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## Class 10 (Marathi)

### Unit 3: Lesson 5

AP word problems# Sequences word problem: growth pattern

Sal finds the equation that describes a growth pattern of shapes made of squares. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- At6:10how did going from one to two get expressed as 2-1?(3 votes)
- In the slope formula, (y2-y1)/(x2-x1), The 2 is the y2 and the 1 is the y1. And, naturally, the 5 is the x2 and the 1 is the x1.

By the way, I'm answering this for anyone else who may have this question as you probably already figured it out!(4 votes)

- at7:35,i have no idea how to get b=-3, the number of blocks between X is 4,so should B=-4?(3 votes)
- You're right - the difference between any 2 consecutive sets in this sequence is 4. But "b" isn't the difference between consecutive terms of this sequence. It's the y intercept of "y = 4x-3", a function that corresponds to our sequence. More completely:

What do you add to the 1st set to get the 2nd? To the 2nd to get the 3rd?

To the 1st to get the 3rd? To the 1st to get the 4th? to the 1st to get the 100th? 1000th? millionth? The xth? If y is the value of the xth term, what's the y as a function of x?

(4 blocks, 4 blocks. 8 blocks, 12 blocks, 4*99 = 396 blocks, 4*999 = 3996 blocks, 4*999,999 = 3,999,996 blocks. 4*(x-1) blocks. y = 1 + 4*(x-1). )

If you distribute 4 across the parentheses and collect like terms, what's y?

( y = 1 + 4x - 4 = 4x - 3 (this is "y = mx + b"). )(0 votes)

- This video helped me a lot but I still need some help. How would you solve this equation?

Determine the value of A when 'n' is 3.

A = 2n + 1

Please help me, I can't continue my math until I figure this out.(1 vote)- What does "2n" mean to you, in the equation? You are being given a possible value for n: n=3. Did you try substituting this value in the equation? What do you come up with?(3 votes)

- im unable to find the vid regarding the summation f series ..(1 vote)
- Series are discussed in the precalculus and calculus playlists of Khan Academy, so you should head over there to find them. But here's one video on the summation of arithmetic series, and one on geometric series:

https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/calculus-series/v/formula-for-arithmetic-series

https://www.khanacademy.org/math/precalculus/seq_induction/infinite-geometric-series/v/infinite-geometric-series

I hope this helps.(3 votes)

- 3:08why do you need to know how many coloums they have ?(3 votes)
- Because the growth pattern each time you increase x by one, is one column. So when X=1 you have zero columns. When x=2 you have 1 column. X=3 you have 2 columns. In general you have (x-1) columns added for any given x value. Each column you add also carries 4 blocks, which is why you must multiply by 4.

4*(x-1). This says for any given x, you are adding (x-1) columns and 4*(x-1) blocks. When x=1, the first group of blocks, we have only one block. This is why you have to add one to your total 4*(x-1)+1 to get the actual number of blocks given your group number X.(3 votes)

- Can somebody solve this

[MIND BURST] [ Trickiest !!] Algebra Problem 2 - NT Maths - YouTube

https://www.youtube.com/watch?v=0WJlvkRPQTc

I am trying this much but not getting it ..........

Please give me any hint if you get(1 vote)- Well, we have an arithmetic progression, 𝐴 = {𝑇(1), 𝑇(2), 𝑇(3), ...}

We know that 𝑇(𝑝) = 𝑇(1) + (𝑝 − 1)𝑑 = 1∕𝑞 ⇔ 𝑇(1) = 1∕𝑞 − (𝑝 − 1)𝑑

Likewise, 𝑇(𝑞) = 1∕𝑝 ⇔ 𝑇(1) = 1∕𝑝 − (𝑞 − 1)𝑑

This gives us: 1∕𝑞 − (𝑝 − 1)𝑑 = 1∕𝑝 − (𝑞 − 1)𝑑 ⇔ 𝑑 = (1∕𝑞 − 1∕𝑝)∕(𝑝 − 𝑞) = 1∕(𝑝𝑞) (given 𝑝 ≠ 𝑞)

So, 𝑇(1) = 1∕𝑝 − (𝑞 − 1)∕(𝑝𝑞) = 1∕𝑝 − 1∕𝑝 + 1∕(𝑝𝑞) = 1∕(𝑝𝑞) = 𝑑

Thereby, 𝑇(𝑘) = 𝑑 + (𝑘 − 1)𝑑 = 𝑘𝑑 = 1 ⇔ 𝑘 = 1∕𝑑 = 𝑝𝑞,

which means that log.𝑘(𝑖) = log.𝑝𝑞(𝑖)

And so, ∑.(𝑖=1, 𝑛) log.𝑘(𝑖) = log.𝑝𝑞(1) + log.𝑝𝑞(2) + log.𝑝𝑞(3) + ... + log.𝑝𝑞(𝑛)

Also, 𝐴 = {𝑑, 2𝑑, 3𝑑, ...}, which means that the product of 𝑛 terms,

𝑃(𝑛) = 𝑑∙2𝑑∙3𝑑∙...∙𝑛𝑑 = 𝑑^𝑛∙𝑛! = (𝑝𝑞)^(−𝑛)∙𝑛!

And so, log.𝑝𝑞(𝑃(𝑛)) = log.𝑝𝑞((𝑝𝑞)^(−𝑛)∙𝑛!) =

= log.𝑝𝑞((𝑝𝑞)^(−𝑛)) + log.𝑝𝑞(1) + log.𝑝𝑞(2) + log.𝑝𝑞(3) + ... + log.𝑝𝑞(𝑛)

So, in the end, ∑.(𝑖=1, 𝑛) log.𝑘(𝑖) − log.𝑝𝑞(𝑃(𝑛)) = −log.𝑝𝑞((𝑝𝑞)^(−𝑛)) = 𝑛

Finally, since 𝑘 = 𝑝𝑞, we have 𝑇(𝑝𝑞) = 𝑇(𝑘) = 1, which means that we can write

∑.(𝑖=1, 𝑛) log.𝑘(𝑖) − log.𝑝𝑞(𝑃(𝑛)) = 𝑛∙𝑇(𝑝𝑞)(2 votes)

- Is there a way to form an equation for the sequence: 1, 3, 6, 10 ?(1 vote)
- I believe the pattern of the sequence is that the first term is 1, then add 2, add 3, add 4, et cetera.

I have a recursive formula for the sequence here:`a(1) = 1`

`a(n) = a(n-1) + n`

Plugging in n = 1 gives 1, n = 2 gives 3, n = 3 gives 6, and so on.(2 votes)

- slope also means gradient?right(1 vote)
- Yes that is correct. Slope and gradient are synonymous.(2 votes)

- is this equation correct?

1+4(n-1)(5 votes)- Yes. It is in fact the same, just written a different way.

1+(4 * (n-1)) is equal to 1 + ((n-1) * 4)

Example:

1st option: 1+4 * (3-1) = 1 + 4 * (2) = 1 + 4*2 = 9

2nd option: 1+(3-1) * 4 = 1 + (2) * 4 = 1 + 2*4 = 9

Hope that helps(3 votes)

- In my class, we use a sub n instead of a sub i. Why is this so?(0 votes)
- I looks to close to a one. it also is used commonly for complex numbers, so n is less confusing.(3 votes)

## Video transcript

Our question asks us, what
equation describes the growth pattern of this sequence
of a block? So we want to figure out, if I
know that x is equal to 10, how many blocks am
I going to have? So let's just look at
this pattern here. So our first term in our
sequence, or our first object, or our first pattern of blocks
right here, we just have 1 block right there. So let me write, the term--
write it up here --so I have the term and, then I'll have
the number of blocks. So in our first term,
we had one block. And then our second term-- I'll
just write this down, just so we have it --what
happened here? So it looks just like our first
term, but we added a column here of four blocks. So it's like 1 plus
4 right there. So we're going to have five
blocks right there. We added 4 to it. Then in our third term
what happened? What happened in
our third term? Well it just looks just like the
second term, but we added another column of four
blocks here. Right? We added this column
right there. If you imagine they were being
added to the left-hand side of the pattern. So we added four more blocks. We have nine blocks now. We have nine blocks, so it looks
like each time we're adding four blocks. And on this fourth
term, same thing. The third term is just
this right here. This right here is what the
third term looked like, and then we added another column
of four blocks right here. So we added four more, so we're
going to have 13 blocks. So our fourth term is 13. So let's see if we can come up
with a formula, either looking at the graphics, or maybe
looking at the numbers themselves. So one way to think about it, so
we start off with-- So when x is equal to 1, let's say that
x is equal to the term, we add just this 1 there. Then when x is equal to 2, we
added one column of four. So this is when x is equal to 2,
we have one column of four. Then when x is equal
to 3, we have two columns of 4, right there. And you could even say when x
is equal to 1, you had zero columns, right? We had no, nothing, no extra
columns of four blocks. We didn't have any. And then when x is equal to
4, we had three columns. We had three columns there,
when x is equal to 4. So what's the pattern here? Or how can we express the number
of blocks we're going to have, given the term
that we have? Well, it looks like we're
always going to have one block, so let me write
it this way. If I write the number of
blocks-- let me write it this way --it looks like
we're always going to have one, right? We have this one right here,
that one right there, that one right there, that
one right there. Looks like we always have one
plus a certain number of columns of four, but how many
columns do we have? When x is equal to 1, we have
no columns of four blocks. When x is equal to 2,
we have one column. When x is equal to 3,
we have two columns. So when x is equal to anything,
it looks like we have one less number
of columns. So it's going to be
x minus 1, right? When x is 2, x minus 1 is 1. When x is 3, x minus 1, so this
right here is x minus 1. x is 2, this is x minus 1. This is x minus 1. This is x minus 1, and x minus
1 will tell us the number of columns we have, right? Here we have one, two,
three columns. Here we have one, two columns. Here we only have one column. Here we have zero columns. So it even works for
the first term. And in every one of these
columns, so this right here, x minus 1 is the number of
columns, and then in each column we have four blocks. So it's the number of columns
times 4, right? For each of these columns,
we have one column. We have one, two, three,
four blocks. So this is the equation that
describes the growth pattern. So let me write this, let me
simplify this a little bit. If I were to multiply 4 times x
minus 1, I get the number of blocks being equal to
1 plus 4 times x. I have to distribute it. 4 times x is 4x, and then
4 times negative 1 is negative 4. So that's equal to the
number of blocks. And we could simplify this. We have a 1 and we have a minus
4, or I guess we're subtracting 4 from it, so this
is going to be equal to 4x minus 3 is the number of blocks
given our x term. So if we're on term 50, it's
going to be 4 times 50, which is 200 minus 3, which
is 197 blocks. Now another way you could have
done it is you could have just said, look, every time we're
adding 4, this is a linear relationship, and you could
essentially find the slope of the line that connects this,
but assume that our line is only defined on integers. And that might be a little bit
more complicated, but the way that you think about it is,
every one, every time we added a block, we added-- or every
time we added a term we added four blocks. So we could write it this way. We could just write change--
so this the triangle right here means change. Delta means change in blocks
divided by change in x. Now you might recognize this. This is slope. And if you don't worry, if slope
is a completely foreign concept to you, you can just
do it the way we did it the first part of this video. And that's a completely
legitimate way, and hopefully it will make some connections
between what slope is. So what is the change in blocks
for a change in x. So when we went from x going
from 1 to 2-- so our change in x here would be 2 minus 1, we
increased by 1 --what was our change in blocks? It would be 4, or 5 minus 1. It's 5 minus 1. And what is this equal to? This is equal to 4 over 1,
which is equal to 4. Let me scroll over
a little bit. So our change in blocks, or
change in x is 4, or our slope is equal to 4. So if you want to do this kind
of the setting up the equation of a line way, you would say
that our equation-- If, well let me write it. Number of blocks are going to
be equal to 4 times the term that we're dealing with,
the term in our pattern, plus some constant. This right here is the
equation of a line. If it's completely foreign to
you, just do it the way we did it earlier in the video. And so, how do we solve
for this constant? Well, we use one of
our terms here. We know that when we had one--
In our first term we only had one block. So let's put that here. So in our first term-- we're
going to have that b right there --we only had one block. So we have 1 is equal
to 4 plus b. If you subtract 4 from both
sides of this equation, so you subtract 4 from both sides,
what do you get? On the left-hand side, 1 minus
4 is negative 3, and that's equal to-- these 4's cancel
out --and and that's equal to b. So another way to get the
equation of a line, we have just solved that b is
equal to negative 3. We said how much do the number
of blocks change for a certain change in x, which is a change
in the number blocks for a change in x, we saw
it's always 4. 4 per change in x. When x changes by 1,
we change by 4. That gave us our slope. And then to solve for-- If you
view this as a line, although this is only defined
on integers, I guess positive integers. In this situation, you could
view this as a y-intercept. To solve for this constant, we
just use one of our terms. You could have used any of them. We used 1 and 1. You could use 3 and 9. You could use anything. We solved b is equal to negative
3, and so if you put b back here, you get four x
minus 3, which is what we got earlier in the video,
right there. Hopefully you found that fun.