# Thread: HELP! Arithmetic/Geometric sequence questions, assignment due tomorrow

1. ## HELP! Arithmetic/Geometric sequence questions, assignment due tomorrow

1. Write an expression, in simplified form, for the general term of the sequence in which the first term is 14 and each term is 5 less than the term preceding it.

2. In an arithmetic sequence, t15=-27 and t10=-12. Find the value of the fist term and the common difference.

3. Write an expression, in simplified form, for the general term of the sequence in which the first term is 16 and each succeeding term is 1/2 the before it.

2. Originally Posted by xkncx
1. Write an expression, in simplified form, for the general term of the sequence in which the first term is 14 and each term is 5 less than the term preceding it.
First, the phrase "each term is 5 less than the term preceding it" is simply saying the common difference is -5. So our sequence looks like this:

14, 9, 4, -1, -6, ...

Note that, for example, the third term is 10 less than where we started. The fourth term is 15 less than where we started, the fifth is 20 less, and so on. Continuing this pattern, the, say, 100th term would be 495 less than what we started with (99 * 5 = 495). So, the nth term is going to be, basically, the first term minus 5*(n-1). So the nth term will be 14 - 5*(n-1), or, if you simplify, 19 - 5n. (Note that this is assuming we consider 14 to be term #1 - sometimes it is considered to be term #0, resulting in the much nicer formula 14 - 5n. Make any adjustments necessary based on your class, but this should get you the general idea.)

Originally Posted by xkncx
2. In an arithmetic sequence, t15=-27 and t10=-12. Find the value of the first term and the common difference.
So the 10th term is -12 and the 15th term is -27. That means they're 15 apart in value, and 5 terms apart. So, in the space of 5 terms, we have to subtract a total of 15, and we have to subtract the same amount. Division (15/5) tells us we should subtract 3 each time, and this checks out - starting with term 10 and ending with term 15:

-12, -15, -18, -21, -24, -27.

So the common difference is -3. Now, we need to work backwards from term 10 to get back to the first term. Since the common difference is -3, but we're working backwards, we need to add 3 each time. The 9th term, then, will be -9, and so on. Keep going back as far as you need to.

Originally Posted by xkncx
3. Write an expression, in simplified form, for the general term of the sequence in which the first term is 16 and each succeeding term is 1/2 the before it.

The second term, then, will be 16 * (1/2), right? (Or, simply, 8).

The third term will be half the second term, or 16 * (1/2) * (1/2) = 16 * (1/2)^2. (Or, simply, 4).

The fourth term will be half the third term, or 16 * (1/2)^2 * (1/2) = 16 * (1/2)^3. See a pattern developing?

Really, the only thing changing is the exponent to 1/2. How does the exponent relate to which term we're on? Once you figure this out, what will the exponent be if the term # is a general one, like n?

Hope this helps! Your text probably has formulas for many of these things, but ultimately understanding WHERE the formulas come from, which is what I tried to explain above, is a lot more useful than just memorizing the formulas, which can be tricky.

3. Hello, xkncx!

You are expected to know a formula for the $\displaystyle n^{th}$ term of an arithmetic sequence:

. . $\displaystyle a_n \:=\:a + (n-1)d,\;\text{ where }a\text{ = first term, }d\text{ = common difference.}$

1. Write an expression for the general term of the sequence in which
the first term is 14 and each term is 5 less than the term preceding it.
We are given: .$\displaystyle a = 14\text{ and }d = -5$

Therefore: .$\displaystyle a_n \;=\;14 + (n-1)(-5) \quad\Rightarrow\quad\boxed{ a_n\;=\;19-5n}$

2. In an arithmetic sequence, $\displaystyle t_{15}=\text{-}27\text{ and }t_{10}= \text{-}12$
Find the first term and the common difference.

Using the formula: .$\displaystyle \begin{array}{cccc} t_{15}\:=\:\text{-}27 & \Rightarrow & a + 14d \:=\:\text{-}27 & {\color{blue}[1]} \\ t_{10}\:=\:\text{-}12 & \Rightarrow & a + 9d \:=\:\text{-}12 & {\color{blue}[2]} \end{array}$

Subtract [2] from [1]: .$\displaystyle 5d \:=\:-15\quad\Rightarrow\quad \boxed{d\:=\:-3}$

Substitute into [2]: .$\displaystyle a + 9(-3) \:=\:-12\quad\Rightarrow\quad\boxed{ a \:=\:15}$

3. Write an expression for the general term of the sequence in which
the first term is 16 and each succeeding term is 1/2 the term before it.
This is a geometric sequence . . . $\displaystyle a_n \;=\;ar^{n-1}$
. . where $\displaystyle a$ = first term, $\displaystyle r$ = common ratio.

We are given: .$\displaystyle a \,=\,16,\;r\,=\,\frac{1}{2}$

Therefore: .$\displaystyle a_n\:=\:16\left(\frac{1}{2}\right)^{n-1}\quad\Rightarrow\quad a_n \:=\:\frac{2^4}{2^{n-1}} \quad\Rightarrow\quad\boxed{a_n\:=\:\frac{1}{2^{n-5}}}$