Arithmetic Sequence: Find Initial Term & Sum to 243

[theakdad]

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

 

[MarkFL]

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]

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

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

$\chi$ $\sigma$

 

[theakdad]

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]

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 $d$?

 

[theakdad]

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]

...How do you know that 3⁵=243 ?

\(\displaystyle 243=3\cdot81=3\cdot9^2=3\cdot3^4=3^5\)

 

[theakdad]

\(\displaystyle 243=3\cdot81=3\cdot9^2=3\cdot3^4=3^5\)

Dont know what to do further...how to deal with such problems!

 

[MarkFL]

Dont know what to do further...how to deal with such problems!

If $r^x=r^y$, then what can we say about $x$ and $y$?

 

[theakdad]

If $r^x=r^y$, then what can we say about $x$ and $y$?

That \(\displaystyle x=y\) ?

 

[MarkFL]

That \(\displaystyle x=y\) ?

Yes good! :D

What does this mean in terms of the given equation:

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

 

[theakdad]

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]

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

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 $a_1$.

 

[theakdad]

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]

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

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 \(\displaystyle a_2\))

 

[theakdad]

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 equation 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]

Yes, that's correct. Now can you use the formula for the sum of the series given by chisigma to answer part b)?

 

[theakdad]

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?

 

[MarkFL]

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.

 

[theakdad]

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]

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

 

[theakdad]

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! I am stiil learning $\LaTeX$ .
So i think this is solved?
Or is there another way to do it?

 

[MarkFL]

Thank you! I am stiil learning $\LaTeX$ .
So i think this is solved?
Or is there another way to do it?

I would say you are done. :D