# Progression

#### Wxlauslen123

##### New member
By writing the rth then of the progression 16,22,29,37...in the form Tr=1+1+2+3+...+(r+4),find

a) the seventh term of the progression
b) the sum of the eighth and ninth terms of the progression

#### Subhotosh Khan

##### Super Moderator
Staff member
By writing the rth then of the progression 16,22,29,37...in the form Tr=1+1+2+3+...+(r+4),find

a) the seventh term of the progression
b) the sum of the eighth and ninth terms of the progression
Please follow the rules of posting in this forum, as enunciated at:

#### Otis

##### Senior Member
… the form Tr=1+1+2+3+...+(r+4) …
Hello Wxlauslen. I'm not sure about that form, but we could write a recursive formula, where r+4 is added to each previous term to get the next term. Does that seem like something you're trying to do? Please show any work that you've done so far.

#### pka

##### Elite Member
By writing the rth then of the progression 16,22,29,37...in the form Tr=1+1+2+3+...+(r+4),find
a) the seventh term of the progression
b) the sum of the eighth and ninth terms of the progression
According to my reading you sequence is: $$\displaystyle t_0=16,~t_1=22,~t_2=29,~t_3=37,\cdots$$ (note the subscripts may be one value off).
Here is a recursive definition for this sequence: $$\displaystyle t_0=16,~t_n=t_{n-1}+(n+5)$$ SEE HERE Scroll down to see table.
That recursive function is the function $$\displaystyle f(n)=\frac{n(n+11)}{2}+16$$ SEE HERE (remember the eight value is when n=7).

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#### Wxlauslen123

##### New member
Hello Wxlauslen. I'm not sure about that form, but we could write a recursive formula, where r+4 is added to each previous term to get the next term. Does that seem like something you're trying to do? Please show any work that you've done so far.

Hello Wxlauslen. I'm not sure about that form, but we could write a recursive formula, where r+4 is added to each previous term to get the next term. Does that seem like something you're trying to do? Please show any work that you've done so far.

I do a bit working found that the common difference between the terms always increase by 1 and the common difference is not the same. The answer for 7th term is 67 and the sum is 171

#### MarkFL

##### Super Moderator
Staff member
I would write something like this, based on what was given:

$$\displaystyle T_r=1+\sum_{k=1}^{r+4}(k)=1+\frac{(r+4)(r+5)}{2}=\frac{r^2+9r+22}{2}$$

#### Wxlauslen123

##### New member
According to my reading you sequence is: $$\displaystyle t_0=16,~t_1=22,~t_2=29,~t_3=37,\cdots$$ (note the subscripts may be one value off).
Here is a recursive definition for this sequence: $$\displaystyle y_0=16,~t_n=t_{n-1}+(n+5)$$ SEE HERE Scroll down to see table.
That recursive function is the function $$\displaystyle f(n)=\frac{n(n+11)}{2}+16$$ SEE HERE (remember the eight value is when n=7).
Can you please explain more about this recursive definition and function, my school dld not teach this one, they only teach arithmetic progression and geometric progression. The question stated wanted to use Tr form to solve the question. I don't think my teacher can accept this working

#### Wxlauslen123

##### New member
I would write something like this, based on what was given:

$$\displaystyle T_r=1+\sum_{k=1}^{r+4}(k)=1+\frac{(r+4)(r+5)}{2}=\frac{r^2+9r+22}{2}$$
Can you please explain more got this? How does Tr form that is given became this?

#### Wxlauslen123

##### New member
Hello Wxlauslen. I'm not sure about that form, but we could write a recursive formula, where r+4 is added to each previous term to get the next term. Does that seem like something you're trying to do? Please show any work that you've done so far.

I did like T7 = 1+1+2+3+4+5+6+7+8+9+10+(7+4)=67

#### MarkFL

##### Super Moderator
Staff member
Can you please explain more got this? How does Tr form that is given became this?
The sum of the first $$n$$ natural numbers is:

$$\displaystyle S_n=\sum_{k=1}^n(k)=\frac{n(n+1)}{2}$$

One simple way to see the is true is to write:

$$\displaystyle S_n=1+2+3+\cdots+(n-2)+(n-1)+n$$

$$\displaystyle S_n=n+(n-1)+(n-2)+\cdots+3+2+1$$

If we add these two equations, we get:

$$\displaystyle 2S_n=(n+1)+(n+1)+(n+1)+\cdots+(n+1)+(n+1)+(n+1)=n(n+1)$$

Hence:

$$\displaystyle S_n=\frac{n(n+1)}{2}$$

#### pka

##### Elite Member
Can you please explain more about this recursive definition and function, my school dld not teach this one, they only teach arithmetic progression and geometric progression. The question stated wanted to use Tr form to solve the question. I don't think my teacher can accept this working
Did you read reply #2? You see this sort of misunderstanding is exactly what happens when you post without telling what you have done, what you know about the topic, and/or what theorems and definitions with which you are working.

#### Otis

##### Senior Member
I do a bit working found … 7th term is 67 and the sum [of 8th and 9th terms] is 171
Hi Wxlauslen. Those answers match what I got, using a recursive method. I'm not familiar with Tr form; please excuse me, as I was thinking about something else.

$$\;$$

#### pka

##### Elite Member
I do a bit working found that the common difference between the terms always increase by 1 and the common difference is not the same. The answer for 7th term is 67 and the sum is 171
Hi Wxlauslen. Those answers match what I got, using a recursive method. I'm not familiar with Tr form; please excuse me, as I was thinking about something else.
If either of you look at this page, the eighth & ninth terms in that table are $$\displaystyle 79~\&~92$$ and sum $$\displaystyle 171$$.
Can either of you explain what the $$\displaystyle Tr$$-form is?

#### Otis

##### Senior Member
… Can either of you explain what the $$\displaystyle Tr$$-form is?
I've posted twice that I don't know it.

#### pka

##### Elite Member
I've posted twice that I don't know it.
Sorry, I missed your posting that. I really hope someone can tell us what that method is. I been around these progressions a long time, but I cannot imagine encountering any concept remotely like such.

P.S Look what I found

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#### Dr.Peterson

##### Elite Member
By writing the rth then of the progression 16,22,29,37...in the form Tr=1+1+2+3+...+(r+4),find

a) the seventh term of the progression
b) the sum of the eighth and ninth terms of the progression
I did like T7 = 1+1+2+3+4+5+6+7+8+9+10+(7+4)=67
To the helpers:
As far as I can tell, "the Tr form" just means the fact given, that Tr = 1+1+2+3+...+(r+4). That is, the rth term is formed by adding r+4 to the previous term, or alternatively, the rth term is 1 + the (r+4)th triangular number. (The notation Tr merely means "the rth term of a sequence called T. It is not some advanced method or concept! The "form" mentioned in the problem is not "Tr", but "1+1+2+3+...+(r+4)".)

To @Wxlauslen123:
I can't tell, without knowing what you have been learning, whether you are supposed to show how to recognize that fact if it were not given, or how you are expected to use it. We really need you to tell us more!

It may be that you are expected to do nothing more than what you did: just list the terms in the sum that defines the 7th term, which ends with 7+4 = 11, and add them up: 1+1+2+3+4+5+6+7+8+9+10+11 = 67. Since that's what you said you did, and since the numbers are not large, I suspect that is all you had to do. You can do the same for part b, just calculating those terms (by adding 8+4 and 9+4, respectively, to terms you already have).

As you've seen, there are many more advanced ways to do this. We who know more than you do see the problem as calling for knowledge of recursive sequences, summation formulas, and so on, but clearly that is not what is expected of you. Those methods would be appropriate if they asked, say, for T1000. You are probably just learning the meaning of sequence notation, and a little about arithmetic progressions and the like. If you have learned the formula for the sum of an arithmetic progression, you could use that. But it isn't really necessary.

Can you please tell us what you are currently learning, so we can set these other ideas aside?

#### Wxlauslen123

##### New member
To the helpers:
As far as I can tell, "the Tr form" just means the fact given, that Tr = 1+1+2+3+...+(r+4). That is, the rth term is formed by adding r+4 to the previous term, or alternatively, the rth term is 1 + the (r+4)th triangular number. (The notation Tr merely means "the rth term of a sequence called T. It is not some advanced method or concept! The "form" mentioned in the problem is not "Tr", but "1+1+2+3+...+(r+4)".)

To @Wxlauslen123:
I can't tell, without knowing what you have been learning, whether you are supposed to show how to recognize that fact if it were not given, or how you are expected to use it. We really need you to tell us more!

It may be that you are expected to do nothing more than what you did: just list the terms in the sum that defines the 7th term, which ends with 7+4 = 11, and add them up: 1+1+2+3+4+5+6+7+8+9+10+11 = 67. Since that's what you said you did, and since the numbers are not large, I suspect that is all you had to do. You can do the same for part b, just calculating those terms (by adding 8+4 and 9+4, respectively, to terms you already have).

As you've seen, there are many more advanced ways to do this. We who know more than you do see the problem as calling for knowledge of recursive sequences, summation formulas, and so on, but clearly that is not what is expected of you. Those methods would be appropriate if they asked, say, for T1000. You are probably just learning the meaning of sequence notation, and a little about arithmetic progressions and the like. If you have learned the formula for the sum of an arithmetic progression, you could use that. But it isn't really necessary.

Can you please tell us what you are currently learning, so we can set these other ideas aside?
Im learning add maths form 5 in school, currently finished chapter 2 progression

#### Wxlauslen123

##### New member
Sorry, I missed your posting that. I really hope someone can tell us what that method is. I been around these progressions a long time, but I cannot imagine encountering any concept remotely like such.

P.S Look what I found
My teacher said that need to use the formula for sum

#### Wxlauslen123

##### New member
If either of you look at this page, the eighth & ninth terms in that table are $$\displaystyle 79~\&~92$$ and sum $$\displaystyle 171$$.
Can either of you explain what the $$\displaystyle Tr$$-form is?
My add maths teacher said the Tr is just a guide, need to use the formula for sum in progression

#### Wxlauslen123

##### New member
My add maths teacher said the Tr is just a guide, need to use the formula for sum in progression