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### Course: Class 10 (Old)ย >ย Unit 5

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?

• At how did going from one to two get expressed as 2-1?
• 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!
• at ,i have no idea how to get b=-3, the number of blocks between X is 4,so should B=-4?
• 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"). )
• 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

(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?
• im unable to find the vid regarding the summation f series ..
(1 vote)
• Can somebody solve this
[MIND BURST] [ Trickiest !!] Algebra Problem 2 - NT Maths - YouTube
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, ๐ด = {๐(1), ๐(2), ๐(3), ...}

We know that ๐(๐) = ๐(1) + (๐ โ 1)๐ = 1โ๐ โ ๐(1) = 1โ๐ โ (๐ โ 1)๐
Likewise, ๐(๐) = 1โ๐ โ ๐(1) = 1โ๐ โ (๐ โ 1)๐
This gives us: 1โ๐ โ (๐ โ 1)๐ = 1โ๐ โ (๐ โ 1)๐ โ ๐ = (1โ๐ โ 1โ๐)โ(๐ โ ๐) = 1โ(๐๐) (given ๐ โ  ๐)

So, ๐(1) = 1โ๐ โ (๐ โ 1)โ(๐๐) = 1โ๐ โ 1โ๐ + 1โ(๐๐) = 1โ(๐๐) = ๐

Thereby, ๐(๐) = ๐ + (๐ โ 1)๐ = ๐๐ = 1 โ ๐ = 1โ๐ = ๐๐,
which means that log.๐(๐) = log.๐๐(๐)
And so, โ.(๐=1, ๐) log.๐(๐) = log.๐๐(1) + log.๐๐(2) + log.๐๐(3) + ... + log.๐๐(๐)

Also, ๐ด = {๐, 2๐, 3๐, ...}, which means that the product of ๐ terms,
๐(๐) = ๐โ2๐โ3๐โ...โ๐๐ = ๐^๐โ๐! = (๐๐)^(โ๐)โ๐!
And so, log.๐๐(๐(๐)) = log.๐๐((๐๐)^(โ๐)โ๐!) =
= log.๐๐((๐๐)^(โ๐)) + log.๐๐(1) + log.๐๐(2) + log.๐๐(3) + ... + log.๐๐(๐)

So, in the end, โ.(๐=1, ๐) log.๐(๐) โ log.๐๐(๐(๐)) = โlog.๐๐((๐๐)^(โ๐)) = ๐

Finally, since ๐ = ๐๐, we have ๐(๐๐) = ๐(๐) = 1, which means that we can write
โ.(๐=1, ๐) log.๐(๐) โ log.๐๐(๐(๐)) = ๐โ๐(๐๐)
• 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.
(1 vote)
• Yes that is correct. Slope and gradient are synonymous.
• In my class, we use a sub n instead of a sub i. Why is this so?