# Lesson: Introduction to Sequences

## Comment on Introduction to Sequences

### Each term in sequence S is

Sir i am having doubt in Solving this problem

Each term in sequence S is determined by multiplying the prior term by 2 and dividing that product by 3. What is the 100th term of the sequence S?

(1) The sum of the first 2 terms is 15
(2) The first term of the sequence is 9

### Hi Rajkumar,

Hi Rajkumar,

I just posted a solution to that question here: https://gmatclub.com/forum/each-term-in-sequence-s-is-determined-by-mult...

Cheers,
Brent

### Sir, but in statement 2,

Sir, but in statement 2,
If we multiply the first term by 2/3 we are getting the second term as 6, and next terms 4,8/3......etc.Here the common difference varies.Hence how it is possible to find 100th term?

### Here's how:

Here's how:

term1 = 9
term2 = (9)(2/3) = 6
term3 = (9)(2/3)(2/3) = 4
term4 = (9)(2/3)(2/3)(2/3) = 8/3
term5 = (9)(2/3)(2/3)(2/3)(2/3) = 16/9
.
.
.
If you recognize that we can keep doing this indefinitely, then you can see that we COULD determine the value of term100

We might also see the pattern and recognize that term100 = (9)(2/3)^99

### Hi Brent,

Hi Brent,
A little stuck on this problem: I don’t know how to finish it after the third step below

A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A?

a) -24(5^12)
b) -23(5^12)
c) -24(5^6)
d) 23(5^12)
5) 24(5^12)

The nth term of an a.p. is given by: a+(n−1)d
a = first term (happen to be looking for A)
n = nth term we are looking for (A is the eighth term)
d = the common difference (4 x 5^12 is the common difference)
a + (8-1)4 x 5^12
a + 7(4 x 5^12)
a + 28 x 5^12

### You're right to say the

You're right to say the difference between any two successive values is 4(5^12). However, I think there's a problem with how you set up your equation.

Here's one approach that doesn't rely on our knowledge of arithmetic progressions:

Given: G and H are equal to 5^12 and 5^13 respectively

5^13 - 5^12 = 5^12(5 - 1) = (5^12)(4) = 4(5^12)

So, the DIFFERENCE between any two successive values is 4(5^12).

In other words, to find the NEXT term in the sequence, we just ADD 4(5^12) to the term before for it.

Conversely, to find the PREVIOUS term in the sequence, we just SUBTRACT 4(5^12).

We have:

H = 5^13
G = 5^12
F = 5^12 - 4(5^12)
E = 5^12 - 4(5^12) - 4(5^12) = 5^12 - 8(5^12)
D = 5^12 - 8(5^12) - 4(5^12) = 5^12 - 12(5^12)
Following this PATTERN, we get:
C = 5^12 - 16(5^12)
B = 5^12 - 20(5^12)
A = 5^12 - 24(5^12)

5^12 - 24(5^12) = 1(5^12) - 24(5^12)
= -23(5^12)

Does that help?

Cheers,
Brent

### A.....(4)(512)..... B.....(4)

A.....(4)(5^12)..... B.....(4)(5^12)..... C....(4)(5^12)... D......(4)(5^12)..... E.....(4)(5^12)..... F.....(4)(5^12).... G.....(4)(5^12)....H

5^12 – (4)( 5^12) x 6 (times 6 because it’s 6 steps to A from F
5^12 – (24)( 5^12)
= -23(512)
Got it! But I Don’t understand why the equation “a+(n−1)d” won’t work? It’s always worked. Is it because it is going backwards (negative)?

I find this on the Forum:

The user is obviously using the equation I have always used but his setup is a little different as he seems to be equating it to 5^12. The part where he is equating it to 5^12 is where I am a little stuck/confused.

G = A + (n-1) d
5^12 = A + (7-1) 4 (5^12)
As n=7, because its the 7th term in the sequence

Therefore,
A = (5^12) - 24 (5^12)
= (5^12) (1-24) Taking 5^12 common
= - 23 (5^12)

### Quote: "I Don’t understand

Quote: "I Don’t understand why the equation “a+(n−1)d” won’t work? It’s always worked. Is it because it is going backwards (negative)?"

The formula will work; it just takes some fiddling at the end.

We know that the difference (d) = 4(5^12)

At this point, we can apply the formula term_n = a + (n−1)d, where a = the first term

If we recognize that G (aka 5^12) is the same as term_7, then we can write:

So, term_7 = a + (7−1)[4(5^12)]

Since term_7 = 5^12, we can write: 5^12 = a + (7−1)[4(5^12)]
Simplify: 5^12 = a + (6)[4(5^12)]
Simplify: 5^12 = a + 24(5^12)
Solve for a to get: a = -23(5^12)

----THAT'S THE EASIER VERSION------------

Notice that it's more work if we use term_8 (aka H)

term_8 = 5^13
So, we can write: 5^13 = a + (8−1)[4(5^12)]
Simplify: 5^13 = a + (7)[4(5^12)]
Simplify: 5^13 = a + 28(5^12)

Now what?

From here, we need to recognize that 5^13 = 5(5^12)

So, we can write: 5(5^12) = a + 28(5^12)
So, a = 5(5^12) - 28(5^12) = -23(5^12)

Cheers,
Brent

### Never mind that last comment.

Never mind that last comment. I was way overthinking it somehow for far too long without realizing/seeing the obvious.The nth term of an a.p. is given by: a+(n−1)d or n = a+(n−1)d. N=N.

### Hey Brent,

Hey Brent,

considering this Q:

https://gmatclub.com/forum/if-sequences-s-has-240-terms-what-is-the-239th-term-of-s-160762.html

If the second stmt gave a formula that wasn´t the same, would it be possible to solve it?

Thanks,

Philipp

Yes. There are some ways to create a statement 2 that would result in an answer of C

For example...
Statement 2) term1 = 5

In this case, the combined statements would be sufficient.

Cheers,
Brent

### Hi Brent,

Hi Brent,

Isn't the formula for the question in the video is:

t(n)=3+7*(n-1)
EXAMPLES
t(1)=3+7*(1-1)=0
t(2)=3+7*(2-1)=10
t(3)=3+7*(3-1)=17
...
t(41)=3+7*(41-1)=283

### Yes, that's the logic we used

Yes, that's the logic we used to find term41. We multiply 7 by 1 less than the term number.

I try not to throw out formulas without explaining the rationale behind them. Instead, I prefer that students recognize the pattern and then make the same conclusion that you made.

Cheers,
Brent

### So, is it always true for

So, is it always true for this type of question to use: last term minus 1st term = n (in this example, 41-1 = 40), multiplied by the increment (in this case, 7) plus the value of the 1st term? Thanks!

### In general, if the first term

In general, if the first term in a sequence is "a" and each subsequent term is equal to the previous term PLUS some constant value (d), we can say that:

term_n = a + d(n-1)

Cheers,
Brent

### This is the same *formula*

This is the same *formula* referred to at 03:45 - correct? Thanks.

That's correct.

### https://gmatclub.com/forum

https://gmatclub.com/forum/the-price-of-darjeeling-tea-in-rupees-per-kilogram-is-100-0-10n-317562.html

### Hi Brent,

Hi Brent,

Can you help in this question please ?

https://gmatclub.com/forum/in-the-sequence-shown-a-n-a-n-1-k-where-2-n-15-and-k-126119.html#p1029520

Thanks,
Karaan

### Hey Brent,

Hey Brent,
I'm stuck on the below. I see the pattern in your explanation but something isn't clicking in regards to the "Each term in the sequence is equal to the SQUARE of term before it". Can you help with this?
https://gmatclub.com/forum/a-sequence-of-numbers-a1-a2-a3-is-defined-as-follows-a1-3-a2-220319.html

We have:
term3 = 15
term4 = 15^2
term5 = 15^4 = (15^2)^2
term6 = 15^8 = (15^4)^2
term7 = 15^16 = (15^8)^2
term8 = 15^32 = (15^16)^2
etc

Notice that term6 = (term5)^2
That is, 15^8 = (15^4)^2

Likewise, term7 = (term6)^2
That is, 15^16 = (15^8)^2

And so on.

Does that help?

### Ohhh! Yes, I see this. After

Ohhh! Yes, I see this. After letting my brain rest I went back and it all made sense. Thank you!

### https://gmatclub.com/forum

https://gmatclub.com/forum/the-table-above-shows-the-results-of-a-survey-of-100-voters-who-each-89187.html

can I get the solution for this question

### What happened... I started at

What happened... I started at t0 instead of t1 for this question and got E.

https://gmatclub.com/forum/in-the-sequence-above-each-term-is-9-more-than-the-previous-term-wha-209261.html

Unless stated otherwise, the first term must be t1.

### https://gmatclub.com/forum

https://gmatclub.com/forum/what-is-the-thousandth-term-of-s-a-certain-sequence-of-numbers-324942.html

How, I don't see how, why is n 1000 in your solution? I can say n = 10 and n^2 = 100, why is it sufficient?

The question asked us to find the "thousandth term of S"
In other words, we want to find the value of term_1000

Statement 1: For every n, the nth term of S is n².
In other words, term_n = n²
For example: term_1 = 1² = 1
term_2 = 2² = 4
term_3 = 3² = 9
term_4 = 4²

Continuing we get term_1000 = 1000² = 1,000,000
Since we can find the precise value of term_1000, statement 1 is sufficient.

Does that help?