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General Term b_n = b_{n-1} * r = a_1 * r^{n-1} b_i is the ith term r is the common ratio b_1 is the first term

I think i spotted a mistake...nothing of great consequence though...

"a" is not defined here...it should be "b"

correct me kindly if i am wrong

also thanks for the post...hugely informative and all-encompassing.
_________________

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

Sequence : It is an ordered list of objects. It can be finite or infinite. The elements may repeat themselves more than once in the sequence, and their ordering is important unlike a set

Arithmetic Progressions

Definition It is a special type of sequence in which the difference between successive terms is constant.

General Term \(a_n = a_{n-1} + d = a_1 + (n-1)d\) \(a_i\) is the ith term \(d\) is the common difference \(a_1\) is the first term

Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP :

\(a_i - a_{i-1} =\) Constant

If you pick any 3 consecutive terms, the middle one is the mean of the other two

For all i,j > k >= 1 : \(\frac{a_i - a_k}{i-k} = \frac{a_j-a_k}{j-k}\)

Summation The sum of an infinite AP can never be finite except if \(a_1=0\) & \(d=0\) The general sum of a n term AP with common difference d is given by \(\frac{n}{2}(2a+(n-1)d)\) The sum formula may be re-written as \(n * Avg(a_1,a_n) = \frac{n}{2} * (FirstTerm+LastTerm)\)

Examples

All odd positive integers : {1,3,5,7,...} \(a_1=1, d=2\)

All positive multiples of 23 : {23,46,69,92,...} \(a_1=23, d=23\)

All negative reals with decimal part 0.1 : {-0.1,-1.1,-2.1,-3.1,...} \(a_1=-0.1, d=-1\)

Geometric Progressions

Definition It is a special type of sequence in which the ratio of consequetive terms is constant

General Term \(b_n = b_{n-1} * r = a_1 * r^{n-1}\) \(b_i\) is the ith term \(r\) is the common ratio \(b_1\) is the first term

Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP :

\(\frac{b_i}{b_{i-1}} =\) Constant

If you pick any 3 consecutive terms, the middle one is the geometric mean of the other two

For all i,j > k >= 1 : \((\frac{b_i}{b_k})^{j-k} = (\frac{b_j}{b_k})^{i-k}\)

Summation The sum of an infinite GP will be finite if absolute value of r < 1 The general sum of a n term GP with common ratio r is given by \(b_1*\frac{r^n - 1}{r-1}\) If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\)

Examples

All positive powers of 2 : {1,2,4,8,...} \(b_1=1, r=2\)

All positive odd and negative even numbers : {1,-2,3,-4,...} \(b_1=1, r=-1\)

All negative powers of 4 : {1/4,1/16,1/64,1/256,...} \(b_1=1/4, r=1/4, sum=\frac{1/4}{(1-1/4)}=(1/3)\)

Harmonic Progressions

Definition It is a special type of sequence in which if you take the inverse of every term, this new sequence forms an AP

Important Properties Of any three consecutive terms of a HP, the middle one is always the harmonic mean of the other two, where the harmonic mean (HM) is defined as : \(\frac{1}{2} * (\frac{1}{a} + \frac{1}{b}) = \frac{1}{HM(a,b)}\) Or in other words : \(HM(a,b) = \frac{2ab}{a+b}\)

APs, GPs, HPs : Linkage

Each progression provides us a definition of "mean" :

Arithmetic Mean : \(\frac{a+b}{2}\) OR \(\frac{a1+..+an}{n}\) Geometric Mean : \(\sqrt{ab}\) OR \((a1 *..* an)^{\frac{1}{n}}\) Harmonic Mean : \(\frac{2ab}{a+b}\) OR \(\frac{n}{\frac{1}{a1}+..+\frac{1}{an}}\)

For all non-negative real numbers : AM >= GM >= HM

In particular for 2 numbers : AM * HM = GM * GM

Example : Let a=50 and b=2, then the AM = (50+2)*0.5 = 26 ; the GM = sqrt(50*2) = 10 ; the HM = (2*50*2)/(52) = 3.85 AM > GM > HM AM*HM = 100 = GM^2

Misc Notes A subsequence (any set of consequutive terms) of an AP is an AP

A subsequence (any set of consequutive terms) of a GP is a GP

A subsequence (any set of consequutive terms) of a HP is a HP

If given an AP, and I pick out a subsequence from that AP, consisting of the terms \(a_{i1},a_{i2},a_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be an AP

For Example : Consider the AP with \(a_1=1, d=2\) {1,3,5,7,9,11,...}, so a_n=1+2*(n-1)=2n-1 Pick out the subsequence of terms \(a_5,a_{10},a_{15},...\) New sequence is {9,19,29,...} which is an AP with \(a_1=9\) and \(d=10\) If given a GP, and I pick out a subsequence from that GP, consisting of the terms \(b_{i1},b_{i2},b_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be a GP

For Example : Consider the GP with \(b_1=1, r=2\) {1,2,4,8,16,32,...}, so b_n=2^(n-1) Pick out the subsequence of terms \(b_2,b_4,b_6,...\) New sequence is {4,16,64,...} which is a GP with \(b_1=4\) and \(r=4\)

The special sequence in which each term is the sum of previous two terms is known as the fibonacci sequence. It is neither an AP nor a GP. The first two terms are 1. {1,1,2,3,5,8,13,...}

In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is [highlight]even.[/highlight] In either case this is also equal to the mean of the first and last terms

Some examples

Example 1 A coin is tossed repeatedly till the result is a tails, what is the probability that the total number of tosses is less than or equal to 5 ?

Solution P(<=5 tosses) = P(1 toss)+...+P(5 tosses) = P(T)+P(HT)+P(HHT)+P(HHHT)+P(HHHHT) We know that P(H)=P(T)=0.5 So Probability = 0.5 + 0.5^2 + ... + 0.5^5 This is just a finite GP, with first term = 0.5, n=5 and ratio = 0.5. Hence : Probability = \(0.5 * \frac{1-0.5^5}{1-0.5} = \frac{1}{2} * \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{31}{32}\)

Example 2 In an arithmetic progression a1,a2,...,a22,a23, the common difference is non-zero, how many terms are greater than 24 ? (1) a1 = 8 (2) a12 = 24

Solution (1) a1=8, does not tell us anything about the common difference, so impossible to say how many terms are greater than 24 (2) a12=24, and we know common difference is non-zero. So either all the terms below a12 are greater than 24 and the terms above it less than 24 or the other way around. In either case, there are exactly 11 terms either side of a12. Sufficient Answer is B

Example 3 For positive integers a,b (a<b) arrange in ascending order the quantities a, b, sqrt(ab), avg(a,b), 2ab/(a+b)

Solution Using the inequality AM>=GM>=HM, the solution is : a <= 2ab/(a+b) <= Sqrt(ab) <= Avg(a,b) <= b

Example 4 For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is

a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4.

Solution The sequence given has first term 1/2 and each subsequent term can be obtained by multiplying with -1/2. So it is a GP. We can use the GP summation formula \(S=b\frac{1-r^n}{1-r}=\frac{1}{2} * \frac{1-(-1/2)^{10}}{1-(-1/2)} = \frac{1}{3} * \frac{1023}{1024}\) 1023/1024 is very close to 1, so this sum is very close to 1/3 Answer is d

Example 5 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

Solution \(a_4+a_12=20\) \(a_4=a_1+3d, a_12=a_1+11d\) \(2a_1+14d=20\) Now we need the sum of first 15 terms, which is given by : \(\frac{15}{2} (2a_1 + (15-1)d) = \frac{15}{2} * (2a_1+14d) = 150\) Answer is (c)

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is

a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4.

Solution The sequence given has first term 1/2 and each subsequent term can be obtained by multiplying with -1/2. So it is a GP. We can use the GP summation formula

1023/1024 is very close to 1, so this sum is very close to 1/3 Answer is d

Can you please elaborate this, I am not able to deduce the outcome. Thanks in advance!

For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is

a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4.

Solution The sequence given has first term 1/2 and each subsequent term can be obtained by multiplying with -1/2. So it is a GP. We can use the GP summation formula

1023/1024 is very close to 1, so this sum is very close to 1/3 Answer is d

Can you please elaborate this, I am not able to deduce the outcome. Thanks in advance!

How often do GP and HP come in the GMAT? I can believe AP is certainly tested. But are GP and HP in the same class as probability and combinatorics which appear only at the 750-800 range?
_________________

If you like it, Kudo it!

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

How often do GP and HP come in the GMAT? I can believe AP is certainly tested. But are GP and HP in the same class as probability and combinatorics which appear only at the 750-800 range?

There are some questions from GMAT Prep for which knowing the properties of GP might be useful (check the post above yours for an example). Though I've never seen the GMAT question testing/mentioning HP.
_________________

Thank you Bunuel. Just wanted to check before going into the relatively arcane stuff!
_________________

If you like it, Kudo it!

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

Sequence : It is an ordered list of objects. It can be finite or infinite. The elements may repeat themselves more than once in the sequence, and their ordering is important unlike a set

Arithmetic Progressions

Definition It is a special type of sequence in which the difference between successive terms is constant.

General Term \(a_n = a_{n-1} + d = a_1 + (n-1)d\) \(a_i\) is the ith term \(d\) is the common difference \(a_1\) is the first term

Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP :

\(a_i - a_{i-1} =\) Constant

If you pick any 3 consecutive terms, the middle one is the mean of the other two

For all i,j > k >= 1 : \(\frac{a_i - a_k}{i-k} = \frac{a_j-a_k}{j-k}\)

Summation The sum of an infinite AP can never be finite except if \(a_1=0\) & \(d=0\) The general sum of a n term AP with common difference d is given by \(\frac{n}{2}(2a+(n-1)d)\) The sum formula may be re-written as \(n * Avg(a_1,a_n) = \frac{n}{2} * (FirstTerm+LastTerm)\)

Examples

All odd positive integers : {1,3,5,7,...} \(a_1=1, d=2\)

All positive multiples of 23 : {23,46,69,92,...} \(a_1=23, d=23\)

All negative reals with decimal part 0.1 : {-0.1,-1.1,-2.1,-3.1,...} \(a_1=-0.1, d=-1\)

Geometric Progressions

Definition It is a special type of sequence in which the ratio of consequetive terms is constant

General Term \(b_n = b_{n-1} * r = a_1 * r^{n-1}\) \(b_i\) is the ith term \(r\) is the common ratio \(b_1\) is the first term

Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP :

\(\frac{b_i}{b_{i-1}} =\) Constant

If you pick any 3 consecutive terms, the middle one is the geometric mean of the other two

For all i,j > k >= 1 : \((\frac{b_i}{b_k})^{j-k} = (\frac{b_j}{b_k})^{i-k}\)

Summation The sum of an infinite GP will be finite if absolute value of r < 1 The general sum of a n term GP with common ratio r is given by [highlight]\(b_1*\frac{r^n - 1}{r-1}\)[/highlight] If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\)

Examples

All positive powers of 2 : {1,2,4,8,...} \(b_1=1, r=2\)

All positive odd and negative even numbers : {1,-2,3,-4,...} \(b_1=1, r=-1\)

All negative powers of 4 : {1/4,1/16,1/64,1/256,...} \(b_1=1/4, r=1/4, sum=\frac{1/4}{(1-1/4)}=(1/3)\)

Harmonic Progressions

Definition It is a special type of sequence in which if you take the inverse of every term, this new sequence forms an AP

Important Properties Of any three consecutive terms of a HP, the middle one is always the harmonic mean of the other two, where the harmonic mean (HM) is defined as : \(\frac{1}{2} * (\frac{1}{a} + \frac{1}{b}) = \frac{1}{HM(a,b)}\) Or in other words : \(HM(a,b) = \frac{2ab}{a+b}\)

APs, GPs, HPs : Linkage

Each progression provides us a definition of "mean" :

Arithmetic Mean : \(\frac{a+b}{2}\) OR \(\frac{a1+..+an}{n}\) Geometric Mean : \(\sqrt{ab}\) OR \((a1 *..* an)^{\frac{1}{n}}\) Harmonic Mean : \(\frac{2ab}{a+b}\) OR \(\frac{n}{\frac{1}{a1}+..+\frac{1}{an}}\)

For all non-negative real numbers : AM >= GM >= HM

In particular for 2 numbers : AM * HM = GM * GM

Example : Let a=50 and b=2, then the AM = (50+2)*0.5 = 26 ; the GM = sqrt(50*2) = 10 ; the HM = (2*50*2)/(52) = 3.85 AM > GM > HM AM*HM = 100 = GM^2

Misc Notes A subsequence (any set of consequutive terms) of an AP is an AP

A subsequence (any set of consequutive terms) of a GP is a GP

A subsequence (any set of consequutive terms) of a HP is a HP

If given an AP, and I pick out a subsequence from that AP, consisting of the terms \(a_{i1},a_{i2},a_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be an AP

For Example : Consider the AP with \(a_1=1, d=2\) {1,3,5,7,9,11,...}, so a_n=1+2*(n-1)=2n-1 Pick out the subsequence of terms \(a_5,a_{10},a_{15},...\) New sequence is {9,19,29,...} which is an AP with \(a_1=9\) and \(d=10\) If given a GP, and I pick out a subsequence from that GP, consisting of the terms \(b_{i1},b_{i2},b_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be a GP

For Example : Consider the GP with \(b_1=1, r=2\) {1,2,4,8,16,32,...}, so b_n=2^(n-1) Pick out the subsequence of terms \(b_2,b_4,b_6,...\) New sequence is {4,16,64,...} which is a GP with \(b_1=4\) and \(r=4\)

The special sequence in which each term is the sum of previous two terms is known as the fibonacci sequence. It is neither an AP nor a GP. The first two terms are 1. {1,1,2,3,5,8,13,...}

In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is even. In either case this is also equal to the mean of the first and last terms

Some examples

Example 1 A coin is tossed repeatedly till the result is a tails, what is the probability that the total number of tosses is less than or equal to 5 ?

Solution P(<=5 tosses) = P(1 toss)+...+P(5 tosses) = P(T)+P(HT)+P(HHT)+P(HHHT)+P(HHHHT) We know that P(H)=P(T)=0.5 So Probability = 0.5 + 0.5^2 + ... + 0.5^5 This is just a finite GP, with first term = 0.5, n=5 and ratio = 0.5. Hence : Probability = \(0.5 * \frac{1-0.5^5}{1-0.5} = \frac{1}{2} * \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{31}{32}\)

Example 2 In an arithmetic progression a1,a2,...,a22,a23, the common difference is non-zero, how many terms are greater than 24 ? (1) a1 = 8 (2) a12 = 24

Solution (1) a1=8, does not tell us anything about the common difference, so impossible to say how many terms are greater than 24 (2) a12=24, and we know common difference is non-zero. So either all the terms below a12 are greater than 24 and the terms above it less than 24 or the other way around. In either case, there are exactly 11 terms either side of a12. Sufficient Answer is B

Example 3 For positive integers a,b (a<b) arrange in ascending order the quantities a, b, sqrt(ab), avg(a,b), 2ab/(a+b)

Solution Using the inequality AM>=GM>=HM, the solution is : a <= 2ab/(a+b) <= Sqrt(ab) <= Avg(a,b) <= b

Example 4 For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is

a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4.

Solution The sequence given has first term 1/2 and each subsequent term can be obtained by multiplying with -1/2. So it is a GP. We can use the GP summation formula \([highlight]S=b\frac{1-r^n}{1-r}[/highlight]=\frac{1}{2} * \frac{1-(-1/2)^{10}}{1-(-1/2)} = \frac{1}{3} * \frac{1023}{1024}\) 1023/1024 is very close to 1, so this sum is very close to 1/3 Answer is d

Example 5 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

Solution \(a_4+a_12=20\) \(a_4=a_1+3d, a_12=a_1+11d\) \(2a_1+14d=20\) Now we need the sum of first 15 terms, which is given by : \(\frac{15}{2} (2a_1 + (15-1)d) = \frac{15}{2} * (2a_1+14d) = 150\) Answer is (c)

Something that should be the same formula is different at different places. Explain! \(b_1*\frac{r^n - 1}{r-1}\) and \(S=b\frac{1-r^n}{1-r}\) Make up your mind!
_________________

Hi, Can some mod take a look into the example 2 provided under Geometric Progression?. It states series - { 1, -2, 3, -4...} as an example with b1 = 1 and r = -1...

However, with the G.P formula, it looks incorrect as the common ratio thing doesn't hold true.

Kindly help.
_________________

Live Life the Way YOU Love It !!

GmatPrep1 [10/09/2012] : 650 (Q42;V38) - need to make lesser silly mistakes. MGMAT 1 [11/09/2012] : 640 (Q44;V34) - need to improve quant pacing and overcome verbal fatigue.

Sequence : It is an ordered list of objects. It can be finite or infinite. The elements may repeat themselves more than once in the sequence, and their ordering is important unlike a set

Arithmetic Progressions

Definition It is a special type of sequence in which the difference between successive terms is constant.

General Term \(a_n = a_{n-1} + d = a_1 + (n-1)d\) \(a_i\) is the ith term \(d\) is the common difference \(a_1\) is the first term

Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP :

\(a_i - a_{i-1} =\) Constant

If you pick any 3 consecutive terms, the middle one is the mean of the other two

For all i,j > k >= 1 : \(\frac{a_i - a_k}{i-k} = \frac{a_j-a_k}{j-k}\)

Summation The sum of an infinite AP can never be finite except if \(a_1=0\) & \(d=0\) The general sum of a n term AP with common difference d is given by \(\frac{n}{2}(2a+(n-1)d)\) The sum formula may be re-written as \(n * Avg(a_1,a_n) = \frac{n}{2} * (FirstTerm+LastTerm)\)

Examples

All odd positive integers : {1,3,5,7,...} \(a_1=1, d=2\)

All positive multiples of 23 : {23,46,69,92,...} \(a_1=23, d=23\)

All negative reals with decimal part 0.1 : {-0.1,-1.1,-2.1,-3.1,...} \(a_1=-0.1, d=-1\)

Geometric Progressions

Definition It is a special type of sequence in which the ratio of consequetive terms is constant

General Term \(b_n = b_{n-1} * r = a_1 * r^{n-1}\) \(b_i\) is the ith term \(r\) is the common ratio \(b_1\) is the first term

Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP :

\(\frac{b_i}{b_{i-1}} =\) Constant

If you pick any 3 consecutive terms, the middle one is the geometric mean of the other two

For all i,j > k >= 1 : \((\frac{b_i}{b_k})^{j-k} = (\frac{b_j}{b_k})^{i-k}\)

Summation The sum of an infinite GP will be finite if absolute value of r < 1 The general sum of a n term GP with common ratio r is given by \(b_1*\frac{r^n - 1}{r-1}\) If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\)

Examples

All positive powers of 2 : {1,2,4,8,...} \(b_1=1, r=2\)

All positive odd and negative even numbers : {1,-2,3,-4,...} \(b_1=1, r=-1\)

All negative powers of 4 : {1/4,1/16,1/64,1/256,...} \(b_1=1/4, r=1/4, sum=\frac{1/4}{(1-1/4)}=(1/3)\)

Harmonic Progressions

Definition It is a special type of sequence in which if you take the inverse of every term, this new sequence forms an AP

Important Properties Of any three consecutive terms of a HP, the middle one is always the harmonic mean of the other two, where the harmonic mean (HM) is defined as : \(\frac{1}{2} * (\frac{1}{a} + \frac{1}{b}) = \frac{1}{HM(a,b)}\) Or in other words : \(HM(a,b) = \frac{2ab}{a+b}\)

APs, GPs, HPs : Linkage

Each progression provides us a definition of "mean" :

Arithmetic Mean : \(\frac{a+b}{2}\) OR \(\frac{a1+..+an}{n}\) Geometric Mean : \(\sqrt{ab}\) OR \((a1 *..* an)^{\frac{1}{n}}\) Harmonic Mean : \(\frac{2ab}{a+b}\) OR \(\frac{n}{\frac{1}{a1}+..+\frac{1}{an}}\)

For all non-negative real numbers : AM >= GM >= HM

In particular for 2 numbers : AM * HM = GM * GM

Example : Let a=50 and b=2, then the AM = (50+2)*0.5 = 26 ; the GM = sqrt(50*2) = 10 ; the HM = (2*50*2)/(52) = 3.85 AM > GM > HM AM*HM = 100 = GM^2

Misc Notes A subsequence (any set of consequutive terms) of an AP is an AP

A subsequence (any set of consequutive terms) of a GP is a GP

A subsequence (any set of consequutive terms) of a HP is a HP

If given an AP, and I pick out a subsequence from that AP, consisting of the terms \(a_{i1},a_{i2},a_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be an AP

For Example : Consider the AP with \(a_1=1, d=2\) {1,3,5,7,9,11,...}, so a_n=1+2*(n-1)=2n-1 Pick out the subsequence of terms \(a_5,a_{10},a_{15},...\) New sequence is {9,19,29,...} which is an AP with \(a_1=9\) and \(d=10\) If given a GP, and I pick out a subsequence from that GP, consisting of the terms \(b_{i1},b_{i2},b_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be a GP

For Example : Consider the GP with \(b_1=1, r=2\) {1,2,4,8,16,32,...}, so b_n=2^(n-1) Pick out the subsequence of terms \(b_2,b_4,b_6,...\) New sequence is {4,16,64,...} which is a GP with \(b_1=4\) and \(r=4\)

The special sequence in which each term is the sum of previous two terms is known as the fibonacci sequence. It is neither an AP nor a GP. The first two terms are 1. {1,1,2,3,5,8,13,...}

In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is even. In either case this is also equal to the mean of the first and last terms

Some examples

Example 1 A coin is tossed repeatedly till the result is a tails, what is the probability that the total number of tosses is less than or equal to 5 ?

Solution P(<=5 tosses) = P(1 toss)+...+P(5 tosses) = P(T)+P(HT)+P(HHT)+P(HHHT)+P(HHHHT) We know that P(H)=P(T)=0.5 So Probability = 0.5 + 0.5^2 + ... + 0.5^5 This is just a finite GP, with first term = 0.5, n=5 and ratio = 0.5. Hence : Probability = \(0.5 * \frac{1-0.5^5}{1-0.5} = \frac{1}{2} * \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{31}{32}\)

Example 2 In an arithmetic progression a1,a2,...,a22,a23, the common difference is non-zero, how many terms are greater than 24 ? (1) a1 = 8 (2) a12 = 24

Solution (1) a1=8, does not tell us anything about the common difference, so impossible to say how many terms are greater than 24 (2) a12=24, and we know common difference is non-zero. So either all the terms below a12 are greater than 24 and the terms above it less than 24 or the other way around. In either case, there are exactly 11 terms either side of a12. Sufficient Answer is B

Example 3 For positive integers a,b (a<b) arrange in ascending order the quantities a, b, sqrt(ab), avg(a,b), 2ab/(a+b)

Solution Using the inequality AM>=GM>=HM, the solution is : a <= 2ab/(a+b) <= Sqrt(ab) <= Avg(a,b) <= b

Example 4 For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is

a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4.

Solution The sequence given has first term 1/2 and each subsequent term can be obtained by multiplying with -1/2. So it is a GP. We can use the GP summation formula \(S=b\frac{1-r^n}{1-r}=\frac{1}{2} * \frac{1-(-1/2)^{10}}{1-(-1/2)} = \frac{1}{3} * \frac{1023}{1024}\) 1023/1024 is very close to 1, so this sum is very close to 1/3 Answer is d

Example 5 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

Solution \(a_4+a_12=20\) \(a_4=a_1+3d, a_12=a_1+11d\) \(2a_1+14d=20\) Now we need the sum of first 15 terms, which is given by : \(\frac{15}{2} (2a_1 + (15-1)d) = \frac{15}{2} * (2a_1+14d) = 150\) Answer is (c)

In the Geometric Progression example #2, I don't understand how the sequence for the 'b_1=1' and 'r=-1' you provided results in a sequence of 1,-2,3,-4,5 I think it just goes 1,-1,1,-1,1,..... Could someone please clarify?

In the Geometric Progression example #2, I don't understand how the sequence for the 'b_1=1' and 'r=-1' you provided results in a sequence of 1,-2,3,-4,5 I think it just goes 1,-1,1,-1,1,..... Could someone please clarify?

Thanks

You are correct. Removed that from the topic.
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