Total Gadha Progression

7/5/13 MBA|CAT|CAT 2011|CAT 2012|CAT Online|MBA 2012|MBA Entrance Exams|CAT Test|Preparation|CAT Questions

totalgadha.com/mod/forum/discuss.php?d=4233 1/15

Home Forums CAT 2012 Quant Lessons Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous

Search forums

< Jump to... >

New Batches at TathaGat Delhi!

Display replies flat, with oldest first

Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Total Gadha - Thursday, 27 November 2008, 12:47 PM

Although the questions on progressions may not come directly in MBA exams, the theory behind progressions isused in every place where we need to sum up numbers. During your CAT 2010 preparations, the formula used inprogressions should become second nature to you as they will save a lot of time. Also, the methods of summingup various types of progressions, arithmetic, geometric or otherwise, should be very clear to you so that you areable to instantly spot the type of series you are facing. Knowing the basic methods of progressions also helps yousimplify a lot of complex series. So let's start with some basic progressions and their properties:

Arithmetic Progression

Numbers are said to be in Arithmetic Progression (A.P.) when the difference between any two consecutive numbers in the progression isconstant i.e. in an Arithmetic Progression the numbers increase or decrease by a constant difference.

Each of the following series forms an Arithmetical Progression:2, 6, 10, 14...10, 7, 4, 1, -2...a, a + d, a + 2d, a + 3d...

Example:

1. If the 7th term of an Arithmetical Progression is 23 and 12th term is 38 find the first term and the common difference.

Answer: 7th term = 23 = a + 6d ---- (1)

12th term = 38 = a + 11d ---- (2)Solving (1) and (2) we get a = 5 and d = 3

2. How many numbers of the series -9, -6, -3 â€¦ should we take so that their sum is equal to 66?Answer: n[-18 + (n - 1)3]/ 2 = 66

n2 - 7n - 44 = 0--> n = 11 or -4.The series is -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21..

We can see that the sum of first 7 terms is 0. The sum of next four terms after 7th terms gives us the sum. Otherwise, if we count 4 termsbackward from -9 we'll get the sum as -66.

3. What is the value of k such that k + 1, 3k - 1, 4k + 1 are in AP?Answer: If the terms are in AP the difference between two consecutive terms will be the same. Hence,(3k - 1) - (k + 1) = (4k + 1) - (3k - 1)2k - 2 = k + 2 --> k = 4.

TO INSERT ARITHMETIC MEANS BETWEEN TWO NUMBERS

Let n arithmetic means m1 , m2 , m3 ... mn be inserted between two numbers a and b. Therefore, a, m1 , m2 , m3 , ... mn, b are in arithmetic

progression.Let d be the common difference.

http://totalgadha.com/

http://totalgadha.com/mod/forum/index.php?id=1

http://totalgadha.com/mod/forum/view.php?f=24

http://totalgadha.com/help.php?module=moodle&file=search.html

http://www.totalgadha.com/

http://totalgadha.com/mod/resource/view.php?id=1160

http://www.statcounter.com/

http://totalgadha.com/user/view.php?id=2&course=1


http://totalgadha.com/cat-2010/



Since b is the (n + 2)th term in the progression, b = a + (n + 1)dWhence d = (b - a)/(n + 1)Hence m1 = a + (b - a)/(n + 1), m2 = a + 2(b - a)/(n + 1).. and so on.

Example:4. If 10 arithmetic means are inserted between 4 and 37, find their sum.First Method:

Let the means be m1 , m2 , m3 ... m10 . Therefore 4, m1 , m2 , m3 ... m10 , 37 are in AP and 37 is the 12th 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 = m1 + m10 = m2 + m9 = m3 + m8 = m4 + m7 = m5 + m6 = 41.

Hence m1 + m2 + m3â€¦ + m9 + m10= (m1 + m10) + (m2 + m9)...+ (m5 + m6)

= 5 x 41 = 205.

Example:5. The sum of three numbers in A.P. is 30, and the sum of their squares is 318. Find the numbers.Answer: Let the numbers be a - d, a, a + dHence a - d + a + a + d = 30 or a = 10The numbers are 10 - d, 10, 10 + d

Therefore, (10 - d)2 + 102 + (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 inthe CBT Club this week.

If you think this article was useful, help others by sharing it with your friends!

You might also like:Absolute Value (Modulus)Simple and Compound Interest

Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby pushpinder rana - Tuesday, 9 December 2008, 11:14 AM

Hi TG !

This is my first post in this forum although I am a regular visitor of this awesome web site (for CAT aspirants) for quiet a some time.What has provoked me to post is, for a couple of days I have been observing that the articles and the study marerial you have shared is notgetting loaded properly. Is this because of some LAN speed or the articles have been moved out of here ?

Regards,

Pushpinder.

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby sumit jamwal - Tuesday, 17 March 2009, 10:41 PM

hi Tg,

nice article.. .it would be nice if you could throw some light on ..topics like averages and mean values...many problems in DI come from same..

Regards,

sumit

Show parent | Reply

Find the sum...by Sam Rox - Tuesday, 21 April 2009, 05:16 PM

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

Show parent | Reply

Re: Find the sum...by Total Gadha - Tuesday, 28 April 2009, 01:03 AM

http://totalgadha.com/course/view.php?id=5

http://totalgadha.com/course/view.php?id=5

http://addthis.com/bookmark.php?v=250&pub=totalgadha

http://totalgadha.com/mod/forum/discuss.php?d=2111

http://totalgadha.com/mod/forum/discuss.php?d=129

http://totalgadha.com/mod/forum/post.php?reply=43871



http://totalgadha.com/mod/forum/discuss.php?d=4233#43871














Hi Sam,

Break it into two series- 1 + 6 + 11... and 4 + 5 + 6..

Total Gadha

Show parent | Reply

pls help.by srinivasan ravi - Tuesday, 9 June 2009, 10:42 AM

Here is an example of five consecutive positive integers whose sum is 1000: 198 + 199 + 200 + 201 + 202 = 1000. Find the largest number ofconsecutive positive integers whose sum is exactly 1000.

a)5b)10c)15d)20e)25pls help me solve this..thanks..

Show parent | Reply

Re: pls help.by vikas sharma - Tuesday, 9 June 2009, 04:44 PM

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?????

Show parent | Reply

Re: pls help.by TG Team - Wednesday, 10 June 2009, 12:14 PM

1000 = 23 x 53 = 5 x 200 = 25 x 40 = 16 x 62.5

1000 = 198 + 199 + 200 + 201 + 202 ( 5 terms)

1000 = 55 + 56 + ... + 60 + 61 + 62 + 63 + 64 + 65 + ...+ 69 + 70 (16 terms)

1000 = 28 + 29 + 30 + ... + 38 + 39 + 40 + 41 + 42 + ... + 50 + 51 + 52 (25 terms)

25 is the largest number of consecutive positive integers whose sum is exactly 1000.

Vikas

I have shown above the three ways to write 1000 as sum of consecutive positive integers.

Show parent | Reply

Re: pls help.by srinivasan ravi - Saturday, 13 June 2009, 09:11 AM

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

Show parent | Reply

Re: pls help.by srinivasan ravi - Saturday, 13 June 2009, 09:13 AM

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

Show parent | Reply

Re: pls help.by gaurav jain - Friday, 3 July 2009, 08:56 AM

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..

Show parent | Reply

Hi ...Please helpby harry sekhon - Wednesday, 15 July 2009, 01:48 PM

Hi TG,

Thanx for the amazing stuff on AP,GP etc.

Hi Everyone,

Please help in solving the problem below:-

How many 3 digit numbers have the property that their digits taken from left to right form an AP or GP.?

Options are: a)15 b)18

c)20 d)None of these

Show parent | Reply

Re: Hi ...Please helpby aashish biala - Thursday, 16 July 2009, 01:18 PM

hi harsimranjeet,

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

Show parent | Reply

Re: pls help.by William Wallace - Thursday, 16 July 2009, 08:50 PM

Nice approach kamal

Show parent | Reply

Re: pls help.by Ronak kabani - Thursday, 6 August 2009, 12:37 PM

Gaurav got ur approach - goodonekamal/William can u plz explain how did u get those three ways of expressing the no 1000is there any generalised way for finding a particular number as the sum of other numbers

Thanks in AdvanceRonak

Show parent | Reply

Doubtby Avishek Chakraborty - Sunday, 9 August 2009, 01:23 AM

TG Sir,

Firstly, I must say that the article is very helpful.

I have one doubt.In example 10 we got p/q as 232/990 and the sum(p+q) is taken as 1222.My question is why are we not minimizing this, Imean shouldn't it be 116/445 and the sum 561? Plz correct me if I am doing any mistake.

Thanx

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Sanjay J - Thursday, 13 August 2009, 04:35 PM

Hi TG / Kamal,

In the following question

five consecutive positive integers whose sum is 1000: 198 + 199 + 200 + 201 + 202 = 1000. Find the largest number of consecutive positive



























integers whose sum is exactly 1000.

Please explain the approach to select only 5, 25 and 16 as no of terms ...

Thanks

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby kumar sourav - Friday, 14 August 2009, 10:22 AM

Hi Sanjay,

actually what kamal did is check out how many times a factor of 1000 can be summated to get 1000 as answer..

so 1000=5*200 means 200 when added 5 times we get 10000..

and we know 199+201=2(200)..

likewise we will check which number hav maximum multiples..here one thing should be kept in mind that like in 25*40...dat 25 number shouldbe smaller tha 40...otherwise it will extend to negative digits

thanks

KS

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby abhishek tripathi - Sunday, 16 August 2009, 09:01 PM

hiiiiiii tg sir actulaay i faced problem undertsanding 2 points can u exaplain dem 1 one is u said tp/tq=(2p-1)/(2q-1).what does dis t signify is itda sum.and second one in da cat 03 question last step is 1+2/7(1+1/7+1/7square+.....)how did u arrive at da ans plz help me out

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Nitin Kumar - Wednesday, 19 August 2009, 01:18 AM

Hi TG Sir,

Tell me how to solve below question:

For how many integral values of P the three terms -1+P-X, P+X and -1+P+X can never be in HP?

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Ankur gupta - Friday, 9 October 2009, 12:27 AM

summation of reciprocal of first N no's?

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby ankit dikshit - Friday, 23 October 2009, 04:32 PM

TG SIR,, I feel my slef comfortable in the quantz n di section.. but u have provided light to some of my weak sections.. I can think Mathematics like amanager... Credit goes to you... I am preparing for cat since aug '08 .. enrolled wid ims full time at cp and time basic test series... but what iget it here in just 1000 bucks.. i have not found such .. even after class room coaching...

this is another gem of articles from TGs factory... I must say.. Kudos to TG Sir and his team..

I wish all success for u... Please launch few more buks on other topics like algebra, logical reasoning, di, modern maths etc.. before cat '09.. sothat we can take benefit from it...

One more suggestion... for DI/LR .. please add more themes on LR/DI in CAt CBT Club... i mean like tournaments/games/alligations .. if u willadd.. caselets/ blood relation/ quantz based reasoning etc.. it will be more benefiacial to all gahdas in TG cat cbt club.

REgards,THE ANKIT DIKSHIT :p

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Neha Sharma - Saturday, 24 October 2009, 05:04 PM

Hi TG,



























Can the method of differences be applied to solve the series whose terms follow this pattern in recurring manner, e.g. 1,6,21,52,105...(200terms)

Here the pattern is:

1 6 21 52 105

5 15 31 53

10 16 22

6 6 so the common difference is 6 after 3rd step.

How to solve this using method of differnces?

Show parent | Reply

Re: Hi ...Please helpby naveen kiran - Sunday, 25 October 2009, 01:39 AM

The following are in AP123,234,345..... 7 numbers135,246,357,.... 5 numbers147,258,369 3 numbers129 1 number

The following are in GP124,139,248 3 numbers111,222,333... 9 numbers

total = 28

hence option "d".

Please correct me if i'm wrong.

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Nikhil Jangi - Sunday, 25 October 2009, 07:02 PM

Hi TG,Please explain the fundas behind harmonic progressions.

Thanks & Regards,Nikhil

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Abinash Sahoo - Monday, 26 October 2009, 11:03 PM

plz helpi have a problem here....in Q-11,GP10+100+1000+...........here r=10n is nt defined.can we put S=ar*-1/r-1 ?i think it's undefined.

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Abinash Sahoo - Monday, 26 October 2009, 11:48 PM

hi TG sir..i have a doubt,plz explain..in Q-15,arithmetico-geometric series,how can the common ratio be 1/7.plz explain

Show parent | Reply

Re: pls help.by nishchai nevrekar - Sunday, 8 November 2009, 08:37 PM

this formula is from the comparison test..which says

(summation of 1/n where n ranges from 1 to 2k) > 1+k/2

here RHS is = 3 right.. so k = 4 ... hence summation ranges from 1 to 24.... hence ans is 16

Show parent | Reply

a doubt....!!by raja k - Tuesday, 10 November 2009, 12:54 PM



























TG sir,

u r doping a freat job and these articles r really use full for beginers like me....!!

i have a doubt sir,

if there exists two different AP series with some common values..

then how can we find the no of common terms in both the series..

hope u would reply soon..!!

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Subhash Medhi - Monday, 26 April 2010, 02:49 AM

Dear sir, Could you kindly write an article on hyper-geometric series too? Please do it for our sake.

Regards,Subhash

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby shail mishra - Thursday, 10 June 2010, 07:53 PM

Hi GuysPlease help in solving this question.

Find the sum of the following series upto infinity.1/1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 ......

It seems to be a very easy question but i don't know why is it not clicking.

RegardsSM

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby TG Team - Friday, 11 June 2010, 01:55 PM

Hi Shail

See the nth term is: 2/n(n + 1) = 2[1/n - 1/n+1]

So the required sum is S = 2[(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6) + (1/6 - 1/7) + ....] = 2.

Good question.

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby siddharth jain - Thursday, 2 September 2010, 04:28 PM

hiii... i m new to this site but can help u out....1x1, 2x3, 3x7, 4x13.... adding 1 to first digit and 2,4,6 to second digit of terms

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Puneet Gulati - Friday, 3 September 2010, 04:41 PM

not so impressive as compared to other articles on total gadha

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby TG Team - Tuesday, 14 September 2010, 04:17 PM

Hi Siddharth

nth term of the second series i.e. 1, 3, 7, 13, ... = n2 - n + 1.

So nth term of the given series is = n(n2 - n + 1) = n3 - n2 + n



























And sum of which is given by = n(n + 1)(3n2 - n + 4)/12.

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby preeti kumari - Thursday, 5 May 2011, 05:19 PM

hicould somebody please help me solve the following:consider the sequence where nth term , tn=n/(n+2), n=1, 2.... The value of t3*t4*t5....*t54 equalsa)2/495b)2/477c)12/55d)1/485e)1/2970

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby destiny unruled - Tuesday, 10 May 2011, 03:12 AM

@ Preeti Kumari

I think it should be t3*t4*t5*.......*t53 not t54

t3*t4*t5*.......*t53 = (3/5)(4/6)(5/7)....(53/55)

We can see that denominator of a term will cancel with numerator of next next term.

So, product = 3*4/(54*55) = 2/495

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby preeti kumari - Tuesday, 10 May 2011, 09:03 AM

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

Show parent | Reply

Re: pls help.by Jitendra Soni - Tuesday, 6 September 2011, 05:29 PM

see if this helps consider the series a-d, a-(d-1)... a-1, a, a+1, a+2, a+ 3...a + (d-1) , a+ d then sum = a(2d+1) = 1000 = 2^3 X 5^3 ... now2d+1 is odd in the LHS, hence it can be one of the 3 factors 5, 25 or 125... this gives d = 2,12,62 & a = 200, 40 & 8 respectively. Now there is acondition that each term should be positive, hence a-d > 0 is satisfied in (a,d) = ( 200,2) & (40, 12). This generates (198, 199, 200, 201, 202)(5 terms) and ( 28, 29.... 40....51, 52) (25 terms). The condition for +ve terms is not satisfied in (a,d) = ( 8,62) so we can take a = 63 at least( so that a-d =1). Now the number of terms will be 1000/63 which gives 15. sumthing :p hence 16 terms in total hence 63 ...70 ( 8 terms) and(55...62) (8 terms) - Jeetu

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby sonnel singh - Wednesday, 21 September 2011, 03:27 AM

can smbdy plz answer this ques..

Ques- Two distinct increasing integer arithematic progressions with same first term equal to 1 are such that the product of their nth neth is2010. find the maximum possible value of n.

1) 20092) 23) 34) 85) 15

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby user332 u - Wednesday, 21 September 2011, 07:04 PM

sir i am not able to solve this problem..so plz help me

7 + 26 + 63 +124 + ......+999

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby sahil madan - Wednesday, 21 September 2011, 07:38 PM

this series can be written as 2^3-1 + 3^3-1 + 4^3-1 + .....+10^3-1































So apply the formula for summation of cubes of n natural no's..I hope you can proceed from here..

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby user332 u - Thursday, 22 September 2011, 03:43 PM

really thanks......

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Assassin CT - Saturday, 1 October 2011, 03:50 PM

I think answer to this Ques- Two distinct increasing integer arithematic progressions with same first term equal to 1 are such that the product of their nth neth is2010. find the maximum possible value of n.

1) 20092) 23) 34) 85) 15

is 3.{n(n-1)(d+D)/2} +2n =2010 is the equation we get at last.

For only n=2,3 d+D is integer.

Highest value is 3 for n.

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby sonnel singh - Sunday, 9 October 2011, 02:33 AM

but how did you get to the following equation

{n(n-1)(d+D)/2} +2n =2010

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Rajasekaran Rajaram - Monday, 10 October 2011, 10:50 PM

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)*d1 = nth term of first series

Series 2 --> 1+(n-1)*d2 = nth term of first series

[1+(n-1)*d1]*[1+(n-1)*d2] = 1+ (n-1)*d1+(n-1)*d2+(n-1)*d1*d2

1+(n-1)[d1+d2+(n-1)*d1d2] = 2010

(n-1)[d1+d2+(n-1)*d1d2] = 2009

2009 = 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.

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby sonnel singh - Friday, 14 October 2011, 03:37 PM

thnk u so much Rajasekaran Rajaram

Show parent | Reply


























Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby n k - Monday, 31 October 2011, 01:04 PM

Hi Everyone,

Can you let me know how to find common terms from two series.

For Ex: How many integers do the following finite arithmetic progressions have in common?1,8, 15, 22, ..,2003 and 2, 13, 24, 35, ., 2004

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby TG Team - Tuesday, 1 November 2011, 03:43 PM

Hi n k

First series has the numbers of the form: 7a + 1

and the second one of the form: 11b + 2

So first common term of the two series is the smallest number which satisfies both the above expressions and i.e. 57 andall other such numbers will be at a gap of 77.

So we need to find the number of terms of an AP whose first term is 57, common difference is 77 and last term is ≤2003.

Kamal Lohia

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby nisha rani - Tuesday, 1 November 2011, 10:20 PM

how do u find sum of following series?

3/5+5/36+7/144+........+21/12100

here tn=2n+1/(n(n+1))^2

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby King Casanov - Wednesday, 2 November 2011, 02:25 AM

i guess u urself have given the clue to solution...

tn=2n+1/ (n(n+1))^2 now (1/n^2) - (1/(n+1)^2)== 2n+1/(n(n+1))^2

The KING

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby King Casanov - Wednesday, 2 November 2011, 11:16 PM

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, butplease do reply ..asap)

P.S. : please tell me, whether my above written questions are grammatically correct??

The KING

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Rahul Kishore - Monday, 7 November 2011, 12:01 PM

Hi,

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

Show parent | Reply



















http://totalgadha.com/mod/forum/view.php?f=24









Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Mohit Sharma - Sunday, 1 April 2012, 11:34 AM

Gr8 Article Sir,Your work is absolutely fantastic. Really a helpful site for MBA Aspirants. Can you please solve this question for sir,

Ques: If x,y,z are in GP and a^x,b^y and c^z are equal, than a,b,c are in which progression?

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby arsh arora - Sunday, 1 April 2012, 12:46 PM

hi mohit,

try taking x y z as 1 2 and 4 and a,b and c as 16 4 and 2 u will see thata b and c dnt follow any pattern thereby no series pattern...

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Mohit Sharma - Tuesday, 3 April 2012, 03:45 PM

Thanks Arsh. for the help

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby arsh arora - Tuesday, 3 April 2012, 05:51 PM

anytime buddy!!!

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby TG Team - Wednesday, 4 April 2012, 12:07 PM

Hi Mohit

Question you are asking is not correct. It should be going like this: If a, b, c are in GP and ax = by = cz, then x, y, z arein which progression?

And the correct answer is : HP

PS: This question was reported in previous year's CAT by many students.

Kamal Lohia

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Mohit Sharma - Wednesday, 4 April 2012, 07:04 PM

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

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby tushar nijhawan - Sunday, 14 October 2012, 01:38 AM

Show parent | Reply

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneousby Gaurav Jotriwal - Sunday, 14 October 2012, 03:53 PM

Dear TG sir,

what will be the solution of:10.11.12.13 + 11.12.13.14 + ....... + 96.97.98.99

RegardsGaurav

Show parent | Reply



































MBA CAT GMAT

You are not logged in. (Login)

http://www.totalgadha.com/

http://totalgadha.com/login/index.php

Documents

Total Gadha Progression