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

arithmetic progresssion geometric progression cat 2010 2011

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:

Arithmetic Progression

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:
2, 6, 10, 14...
10, 7, 4, 1, -2...
a, a + d, a + 2d, a + 3d...

           arithmetic progression geometric progression cat 2007 2008

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 … 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 terms backward 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.
         arithmetic progression geometric progression cat 2007 2008  

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.
Since b is the (n + 2)th term in the progression, b = a + (n + 1)d
Whence 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 = 3
Therefore 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 + m3… + m9 + m10= (m1 + m10) + (m2 + m9)...+ (m5 + m6)
= 5 x 41 = 205.

               cat 2007 2008 arithmetic progression geometric progression

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 + d
Hence a - d + a + a + d = 30 or a = 10
The numbers are 10 - d, 10, 10 + d
Therefore, (10 - d)2 + 102 + (10 + d)2 = 318
Or d = 3, therefore the numbers are 7, 10, and 13.

arithmetic progression geometric progression cat 2007 2008

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.

 

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Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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 not getting loaded properly. Is this because of some LAN speed or the articles have been moved out of here ?

Regards,

Pushpinder.

 

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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,
sumitcool
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 regardscool
Re: Find the sum...
by Total Gadha - Tuesday, 28 April 2009, 01:03 AM
  Hi Sam,

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

Total Gadha
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 of consecutive positive integers whose sum is exactly 1000.

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

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?????

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. smile

Vikas smile

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

 

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??
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
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..
Hi ...Please help
by 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

Re: Hi ...Please help
by 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.

Re: pls help.
by William Wallace - Thursday, 16 July 2009, 08:50 PM
  Nice approach kamal
Re: pls help.
by Ronak kabani - Thursday, 6 August 2009, 12:37 PM
  Gaurav got ur approach - goodone
kamal/William can u plz explain how did u get those three ways of expressing the no 1000
is there any generalised way for finding a particular number as the  sum of other numbers

Thanks in Advance
Ronak
Doubt
by 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, I mean shouldn't it be 116/445 and the sum 561? Plz correct me if I am doing any mistake.

Thanx

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

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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 should be smaller tha 40...otherwise it will extend to negative digits

thanks

KS

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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 it da 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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?

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by Ankur gupta - Friday, 9 October 2009, 12:27 AM
  summation of reciprocal of first N no's?
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by ankit dikshit - Friday, 23 October 2009, 04:32 PM
  TG SIR,, smile
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 a manager... 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 i get 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..smile

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.. so that 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 will add.. 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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...(200 terms)

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?

Re: Hi ...Please help
by naveen kiran - Sunday, 25 October 2009, 01:39 AM
  The following are in AP
123,234,345..... 7 numbers
135,246,357,.... 5 numbers
147,258,369 3 numbers
129 1 number

The following are in GP
124,139,248 3 numbers
111,222,333... 9 numbers

total = 28

hence option "d".

Please correct me if i'm wrong.
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by Nikhil Jangi - Sunday, 25 October 2009, 07:02 PM
  Hi TG,
Please explain the fundas behind harmonic progressions.

Thanks & Regards,
Nikhil
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by Abinash Sahoo - Monday, 26 October 2009, 11:03 PM
  plz help
i have a problem here....
in Q-11,GP
10+100+1000+...........
here r=10
n is nt defined.
can we put
S=ar*-1/r-1    ?
i think it's undefined.
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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
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
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..!! 

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by shail mishra - Thursday, 10 June 2010, 07:53 PM
  Hi Guys
Please 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.

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

Hi Shailsmile

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. smile

Good question.

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by Puneet Gulati - Friday, 3 September 2010, 04:41 PM
  not so impressive as compared to other articles on total gadha
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by TG Team - Tuesday, 14 September 2010, 04:17 PM
 

Hi Siddharth smile

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. smile 

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by preeti kumari - Thursday, 5 May 2011, 05:19 PM
  hi
could 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 equals
a)2/495
b)2/477
c)12/55
d)1/485
e)1/2970
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by preeti kumari - Tuesday, 10 May 2011, 09:03 AM
  thanx.. smile
and yes it is t53, not t54...my mistake!!
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 ... now 2d+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 a condition 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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 is 2010. find the maximum possible value of n.

1) 2009
2) 2
3) 3
4) 8
5) 15
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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

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

really thanks......

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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 is 2010. find the maximum possible value of n.

1) 2009
2) 2
3) 3
4) 8
5) 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.
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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.

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by sonnel singh - Friday, 14 October 2011, 03:37 PM
  thnk u so much Rajasekaran Rajaram
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by TG Team - Tuesday, 1 November 2011, 03:43 PM
 

Hi n ksmile

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 and all 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

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

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

Hi,

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

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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?
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by 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...

Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by Mohit Sharma - Tuesday, 3 April 2012, 03:45 PM
  Thanks Arsh. for the help
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by arsh arora - Tuesday, 3 April 2012, 05:51 PM
  anytime buddy!!!
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by TG Team - Wednesday, 4 April 2012, 12:07 PM
 

Hi Mohit smile

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 are in which progression?

And the correct answer is : HP

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

Kamal Lohia  

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

Regards
Gaurav
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by ankit SAHAY - Sunday, 22 September 2013, 09:31 PM
  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

?
Re: Methods of Summation- Arithmetic Progression, Geometric Progression and Miscellaneous
by Pradeep Kumar - Tuesday, 1 October 2013, 10:12 AM
 

Hi Ankit,

You can group the odd and even (negative and positive respectively) terms separately. It would be of the the form:

(2n) ^ 3 - (2n -1) ^ 3

And here n will be 1,2...8 and not sixteen. Use the sum of cubes formula and you should be able to arrive at the solution.