? |

Re: pls help. | |

thanks kamal..but how did u find those 3 ways..is there any method?? |

Re: pls help. | |

help me solve this problem How many terms, at the minimum of sequence 1,1/2,1/3,1/4...., must be added together for their sum to be not less than 3? (1) 16 (2) 17 (3) 18 (4) 20 |

Re: pls help. | |

hi ravi. u can solve the question with the help of the choices given. 1000=n/2 [2a +n-1] since d=1 when n=25 a is an integer.. hence thats the right answer.. |

Re: Hi ...Please help | |

hi harsimranjeet, i think the answer to your question is 38 and hence, the correct option would be none of these. |

Re: pls help. | |

Nice approach kamal |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

summation of reciprocal of first N no's? |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

Hi TG, Please explain the fundas behind harmonic progressions. Thanks & Regards, Nikhil |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

not so impressive as compared to other articles on total gadha |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

thanx.. and yes it is t53, not t54...my mistake!! |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

sir i am not able to solve this problem..so plz help me 7 + 26 + 63 +124 + ......+999 |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

really thanks...... |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

but how did you get to the following equation {n(n-1)(d+D)/2} +2n =2010 |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

sonnel singh, I would go with option 4) 8. Consider two series with different difference between them i.e. d1 and d2. Series 1 --> 1+(n-1)*d _{1} = nth term of first seriesSeries 2 --> 1+(n-1)*d _{2} = nth term of first series[1+(n-1)*d _{1}]*[1+(n-1)*d_{2}] = 1+
(n-1)*d_{1}+(n-1)*d_{2}+(n-1)*d_{1}*d_{2}1+(n-1)[d _{1}+d_{2}+(n-1)*d_{1}d_{2}] = 2010(n-1)[d _{1}+d_{2}+(n-1)*d_{1}d_{2}] = 20092009 = 41*7*7*1 n-1 can be 41 or 7 or 1 only. So n can be 42,8 or 2. 2 and 8 are in options. For maximum value i would go with 8. |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

thnk u so much Rajasekaran Rajaram |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

how do u find sum of following series? 3/5+5/36+7/144+........+21/12100 here tn=2n+1/(n(n+1))^2 |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

TG Team, Are the fundas (and topics) covered in CAT 2011 Quant Lessons complete as far as whole "important topics" of CAT are concerned?? And does a good understanding of above mentioned lessons help me in getting into one of"II" ?(yes I know its a very vague question, but please do reply ..asap) P.S. : please tell me, whether my above written questions are grammatically correct?? The KING |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

Hi, How to find the n'th term of a random series. |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

Thanks Arsh. for the help |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

anytime buddy!!! |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

Thanks sir. I got this question from Amit Sharma. I think wat u r saying must b correct. |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

Dear TG sir, what will be the solution of: 10.11.12.13 + 11.12.13.14 + ....... + 96.97.98.99 Regards Gaurav |

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous | |

help solving this pls; If S = - 1^3 + 2^3 - 3^3 + 4^3 - ..... + 14^3 - 15^3 + 16^3, then the value of S is ? |