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Tips and Tricks for Sequences Questions

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Tips and Tricks for Sequences Questions [#permalink]

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New post 04 Oct 2017, 12:32
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Guys, lets start with few basic rules.

Rule 1: Positions in an AP or GP are relative to each other.

Example: Given \(T_2\)=p,\(T_4\)=q find \(T_8\), where \(T_n\)is the n-th term in the AP.

Think of \(T_4\)−\(T_2\)=q−p as one "step".
\(T_8\) is 2 steps away from \(T_4\) (think \(T_2,T_4,T_6,T_8\)).
Hence \(T_8\)=q+2(q−p)=3q−2p

In this way we avoid having to first work out a and d by solving simultaneous equations and then applying the formula for \(T_8\).

Rule 2: We can apply the concept of a zero or negative number-of-term index to provide symmetry to our variables . We can even consider shifting all the index by a constant number.

Example: Three consecutive terms in an AP have a sum of x and product of y. Find the three terms.
its always better to assume the three terms be a−d,a,a+d.

This is much faster than solving for the terms a,a+d,a+2d

We can understand the concepts quickly by looking at the problems below. Lets see how we can solve few 700-level problems using these.


Problem 1:


If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence?
A. 2
B. 3
C. 4
D. 5
E. 6

The usual way to solve the problem is to frame 2 equations and solve them simultaneously.
However, we can use our trick to solve it.

Given, \(a_4\)+\(a_7\)=40
We can take \(a_4\)=a and \(a_7\)= a+3d.
But this complicates our equation. (See trick 2)

So, let’s take the mid index of \(a_4\) and\(a_7\) which is (\(a_{5.5})\)
Note: For our purpose, let’s assume that fractional index exists.
Let \((a_{5.5})\)= a
Now, \(a_7\)=a+3d
Why we took 3d? Why not 1.5 d? Because the distance \(a_{5.5}\) and \(a_7\) = 1.5 = 3(0.5) = 3 index units

Similarly, \(a_4\)= a-3d
But \(a_4\)+\(a_7\)= 40 (given)
Simple substitution gives: (a-3d)+ (a+3d)=40
=>2a= 40
=>a=20

Now, we have asked to find the value of \(a_1\)= a-9d=20-9d
In the GMAT world, d will never be equal to 0.6 or 0.7
Thus, d can only be either 1 or 2 so that a is a positive number. (Why only positive?? See the options)
Thus, a can be either 20-9(1) or 20-(9)2
Thus, a can be either 11 or 2.
Since 11 is not in the option, answer is Option (A).

Note: We solved the problem without even forming any second equation ( data given about GP).

Lets do another 700 level question.


Problem 2:


The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?
A. 300
B. 120
C. 150
D. 170
E. 270

For the symmetry reasons mentioned above, lets consider our base to be the midpoint of \(a_4\) and \((a_{12})\) which is \(a_8\)
Let \(a_8\)=a
Since we are dealing with 4th term, 8th term and the 12th term. Let’s assume common difference of 4d.
I mean let’s say 4d=D= our common difference for this question

Thus, \(a_4\)= a-D and \(a_{12}\)= a+D
Hence, \(a_4\)+\(a_{12}\)= 2a

According to the question, 2a=20
=> a=10

Now lets apply the alternate formula for finding the sum, which is:
Sum of the AP = Middle term of AP x No of terms = \(a_8\) * 15= 10*15=150

Lets do one last question.


Problem 3:


The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?
(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Let the 4 consecutive odd numbers be b-3, b-1, b+1, b+3.
Since, b-1 is odd => b is even.
Let the 3 consecutive even numbers be a-2, a, a+2. => a is even

Now, sum of four consecutive odd numbers= the sum of 3 consecutive even numbers
=> 4b= 3a
=> 4(even)= 3a
=> 8k= 3a ( 4 times an even number will be a multiple of 8)
=> a is a multiple of 8.

Now we can find the number of multiples of 8 between 101 and 200.
101 <x< 200
=> 8(13)<x<8(25)

Thus, the total number of multiples of 8 will be total number of natural numbers between 13 and 24.
=> 24-13+1= 12 Option (A)

Thus, we see that the rules are simple but powerful and we can solve a lot of time if we apply these rules in any Sequence questions. Please let me know if this helped you in any way.
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Re: Tips and Tricks for Sequences Questions [#permalink]

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New post 04 Oct 2017, 12:42
vivek6199 wrote:
Guys, lets start with few basic rules.

Rule 1: Positions in an AP or GP are relative to each other.

Example: Given \(T_2\)=p,\(T_4\)=q find \(T_8\), where \(T_n\)is the n-th term in the AP.

Think of \(T_4\)−\(T_2\)=q−p as one "step".
\(T_8\) is 2 steps away from \(T_4\) (think \(T_2,T_4,T_6,T_8\)).
Hence \(T_8\)=q+2(q−p)=3q−2p

In this way we avoid having to first work out a and d by solving simultaneous equations and then applying the formula for \(T_8\).

Rule 2: We can apply the concept of a zero or negative number-of-term index to provide symmetry to our variables . We can even consider shifting all the index by a constant number.

Example: Three consecutive terms in an AP have a sum of x and product of y. Find the three terms.
its always better to assume the three terms be a−d,a,a+d.

This is much faster than solving for the terms a,a+d,a+2d

We can understand the concepts quickly by looking at the problems below. The rules are simple but powerful. Lets see how we can solve few 700-level problems using these.

Problem 1:


If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence?
A. 2
B. 3
C. 4
D. 5
E. 6

The usual way to solve the problem is to frame 2 equations and solve them simultaneously.
However, we can use our trick to solve it.

Given, \(a_4\)+\(a_7\)=40
We can take \(a_4\)=a and \(a_7\)= a+3d.
But this complicates our equation. (See trick 2)

So, let’s take the mid index of \(a_4\) and\(a_7\) which is (\(a_{5.5})\)
Note: For our purpose, let’s assume that fractional index exists.
Let \((a_{5.5})\)= a
Now, \(a_7\)=a+3d Why 3d? Why not 1.5 d? Because the distance \(a_{5.5}\) and \(a_7\) = 1.5 = 3(0.5) = 3 index units

Similarly, \(a_4\)= a-3d
But \(a_4\)+\(a_7\)= 40 (given)
Simple substitution gives: (a-3d)+ (a+3d)=40
=>2a= 40
=>a=20

Now, we have asked to find the value of \(a_1\)= a-9d=20-9d
In the GMAT world, d will never be equal to 0.6 or 0.7
Thus, d can only be either 1 or 2 so that a is a positive number. (Why only positive?? See the options)
Thus, a can be either 20-9(1) or 20-(9)2
Thus, a can be either 11 or 2.
Since 11 is not in the option, answer is Option (A).

Note: We solved the problem without even forming any second equation ( data given about GP).

Lets do another 700 level question.


Problem 2:


The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

A. 300
B. 120
C. 150
D. 170
E. 270

For the symmetry reasons mentioned above, lets consider our base to be the midpoint of \(a_4\) and \((a_{12})\) which is \(a_8\)
Let \(a_8\)=a
Since we are dealing with 4th term, 8th term and the 12th term. Let’s assume common difference of 4d.
I mean let’s say 4d=D= our common difference for this question
Thus, \(a_4\)= a-D and \(a_{12}\)= a+D
Hence, \(a_4\)+\(a_{12}\)= 2a

According to the question, 2a=20
=> a=10

Now lets apply the alternate formula for finding the sum, which is:
Sum of the AP = Middle term of AP x No of terms = \(a_8\) * 15= 10*15=150

Lets do one last question.


Problem 3:


The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

(A) 12
(B) 17
(C) 25
(D) 33
(E) 50

Let the 4 consecutive odd numbers be b-3, b-1, b+1, b+3.
Since, b-1 is odd => b is even.
Let the 3 consecutive even numbers be a-2, a, a+2. => a is even

Now, sum of four consecutive odd numbers= the sum of 3 consecutive even numbers
=> 4b= 3a
=> 4(even)= 3a
=> 8k= 3a ( 4 times an even number will be a multiple of 8)
=> a is a multiple of 8.

Now we can find the number of multiples of 8 between 101 and 200.
101 <x< 200
=> 8(13)<x<8(25)
Thus, the total number of multiples of 8 will be total number of natural numbers between 13 and 24.
=> 24-13+1= 12 Option (A)


First question is discussed here: https://gmatclub.com/forum/if-the-first ... 37480.html
Second question is discussed here: https://gmatclub.com/forum/the-sum-of-t ... 82035.html
Third question is discussed here: https://gmatclub.com/forum/the-sum-of-f ... 20930.html

12. Sequences



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Re: Tips and Tricks for Sequences Questions   [#permalink] 04 Oct 2017, 12:42
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Tips and Tricks for Sequences Questions

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