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# toughest progression questions

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Manager
Joined: 29 Dec 2009
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toughest progression questions [#permalink]  18 Aug 2010, 11:49
6
KUDOS
ARITHEMATIC PROGRESSION

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 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.

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.

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.
Intern
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Re: toughest progression questions [#permalink]  19 Aug 2010, 21:18
awesome...thanks.
Manager
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Kudos [?]: 13 [0], given: 10

Re: toughest progression questions [#permalink]  19 Aug 2010, 23:40
if ppl like the post then i can post most diificult questions of GP,POWERS AND MOD. After solving these questions ppl can handle any question on GMAT.
CEO
Status: Nothing comes easy: neither do I want.
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Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
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Kudos [?]: 817 [1] , given: 232

Re: toughest progression questions [#permalink]  20 Aug 2010, 03:03
1
KUDOS
in the fourth example

total sum after insertion = n/2 * (a+l) where a = first term and l = last term
n becomes 2+10 = 12
thus sum = 12/2 * ( 4+37 ) = 246

sum of 10 means = 246 - 4 - 37 = 205
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VP
Status: Current Student
Joined: 24 Aug 2010
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WE: Sales (Consumer Products)
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Re: toughest progression questions [#permalink]  13 Sep 2010, 12:01
Would you mind walking through the steps of the algebra in example 1. How do we solve for (1) and (2). Sorry, to ask such basic questions, but I'm not too good at solving equations with multiple variables. Thank you.

ETA: Also could you please explain how you came up with the formula to solve example 2. Thanks!
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Manager
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Re: toughest progression questions [#permalink]  14 Sep 2010, 23:04
hi jatt86 can you post the questions of GP,POWERS AND MOD. I am waiting for those !!
Senior Manager
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Schools: UC Berkley, UCLA
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Re: toughest progression questions [#permalink]  13 Feb 2011, 16:15
Great Compilation!!!
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Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela

Manager
Joined: 17 Feb 2011
Posts: 202
Concentration: Real Estate, Finance
Schools: MIT (Sloan) - Class of 2014
GMAT 1: 760 Q50 V44
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Kudos [?]: 445 [0], given: 70

Re: toughest progression questions [#permalink]  17 Feb 2011, 16:58
Can you please post the GP questions? I have problems with tough GP questions. Thanks!
Intern
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Re: toughest progression questions [#permalink]  20 Feb 2011, 01:51
Great Set of qs

In question 2 could someone tell me why the first term is not -9 and instead -18

Thanks
Manager
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Kudos [?]: 11 [0], given: 46

Re: toughest progression questions [#permalink]  01 Jul 2011, 13:55
cheetarah1980 wrote:
Would you mind walking through the steps of the algebra in example 1. How do we solve for (1) and (2). Sorry, to ask such basic questions, but I'm not too good at solving equations with multiple variables. Thank you.

ETA: Also could you please explain how you came up with the formula to solve example 2. Thanks!

here...q1.

(a7)=23=(a1)+(7-1)*d=(a1)+6d
(a12)=38=(a1)+(12-1)*d=(a1)+11d

(a1) + 11d = 38
-(a1) - 6d = -23
=> 5d = 15 ie d=3

then substitute in any of the above equations and get a = 5

clear now? hope it helped

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How charged with punishments the scroll,
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I am the captain of my soul.
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Manager
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Re: toughest progression questions [#permalink]  01 Jul 2011, 13:57
Pixy wrote:
Great Set of qs

In question 2 could someone tell me why the first term is not -9 and instead -18

Thanks

the formula for sum of n terms is

S=(n/2)(2a+(n-1)d)
or
S=(n/2)(an+a1) but since an=a+(n-1)d it translates to the same formula!

Hope it helped.
_________________

It matters not how strait the gate,
How charged with punishments the scroll,
I am the master of my fate :
I am the captain of my soul.
~ William Ernest Henley

Intern
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Re: toughest progression questions [#permalink]  01 Jul 2011, 17:53
These are great. Kudos to you!
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Re: toughest progression questions [#permalink]  06 Dec 2013, 22:31
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Re: toughest progression questions   [#permalink] 06 Dec 2013, 22:31
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