# Arithmetic sequence

#### wishmaster

##### Active member
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

Staff member
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
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

$\chi$ $\sigma$

#### wishmaster

##### Active member
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

Staff member
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
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?

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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

Staff member
...How do you know that 35=243 ?
$$\displaystyle 243=3\cdot81=3\cdot9^2=3\cdot3^4=3^5$$

#### wishmaster

##### Active member
$$\displaystyle 243=3\cdot81=3\cdot9^2=3\cdot3^4=3^5$$
Dont know what to do further.....how to deal with such problems!

#### MarkFL

Staff member
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 $y$?

#### wishmaster

##### Active member
If $r^x=r^y$, then what can we say about $x$ and $y$?
That $$\displaystyle x=y$$ ?

#### MarkFL

Staff member
That $$\displaystyle x=y$$ ?
Yes good!

What does this mean in terms of the given equation:

$$\displaystyle 3^{a_1+a_2}=243$$

#### wishmaster

##### Active member
Yes good!

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

Staff member
$$\displaystyle \sqrt[a_1+a_2]{243}=3$$
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
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
$$\displaystyle a_1+d=5$$
$$\displaystyle a_1+2d=6$$ ?
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
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

Staff member
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
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?

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#### MarkFL

Staff member
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:

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

$$\displaystyle n^2+3n-154=0$$

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
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:

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

$$\displaystyle n^2+3n-154=0$$

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

Staff member
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 $\mp$.

#### wishmaster

##### Active member
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 $\mp$.
Thank you! Im stiil learning $\LaTeX$ .
So i think this is solved?
Or is there another way to do it?

#### MarkFL

Thank you! Im stiil learning $\LaTeX$ .