GRE Sequences and Series

A GRE quant problem varies from vapid arithmetic problem to maverick second-degree algebra problem. In our article on GRE Sequence and series we have almost everything beginning from formulas to tedious problems. After going through our article on GRE Sequences and series, solving a GRE Sequences and series will be an easy task for you.

A sequence is a collection of number in a set order. Each term of a sequence is determined by a RULE. Here are examples of sequence RULES:

\({ A }_{ n }=9n+3\)

The \({ n }^{ th }\) term of this sequence is defined by the rule 9n+3, for integer . For example, the fourth term in this sequence = 9(4) + 3 = 39.

Let’s take one more example:

\({ Q }_{ n }={ n }^{ 2 }+4\)

The \({ n }^{ th }\) term of this sequence is defined by the rule \({ n }^{ 2 }+4\), for integers \(n\ge 1\). The first term in this sequence is

\({ 1 }^{ 2 }+4=5\)

In the above cases, each term of the sequence is defined as a function of n, the place in which the term occurs in the sequence. This is a direct definition of a sequence formula.

Questions in GRE mostly appear from Arithmetic propagation from Geometric progression.

 

Arithmetic progressions:

Number are said to be in arithmetic progression when they increase or decrease by a common difference. Thus each of the following series forms an arithmetic progression:

3, 7, 11, 15………

8, 2, -4, -10………

a, a+d, a+2d, a+3d………

The common difference is found by subtracting any term of the series from the next term.

Common difference of an A.P. =

\({ N }^{ th }term\quad -\quad { (N-1) }^{ th }term\)

If we examine the series a, a+d, a+2d, a+3d……… we notice that in any term the coefficient of d is always less by one than the position of that term in the series.

Thus the \({ N }^{ th }term\) term of an arithmetic progression is given by \({ N }^{ th }term\) term = a+(N-1)d.

To find the sum of the given number of terms in an Arithmetic progression:

Let a denote the first term, d be the common difference,

Sum of term upto N terms =

\(=\quad \frac { N }{ 2 } (2a\quad +\quad (N-1)d)\)

Geometric Progression:

Quantities are said to be in Geometric Progression when they increase or decrease by a constant factor.

The constant factor is also called the common ratio & it is founded by dividing any term by the term immediately preceding it.

If we examine the series

\(a,\quad ar,\quad a{ r }^{ 2 },\quad a{ { r }^{ 3 } }\_ \quad \_ \quad \_ a{ r }^{ n-1 }\quad\)

We notice that in any term the index of r is always less by one than the number of terms in the series.

\({ n }^{ th }term\quad\) term of the sequence =

\(a{ r }^{ n-1 }\quad\)

To find the sum of a Number of terms in a Geometric Progression:

Let a be the first term, r be the common ratio,

Sum of numbers upto n terms

\(=\quad \frac { a(1-{ r }^{ n }) }{ (1-r) } \quad (If\quad r>1)\quad \\ =\quad \frac { a }{ (1-r) } \quad (If\quad r<1)\)

The GRE also uses recursive formulas to define sequences with direct formulas, the values of each term in a sequence is defined in term of its term number in the sequence. With recursive formulas, each item number of a sequence is defined in terms of the value of PREVIOUS ITEM in the sequence.

A recursive formula looks like this:

\({ A }_{ n }={ A }_{ n-1 }+9\)

This formula simply means “THIS term \({ A }_{ n }\) equals the PREVIOUS term \({ A }_{ n-1 }\) plus 9”.

Whenever you look at a recursive formula, articulate its meaning in words in your mind.

As we have discussed, sequence problems generally involve finding patterns among the terms in a sequence.

Let’s take an example:

If \({ A }_{ n }={ 3 }^{ n }\), what is the unit digit of \({ A }_{ 65 }\)?

Clearly, you cannot be expected to multiply out \({ 3 }^{ 65 }\) on the GRE, even with a calculator. Therefore, you must look for a patterns in the powers of three.

\({ 3 }^{ 1 }=3\\ { 3 }^{ 2 }=9\\ { 3 }^{ 3 }=27\\ { 3 }^{ 4 }=81\\ { 3 }^{ 5 }=243\\ { 3 }^{ 6 }=729\\ { 3 }^{ 7 }=2187\\ { 3 }^{ 8 }=6561\)

You can see that the unit’s digits of powers of 3 follow the pattern “3, 9, 7, 1”. As 64 is divisible by 4. Unit digit of \({ 3 }^{ 64 }\) is 1 and Unit digit of \({ 3 }^{ 65 }\) is 3.

Problem 1) The sequence A is defined by

\({ A }_{ n }\quad =\quad { A }_{ n-1 }+{ A }_{ n-2 }+{ A }_{ n-3 }-5\) for each integer \(n\ge 4\). If \({ A }_{ 1 }=4,\quad { A }_{ 2 }=0\quad and\quad { A }_{ 4 }\quad =\quad -4\), what is the value of \({ A }_{ 6 }\)?

Sol.

\({ A }_{ n }\quad ={ \quad A }_{ n-1 }+{ A }_{ n-2 }+{ A }_{ n-3 }-5\quad (let\quad n=4)\\ { A }_{ 4 }\quad =\quad { A }_{ 3 }+{ A }_{ 2 }+{ A }_{ 1 }-5\\ -4\quad =\quad { A }_{ 3 }+0+4-5\quad =\quad { A }_{ 3 }-1\\ -4\quad =\quad { A }_{ 3 }-1\\ { A }_{ 3 }\quad =\quad -4+1\quad =\quad -3\\ { A }_{ n }\quad =\quad { A }_{ n-1 }+{ A }_{ n-2 }+{ A }_{ n-3 }-5\quad (let\quad n=5)\\ { A }_{ 5 }={ A }_{ 4 }+{ A }_{ 3 }+{ A }_{ 2 }-5\\ \quad \quad =\quad -4+(-3)+0-5\quad =\quad -4-3-5\quad =\quad -12\\ { A }_{ 6 }\quad =\quad { A }_{ 5 }+{ A }_{ 4 }+{ A }_{ 3 }-5\\ { A }_{ 6 }\quad =\quad -12\quad +\quad (-4)\quad +\quad (-3)\quad -\quad 5\quad =\quad -24\\ Therefore,\quad { A }_{ 6 }\quad =\quad { (-24) }^{ Ans }\)

 

Problem 2) The sequence P is defined by

\({ P }_{ n }\quad =\quad 10({ P }_{ n-1 })-2\) for each integer \(n\ge 2\). If \({ P }_{ 1 }=2\), what is the value of \({ P }_{ 4 }\)?

Sol.

\({ P }_{ n\quad }=\quad 10({ P }_{ n-1 })-2\quad (let\quad n=2)\\ { P }_{ 2\quad }=\quad 10{ P }_{ 1 }-2\quad =\quad 10(2)-2\quad =\quad 18\\ { P }_{ n }\quad =\quad 10{ P }_{ n-1 }-2\quad (let\quad n=3)\\ { P }_{ 3 }\quad =\quad 10{ P }_{ 2 }-2\\ \quad \quad \quad =\quad 10(18)-2\quad =\quad 180-2\quad =\quad 178\\ { P }_{ n }\quad =\quad 10({ P }_{ n-1 })-2\quad (let\quad n=4)\\ { P }_{ 4 }\quad =\quad 10({ P }_{ 3 })-2\\ \quad \quad \quad =\quad 10(178)-2\\ \quad \quad \quad =\quad { (1778) }^{ Ans }\)

 

Problem 3) The sequence S is defined by

\({ S }_{ n }\quad =\quad 2({ S }_{ n-1 })-4\) for each integer \(n\ge 2\). If \({ S }_{ 1 }=6\), what is the value of \({ S }_{ 5 }\)?

Sol.

\({ S }_{ n }=2{ S }_{ n-1 }-4\quad (let\quad n=2)\\ { S }_{ 2 }=2{ S }_{ 1 }-4\\ { S }_{ 2 }=2{ S }_{ 1 }-4\quad ({ S }_{ 1 }=6)\\ \quad =\quad 2(6)-4\quad =\quad 8\\ { S }_{ n }=2{ S }_{ n-1 }-4\quad (let\quad n=3)\\ { S }_{ 3 }=2{ S }_{ 2 }-4\\ \quad \quad \quad =2(8)-4\quad =\quad 12\\ { S }_{ n }\quad =\quad 2{ S }_{ n-1 }-4\quad (let\quad n=4)\\ { S }_{ 4\quad }=\quad 2{ S }_{ 3 }-4\quad =\quad 2(12)-4\quad =\quad 24-4\quad =\quad 20\\ { S }_{ n }\quad =\quad 2{ S }_{ n-1 }-4\quad (let\quad n=5)\\ { S }_{ 5 }=\quad 2{ S }_{ 4 }-4\\ \quad \quad =\quad 2(20)-4\quad =\quad { (36) }^{ Ans }\)

 

Problem 4) Find the value of the expression 1-6+2-7+3-8……… to 100 terms

a) -250

b) -500

c) -450

d) -300

Sol.  The series (1-6+2-7+3-8……… to 100 terms) can be written as:

\(=\quad (1+2+3+4———to\quad 50\quad terms)\quad +\\ (-6-7-8———to\quad 50\quad terms)\\\)

For sequence (1+2+3 ……… to 50 terms)

Using formula for sum of Airthmetic Progression

\({ S }_{ n }=\frac { n }{ 2 } (2a+(n-1)d)\\ { S }_{ 50 }=\quad \frac { 50 }{ 2 } (2(1)+(50-1)1)\\ \quad \quad =\quad 25(2+49)\\ \quad \quad =\quad 51\times 25\quad =\quad { (1275) }^{ Ans }\\\)

For sequence (6+7+8 ……… to 50 terms)

Using formula for sum of n terms of Arithmetic progression

\({ S }_{ n }=\frac { n }{ 2 } (2a+(n-1)d)\\ { S }_{ 50 }=\quad \frac { 50 }{ 2 } (2(6)+(50-1)1)\\ \quad \quad =\quad 25(12+49)\\ \quad \quad =\quad 61\times 25\quad =\quad { (1525) }^{ Ans }\\\)

Value of expression

\(=\quad (1+2+3——-to\quad 50\quad terms)\\ -(6+7+8——–to\quad 50\quad terms)\\ =\quad 1275\quad -\quad 1525\\ =\quad { (-250) }^{ Ans }\)

 

Problem 5) The first three terms in an arithmetic sequence are 30, 33 & 36. What is the 80th term?

Sol.  For arithmetic progression: 30, 33, 36, ………

a=30

Common difference = 33-30=3

Using formula for nth term of an arithmetic progression:

\({ a }_{ N }\quad =\quad a\quad +\quad (N-1)d\\ \quad =\quad 30\quad +\quad (N-1)(3)\\ { a }_{ 80\quad }=\quad 30\quad +\quad (80-1)(3)\\ \quad \quad \quad =\quad 30\quad +\quad 79\times 3\\ \quad \quad \quad =\quad { (267) }^{ Ans }\)

 

Problem 6) In a certain sequence, the term \({ a }_{ n }\) is defined by the formula \({ a }_{ n }\quad =\quad { a }_{ n-1 }+10\)

for each integer \(n\ge 2\). What is the positive difference between \({ a }_{ 10 }\) and \({ a }_{ 15 }\)?

Sol.

\({ a }_{ n }\quad =\quad { a }_{ n-1 }+10\quad (let\quad n=2)\\ { a }_{ 2 }\quad =\quad { a }_{ 1 }+10\)

Therefore, this is an arithmetic sequence where the difference between successive terms is always +10.

The difference between, for example,\({ a }_{ 10 }\)  & \({ a }_{ 11 }\), is exactly 10, regardless of the actual values of the two terms.

Starting from \({ a }_{ 10 }\), there is a sequence of 5 terms to get to \({ a }_{ 15 }\).

Therefore, the difference between \({ a }_{ 10 }\) & \({ a }_{ 15 }\) is 10×5 = 50.

 

Problem 7) The sequence A is defined by \({ A }_{ n }={ A }_{ n-1 }+2\) for each integer \(n\ge 2\), & \({ A }_{ 1 }=45\). What is the sum of the first 100 terms in sequence A?

Sol.

\({ A }_{ 1 }=\quad first\quad term\quad =\quad 45\\ { A }_{ n }\quad =\quad { A }_{ n-1 }+2\quad (let\quad n=2)\\ { A }_{ 2 }\quad =\quad { A }_{ 1 }+2\quad =\quad 45+2\quad =\quad 47\\ { A }_{ 3 }\quad =\quad { A }_{ 2 }+2\quad =\quad 47+2\quad =\quad 49\)

The first few terms of the sequence are 45, 47, 49, 51 ………

This is an arithmetic progression with a=45 & common progression (d) = 47-45 = 2.

Using formula for sum of n terms of Arithmetic progression:

\({ S }_{ n }=\frac { n }{ 2 } (2a\quad +\quad (n-1)d)\)

Sum of first 100 terms in sequence A

\(=\frac { 100 }{ 2 } (2(45)\quad +\quad (100-1)2)\\ =50(90\quad +\quad 99\times 2)\\ =50(90+198)\\ ={ (14400) }^{ Ans }\)

 

Problem 8) The sequence S is defined by \({ S }_{ n-1 }=\frac { 1 }{ 4 } { S }_{ n }\) for each integer \(n\ge 2\). If \({ S }_{ 1 }=-4\), what is the value of \({ S }_{ 4 }\)?

a) -256

b) -64

c) \(\frac { -1 }{ 16 }\)

d) \(\frac { 1 }{ 16 }\)

Sol.

\({ S }_{ n-1 }=\frac { 1 }{ 4 } { S }_{ n }\)

(let n=2)

\({ S }_{ 1 }=\frac { 1 }{ 4 } { S }_{ 2 }\\ { S }_{ 2\quad }=\quad 4{ S }_{ 1 }=\quad 4(-4)\quad =\quad -16\\ { S }_{ 2 }=\frac { 1 }{ 4 } { S }_{ 3 }\\ { S }_{ 3 }=4{ S }_{ 2 }=4(-16)=-64\)

First few terms of sequence are: -4, -16, -64

We can easily infer that this is a geometric progression with a=-4,

r (common ratio) =4

\({ S }_{ n }({ n }^{ th }\quad term\quad of\quad sequence)\quad =\quad a{ r }^{ n-1 }\\ Value\quad of\quad { S }_{ 4 }\quad =\quad (-4){ (4) }^{ n-1 }\\ =\quad (-4){ (4) }^{ 4-1 }=\quad (-4){ (4) }^{ 3 }={ (-256) }^{ Ans }\)

 

Problem 9) \({ A }_{ n }={ 2 }^{ n }-1\) for all integer \(n\ge 1\)

Quantity A:         The unit digit of \({ A }_{ 26 }\)

Quantity B:         The unit digit of \({ A }_{ 34 }\)

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Ans. 

\({ A }_{ 26 }={ 2 }^{ 26 }-1,\quad { A }_{ 34 }={ 2 }^{ 34 }-1\)

Clearly, we are not expected to multiply out \({ 2 }^{ 26 }\quad or{ \quad 2 }^{ 34 }\) on the GRE, even with a calculator.

Therefore, we must look for a pattern in the powers of 2.

\({ 2 }^{ 1 }=2\\ { 2 }^{ 2 }=4\\ { 2 }^{ 3 }=8\\ { 2 }^{ 4 }=16\\ { 2 }^{ 5 }=32\\ { 2 }^{ 6 }=64\\ { 2 }^{ 7 }=128\\ { 2 }^{ 8 }=256\)

We can see that the unit’s digits of powers of 2. follow the pattern “2, 4, 8, 6, 2, 4, 8, 6, ………”

Unit digit of \({ 2 }^{ 26 }=4\)

Unit digit of \({ 2 }^{ 34 }=4\)

Unit digit of \(({ 2 }^{ 26 }-1)=3\)

Unit digit of \(({ 2 }^{ 34 }-1)=3\)

Therefore, two quantities are equal.

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