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Arithmetic sequence

wishmaster

Active member
Oct 11, 2013
211
In arithmetic sequence we know that \(\displaystyle a_1+a_3 = 6\) and \(\displaystyle 3^{a_1+a_2}=243\)

a) Find the initial term of the sequence
b) Calculate,how much members of the sequence we have to add \(\displaystyle (a_1+a_2+...a_n)\) that we get the result 243?
Have no idea where to start :confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Can you state the $n$th term of the sequence in terms of the first term $a_1$ and the common difference $d$? Is $243$ a power of $3$?
 

chisigma

Well-known member
Feb 13, 2012
1,704
In arithmetic sequence we know that \(\displaystyle a_1+a_3 = 6\) and \(\displaystyle 3^{a_1+a_2}=243\)

a) Find the initial term of the sequence
b) Calculate,how much members of the sequence we have to add \(\displaystyle (a_1+a_2+...a_n)\) that we get the result 243?
Have no idea where to start :confused:
You know that...

$\displaystyle a_{1}+ a_{3} = 6$

$\displaystyle \log_{3} 243 = a_{1} + a_{2} = 5\ (1)$

Now the general expression of an arithmetic sequence is...

$\displaystyle a_{n+1} = a_{n} + r\ (2)$

... and the (1) and (2) allow You to find $a_{1}$ and $r$ because is...

$\displaystyle a_{1} + r = 3$

$\displaystyle 2\ a_{1} + r = 5\ (3)$

... so that is $a_{1}=2$ and $r=1$. The sum of the first n terms of an arithmetic sequence is...

$\displaystyle S_{n} = \frac {n}{2}\ \{2\ a_{1} + (n-1)\ r\}\ (4)$

In Your opinion, setting in (4) $a_{1}=2$ and $r=1$ does it exist some n for which is $S_{n} = 243$?...

Kind regards

$\chi$ $\sigma$
 

wishmaster

Active member
Oct 11, 2013
211
Can you state the $n$th term of the sequence in terms of the first term $a_1$ and the common difference $d$? Is $243$ a power of $3$?
Dont know how to start....

\(\displaystyle 3^{a_1+a_2} = 243 \)so 3 on power/exponent of \(\displaystyle 3^{a_1+a_2}\) is 243
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Dont know how to start....

\(\displaystyle 3^{a_1+a_2} = 243 \)so 3 on power/exponent of \(\displaystyle 3^{a_1+a_2}\) is 243
The $n$th term of an arithmetic sequence , or arithmetic progression (AP) is given by:

\(\displaystyle a_n=a_1+(n-1)d\)

Where $d$ is the common difference between successive terms.

Also \(\displaystyle 3^5=243\)

Can you put this together to obtain a linear 2X2 system of equations in $a_1$ and $d$?
 

wishmaster

Active member
Oct 11, 2013
211
You know that...

$\displaystyle a_{1}+ a_{3} = 6$

$\displaystyle \log_{3} 243 = a_{1} + a_{2} = 5\ (1)$

Now the general expression of an arithmetic sequence is...

$\displaystyle a_{n+1} = a_{n} + r\ (2)$

... and the (1) and (2) allow You to find $a_{1}$ and $r$ because is...

$\displaystyle a_{1} + r = 3$

$\displaystyle 2\ a_{1} + r = 5\ (3)$

... so that is $a_{1}=2$ and $r=1$. The sum of the first n terms of an arithmetic sequence is...

$\displaystyle S_{n} = \frac {n}{2}\ \{2\ a_{1} + (n-1)\ r\}\ (4)$

In Your opinion, setting in (4) $a_{1}=2$ and $r=1$ does it exist some n for which is $S_{n} = 243$?...

Kind regards

$\chi$ $\sigma$
Can it be done without logarithms?

- - - Updated - - -

The $n$th term of an arithmetic sequence , or arithmetic progression (AP) is given by:

\(\displaystyle a_n=a_1+(n-1)d\)

Where $d$ is the common difference between successive terms.

Also \(\displaystyle 3^5=243\)

Can you put this together to obtain a linear 2X2 system of equations in $a_1$ and $d$?

How do you know that 35=243 ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
\(\displaystyle 243=3\cdot81=3\cdot9^2=3\cdot3^4=3^5\)
Dont know what to do further.....how to deal with such problems!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
Yes good! :D

What does this mean in terms of the given equation:

\(\displaystyle 3^{a_1+a_2}=243\)
\(\displaystyle \sqrt[a_1+a_2]{243}=3\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \sqrt[a_1+a_2]{243}=3\)
While that is true, I meant:

\(\displaystyle 3^{a_1+a_2}=243-3^5\)

Hence:

\(\displaystyle a_1+a_2=5\)

See how we got the same base and then equated exponents?

So, now you have the above and the given:

\(\displaystyle a_1+a_3=6\)

Now can you rewrite $a_2$ and $a_3$ using the formula I provided above for the $n$th term?

Once you do this, you will have two linear equations in two unknowns. At this point you can use elimination or substitution to get the value of $a_1$.
 

wishmaster

Active member
Oct 11, 2013
211
While that is true, I meant:

\(\displaystyle 3^{a_1+a_2}=243-3^5\)

Hence:

\(\displaystyle a_1+a_2=5\)

See how we got the same base and then equated exponents?

So, now you have the above and the given:

\(\displaystyle a_1+a_3=6\)

Now can you rewrite $a_2$ and $a_3$ using the formula I provided above for the $n$th term?

Once you do this, you will have two linear equations in two unknowns. At this point you can use elimination or substitution to get the value of $a_1$.
\(\displaystyle a_1+d=5\)
\(\displaystyle a_1+2d=6\) ?
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
\(\displaystyle a_1+d=5\)
\(\displaystyle a_1+2d=6\) ?
Not quite. Remember that \(\displaystyle a_2 = a_1 + d\) and, more generally, \(\displaystyle a_n = a_1 + (n-1)d\). You seem to have left out the additional \(\displaystyle a_1\) terms when doing your working

For the first equation you have \(\displaystyle a_1 + a_2 = 5\)

Because \(\displaystyle a_2 = a_1 + d\) then you get \(\displaystyle a_1 + (a_1+d) = 5\) which can be simplified (the expression in the brackets is an expression for \(\displaystyle a_2\))
 
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wishmaster

Active member
Oct 11, 2013
211
Not quite. Remember that \(\displaystyle a_2 = a_1 + d\) and, more generally, \(\displaystyle a_n = a_1 + (n-1)d\). You seem to have left out the additional \(\displaystyle a_1\) terms when doing your working

For the first equation you have \(\displaystyle a_1 + a_2 = 5\)

Because \(\displaystyle a_2 = a_1 + d\) then you get \(\displaystyle a_1 + (a_1+d) = 5\) which can be simplified (the expression in the brackets is an expression for \(\displaystyle a_2\))
So then is so i believe:

\(\displaystyle 2a_1+d=5\)
\(\displaystyle 2a_1+2d=6\)
so then i multiply first eqation by (-1) and i get:

\(\displaystyle -2a_1-d=-5\)
\(\displaystyle 2a_1+2d=6\)

i add the equations and i get \(\displaystyle d=1\)

and from here that \(\displaystyle a_1=2\) ,\(\displaystyle a_2=3\) and \(\displaystyle a_3\) is 4

Am i on the right path?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, that's correct. Now can you use the formula for the sum of the series given by chisigma to answer part b)?
 

wishmaster

Active member
Oct 11, 2013
211
Yes, that's correct. Now can you use the formula for the sum of the series given by chisigma to answer part b)?
I have used the formula,and further solved it with quadratic formula:

\(\displaystyle \frac{n(2a_1+(n-1)d)}{2}=77\)
From here i got:

\(\displaystyle n^2+3n-154=0\)

So:

\(\displaystyle a=1, b=3, c=154\)

\(\displaystyle \frac{-b+/-\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x_1= 11, x_2= -14\) Cant be solution.

So my solution is 11! We have to add 11 members of the sequence to get 77.

Im wondering if there is shorter way to calculate this?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I thought the sum needed to be 243?

We could use the formula:

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

For a sum of 77, we could use the well-known formula \(\displaystyle \sum_{k=1}^n(k)=\frac{n(n+1)}{2}\) to state:

\(\displaystyle S_n=\frac{n(n+1)}{2}+n=77\)

Arrange in standard quadratic form:

\(\displaystyle n^2+3n-154=0\)

Factor:

\(\displaystyle (n-11)(n+14)=0\)

Discarding the negative root, we find:

\(\displaystyle n=11\)

However, I recommend the formula given by chisigma as it is more general.
 

wishmaster

Active member
Oct 11, 2013
211
I thought the sum needed to be 243?

We could use the formula:

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

For a sum of 77, we could use the well-known formula \(\displaystyle \sum_{k=1}^n(k)=\frac{n(n+1)}{2}\) to state:

\(\displaystyle S_n=\frac{n(n+1)}{2}+n=77\)

Arrange in standard quadratic form:

\(\displaystyle n^2+3n-154=0\)

Factor:

\(\displaystyle (n-11)(n+14)=0\)

Discarding the negative root, we find:

\(\displaystyle n=11\)

However, I recommend the formula given by chisigma as it is more general.
The second question was: How many members of the sequence do you need to get the sum of 77.
I think i have done it with formula given by chisigma.....or maybe i have missed something.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The second question was: How many members of the sequence do you need to get the sum of 77.
I think i have done it with formula given by chisigma.....or maybe i have missed something.
Yes, you applied it correctly.

In your statement of the quadratic formula, there is a $\LaTeX$ command to display the plus/minus character. The code \pm produces $\pm$. And in case you ever need it, the code \mp produces $\mp$. :D
 

wishmaster

Active member
Oct 11, 2013
211
Yes, you applied it correctly.

In your statement of the quadratic formula, there is a $\LaTeX$ command to display the plus/minus character. The code \pm produces $\pm$. And in case you ever need it, the code \mp produces $\mp$. :D
Thank you! Im stiil learning $\LaTeX$ .
So i think this is solved?
Or is there another way to do it?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775