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Eleventh grader Kausar asks, "I had this question several times but i do not know how to do it. You are given the 5th and 7th terms of a geometric sequence. Can you determine the 29th term without finding the general term?"

Hi Kausar, the answer to your question is, "Yes, you can." Having said that, I'd like to add that if you'd been asked for the 28th term, or the 30th term, the answer would be "No, you can't." Let's take a look at why.

Let's say the fifth term is 32, and the seventh term is 8. Then we can reason as follows:

"The sixth term is the fifth term times the common ratio, and the seventh term is the sixth term times the common ratio. Therefore, the seventh term is the fifth term times the common ratio squared."

Algebraically, this reasoning looks like this:

a_{6} = a_{5}r; a_{7} = a_{6}r; a_{7} = (a_{5}r)r = a_{5}r^{2}. Thus, 8 = 32r^{2}, which leads to r^{2} = 1/4.

Now, without bothering to get the general term, we ask, "How many times would I have to multiply the seventh term by r^{2} to get the 29th term? And the answer is: (29 - 7)/2 = 11.

So a_{29} = a_{7}r^{11} = 8(1/4)^{11}, which is approximately 1.9x10^{-6}.

So why does this work for the 29th term, but not the 28th? Because (28 - 7)/2 is not an integer, so we can't get there by multiplying by r^{2}. We'd have to find r, and multiply by r (28 - 7) times.

So why can't we do that? Well...because the equation r^{2 }= 1/4 doesn't have just one solution; it has two. r = 1/2 or r = -1/2. So you have two possible values for the 28th term.

I occasionally get hired to write problems for math competititions, and one of the "tricks" I'll occasionally pull on students is to give a sequence like this, where students are likely to end up with an equation like r^{2} = 1/4, and then forget that this has two solutions instead of just one. It's amazing how often a simple trick like that will trip students up!

One student asks for an explanation of central angles - how is a central angle different from an inscribed angle?

The best answer to this question is probably a picture (or, rather, a pair of pictures). The first picture shows a circle with center at X. The angle has its center at the center of the circle. Therefore, it's a central angle.

The second picture shows an additional angle, marked in green. This angle is an inscribed angle, because its vertex lies on the circle.

In both cases, the angles cut off the same arc - arc AB. So how are these two angles different? Well, if you look at the picture, you can see pretty easily that an inscribed angle is smaller than a central angle that cuts off the same arc. I think we can prove that the inscribed angle is half the size of the central angle. Want to give it a try?

Let's construct a segment from Z to X, which splits the quadrilateral AXBZ into two triangles: AZX and BZX. Both of these triangles are isosceles (because two of their sides are radii of the circle). If we let m be the measure of angle AZX, n be the measure of angle BZX, and k be the measure of the central angle, then we can write the following equation for the sum of all the interior angles of the quadrilateral AXBZ:

(measure of AZB) + (measure of B) + (measure of A) + (measure of X) = 360

(m + n) + (n) + (m) + (360 - k) = 360

2m + 2n + 360 - k = 360

k = 2(m + n)

And since m + n is the measure of AZB, we've shown that the measure of a central angle has twice the measure of any inscribed angle that cuts the same arc. I hope that's helpful!

*"Good day sir, pls sir I need explanation on how to solve an equation like this x ^{4 }- 3x^{2} + 2. although I don't have problem on factorization but I don't know what to do when the power of "X" is greater than 2. looking for your reply Sir"*

I can give you several answers to this question. One is a short quick answer that works for this problem. The other answers are more detailed, and give you more information than you need to solve this specific problem. Let's start with the short answer.

## Short Answer

Notice that all the exponents in this expression are even. This means that we could do a substitution like this: y = x^{2}. Doing this substitution changes the expression to:

y^{2} - 3y +2

As you said, you know what to do when it's a quadratic (the highest exponent is 2). This factors as follows:

(y - 1)(y - 2)

But wait - you're not done, because the problem had x in it, not y, so you have to put the x's back in:

(x^{2} - 1)(x^{2} - 2)

Now that you've substituted x back in, you should realize that one of those can be factored:

(x - 1)(x + 1)(x^{2 }- 2). And that's your factorization. Technically, that's your factorization over rationals; if you're including irrationals, the second binomial can be factored as well:

(x - 1)(x + 1)(x - √2)(x + √2)

## Longer Answer

The method shown above doesn't just work if the exponents are 4 and 2; they also work if the exponents are 6 and 3, or 10 and 5, or any other such combination. If you wanted to get really ugly, I suppose it would also work if you had fractional exponents like 5 and 5/2. The method works if it's a trinomial (three terms) and the exponent of the highest degree term is twice the exponent of the middle term, and the lowest degree term is a constant. For example, consider this one:

x^{6} + 8x^{3} + 12

We'll substitute y = x^{3}.

y^{2} + 8y + 12

(y + 6)(y + 2)

(x^{3} + 6)(x^{3} + 2)

Of course, if you were factoring over reals, and you happen to know the rule for factoring a sum of cubes, you could factor this further. But since you mentioned not knowing what to do for exponents larger than two, we can save that for another day!

Similarly, if it was x^{10} +8x^{5} + 12, you would do the substitution y = x^{5}.

## Longest Answer

Again, this only works for trinomials under certain circumstances. But what if the expression doesn't match those criteria? Let's take an example:

x^{3} + 5x^{2} - 2x - 10

In this case, we can't do a substitution to simplify it, so we'll try doing some *grouping*. Since there are four terms, I decide to try breaking it into two groups of two terms each:

(x^{3} + 5x^{2}) - (2x + 10)

Now we can factor each group:

x^{2}(x + 5) - 2(x + 5)

Oh! Since both binomials in parentheses are the same, we can rewrite this as:

(x^{2} - 2)(x + 5).

This technique *sometimes* works. Sometimes grouping works only if you get creative in your groupings. Sometimes you need to rearrange the terms. Sometimes you might need to group them into groups of three terms, or groups of four terms.

Here's an example of rearranging terms:

x^{3} + 3x^{2} + 3x + 1

(x^{3} + 1) + (3x^{2} + 3x)

(x + 1)(x^{2} - x + 1) + 3x(x + 1)

(x + 1)(x^{2} + 2x + 1)

(x + 1)^{3}

## Confession Time

When I first saw your problem I did *not* come up with the simple solution. I wasn't paying attention, I guess, and my first thought was: I'll do this by grouping. I know. I made it harder than it really is. But even when you make a problem harder than it really is, sometimes you can come up with something interesting. In this case, I looked at your expression and rewrote it splitting the middle term into two pieces* like this:

x^{4 }- 3x^{2} + 2

x^{4 }- x^{2} - 2x^{2 }+ 2

(x^{4}^{ }- x^{2}) - (2x^{2 }- 2)

x^{2}(x^{2 }- 1) - 2(x^{2} - 1)

(x^{2} - 2)(x^{2} - 1)

(x^{2} - 2)(x - 1)(x + 1)

And I got the same answer, even though I took a silly, round-about method.

Hope all of this was helpful - or at least interesting!

* And if you're wondering, "Why in the world did you split it in that particular way?" The answer is, "I don't really know. I just had a gut feeling it would be helpful."

*"Sir , I saw your post about how to convert from alphabet to binary , in which I agree about the post. But there was a question I saw I'm my textbook which state that convert the hexadecimal number 4B3.3 to decimal. In your post here https://www.theproblemsite.com/ask/2016/01/converting-letters-to-binary, I notice that A= 65 and B =66. then I equate 66 to represent B. but after solving it there answer is different from mine .they got 1203.1875, and I got 18019.1875. How come sir? But I notice they equate there B = 11. Sir, I need your highlight on it. Thanks for talking time to read it, waiting for ur reply."*

Hi Abolade, Thanks for asking this question. To understand the answer to that, I need to talk for just a moment about graphemes. A grapheme, according to one dictionary, is "The smallest meaningful contrastive unit in a writing system." That's a fancy way of saying that a grapheme is a symbol that represents something meaningful in a writing system. For example, "7" is a grapheme for the number seven. It's a symbol, and whenever you see it, you automatically know that it represents this many things:

X X X X X X X.

On the other hand, "72" is not a grapheme because it's not the *smallest* graphical unit. "72" is actually two graphemes: the grapheme for the number seven, and the grapheme for the number two.

Graphemes are also used to represent letters. "A" is the grapheme we use to represent the first vowel in the alphabet, "B" is the grapheme we use to represent the first consonant, and so forth. Of course, the graphemes for these letters might look different if you were writing a different language, such as Greek: α β γ δ ε...

So here is where things get confusing. In base ten (our normal counting system), we have ten digits. And therefore, we have ten graphemes: 0 1 2 3 4 5 6 7 8 9. Perfect! We have just enough numerical graphemes! And if we're in base eight (also known as "octal"), we have eight graphemes: 0 1 2 3 4 5 6 7. We have more than enough graphemes (the graphemes for eight and nine don't get used in this base).

In fact, for any base less than ten, we have more than enough graphemes. The problem is when we start talking about bases *greater* than ten. Then we don't have enough graphemes!

So instead of inventing *new *graphemes to represent these digits, mathematicians said, "Why waste the effort developing new symbols, when we've got all these other graphemes lying around not being used?" Specifically, we're talking about the alphabet graphemes. So if you are in base eleven, your graphemes are: 0 1 2 3 4 5 6 7 8 9 A, where A is the grapheme for the number ten.

Similarly, if you are in base sixteen (also known as "hexadecimal"), you have sixteen graphemes, and six of them are stolen from the alphabet: 0 1 2 3 4 5 6 7 8 9 A B C D E F.

It's very important to note that in this context, A, B, C, D, E, and F, even though they look like the graphemes for LETTERS, have a very different meaning; they are now graphemes for NUMBERS!

A is the grapheme for the number ten

B is the grapheme for the number eleven

C is the grapheme for the number twelve

D is the grapheme for the number thirteen

E is the grapheme for the number fourteen

F is the grapheme for the number fifteen

The mathematicians very selfishly pirated letters from the alphabet to take on a completely different meaning. I mean, why not? It's not like there would be any confusion, right? Shakespeare was never going to write a sonnet about the hexadecimal number system, so there's no chance there would ever have an overlap of meaning where we were unsure whether A represents a number or a letter, right?

Wrong! Welcome to the world of computers! In a computer system, even though you can *type* graphemes and a computer can *display* graphemes, the computer does not *understand *graphemes. Computers "think" strictly in terms of numbers, not graphemes. Which means that people who designed computers had to come up with a system of converting graphemes into numbers, so the computer would be able to handle them.

How many graphemes are there? Well, don't forget that not only are there number graphemes and letter graphemes, there are also punctuation graphemes, mathematical operation graphemes, and special symbols like the pipe character, the tilde, etc.

So somewhere along the way, someone (or a committee of someones) had to develop a conversion chart so that whenever someone typed a character on the keyboard, there was a *number* that corresponded to whatever they typed. For example:

33 is the number that represents the exclamation mark grapheme

34 is the number that represents the quotation mark grapheme

43 is the number that represents the addition grapheme

65 is the number that represents the upper case "A" grapheme

66 is the number that represents the upper case "B" grapheme

97 is the number that represents the lower case "a" grapheme

Uh oh...now we have a problem! The grapheme "A" is a representation for the number 10, but 65 is the *computer's* value for the "A" grapheme. And this is the source of the confusion. Consider the following:

BAD = a hexadecimal number

BAD = of poor quality or low standard

Both of these statements are true. BAD is a hexadecimal *number*, but it's also a *word!* In one case, B, A, and D are graphemes for numbers, and in the other case, they are graphemes for letters! If you wanted to convert BAD (the number) into base ten, you would do this: 11 x 16^{2} + 10 x 16 + 4 = 2980. But if you wanted to convert BAD (the text) into numbers, so the computer could deal with it, you would do this: B = 66, A = 65, D = 68.

So if you're dealing with numbers (as in your textbook), A = 10, B = 11, etc. If you're dealing with text (words), A = 65, B = 66, etc.

If only those lazy mathematicians had invented their own graphemes instead of just stealing from the alphabet, we wouldn't have this confusion!

Thanks for asking, and I hope there was something helpful to you in this lengthy explanation!

Professor Puzzler

An anonymous reader asks, "How do you get good at 'listening' for iambic pentameters?"

Poetry, like music, has rhythm. In music, the rhythm is established using a few different intstruments. In a band, the drummer and the bass guitarist are the primary rhythm-holders. Those deep, bassy sounds like the guitar and the kick (bass) drum set our subwoofers to vibrating, and mimic our own heart rhythms. In some music, the rhythm is prominently defined by the kick drum and the snare drum. A GEICO ad from awhile back made this point with a pig singing "Boots and pants and boots and pants and..."

The words "boots" and "pants" represent the kick drum*, and the word "and" represents the snare. The kick drum provides the primary driving beat, and the snare represents what we call the "off-beat".

So the pig is actually giving you the beat of a song.

You can feel the rhythm of that if you repeat it over and over, and if you do it just right, it's almost indistinguishable from a techno band.

The rhythm of a poem (in poetry, we call it the "meter" of the poem) can be understood in terms of kick drums and snares, which means it can also be understood in terms of phrases like "boots-and-pants-and". In poetry, we have stressed syllables and unstressed syllables. Stressed syllables correspond to the driving "kick" beat, and the unstressed syllables correspond to the off-beat of the snare. So what does iambic pentameter look like?

and-BOOTS-and-PANTS-and-BOOTS-and-PANTS-and-BOOTS.

Notice that we start with the snare, or the off-beat, and we end with the kick. Also notice that we have 5 kick beats. This is why it's called penta-meter; "penta" is five and "meter" is rhythm, so pentameter is a five-rhythm.

The nice thing about having five kicks in a line is that you've got five fingers on one hand, so while you are saying "and boots and pants..." each time you say either "boots" or "pants" you can tap a finger on the table, and when you've tapped all five fingers, you've completed a line of the rhythm. So if you want to get good at listening for iambic pentameter, do this repeatedly. Say the "and-boots" pattern while tapping out the kick beat on your fingers. If you want, after you do the fifth finger, give a little pause, and then start all over again:

and-BOOTS-and-PANTS-and-BOOTS-and-PANTS-and-BOOTS [pause] and-BOOTS-and-PANTS-and-BOOTS-and-PANTS-and-BOOTS [pause] etc.

After awhile, that rhythm will get ingrained in you, and you'll be doing it without even thinking about it. Now let's take a look at something that's written in iambic pentameter:

The quality of mercy is not strain'd,

It droppeth as the gentle rain from heav'n

Upon the place beneath: it is twice blest;

It blesseth him that gives and him that takes. (Shakespeare, *The Merchant of Venice*)

If you've got that "boots-and-pants" thing going through your mind, it'll be easier to see that this is iambic pentameter:

the QUAL i TY of MER cy IS not STRAIN'D

it DROPP eth AS the GEN tle RAIN from HEAV'N

u PON the PLACE be NEATH: it IS twice BLEST;

it BLESS eth HIM that GIVES and HIM that TAKES.

Can you feel the rhythm of it?

## Important Disclaimers

- Thinking "and-boots-and-pants" may be a great way to recognize and understand the rhythm of poetry, but it is NOT a good way to
*read*poetry; nobody wants to hear Shakespeare sounding like techno music**. Instead, you should use boots-and-pants to help you recognize where the beats are, but then once you've understood which syllables get the stress, you're going to read this in your normal reading voice, without giving unnecessary*extra*accent to any syllables. - Also, even though when you practiced the rhythm, you left a pause between lines, you won't necessarily do that when reading a poem. Focus on the meaning of the lines; if a line does not complete a thought, you're not going to pause at the end of it. And if a thought is completed
*within*a line, you're going to pause there. Remember that meaning is most important, and rhythm serves the poem, not the other way around! So with this in mind, the Shakespeare piece would be read like this:

The quality of mercy is not strain'd, [pause]

It droppeth as the gentle rain from heav'n upon the place beneath; [pause]

It is twice blest; [pause]

It blesseth him that gives and him that takes.

## Other Meters

The "boots-and-pants" idea can be used to recognize other meters as well. For example, if you want to get the feel for anapestic tetrameter, you should start thinking:

and-the-BOOTS-and-the-PANTS-and-the-BOOTS-and-the-PANTS.

Notice that we've added the word "the" after the word "and" each time. This gives the meter a "triplet" feel. Also note that we only hit four driving beats in a line instead of five. So if you're counting this on your fingers, leave out your thumb.

You can see this rhythm in the well known Christmas poem "Twas the Night Before Christmas":

twas the NIGHT be fore CHRIST mas and ALL through the HOUSE

not a CREA ture was STIRR ing not E ven a MOUSE

* Technically, "boo" and "pan" represent the kick, and the "ts" represents the hi-hat. Thus, we really break it down like this: BOO-ts-and-PAN-ts-and.

** Well, okay, maybe "techno-shakespeare" should be a thing. But I'm not sure your poetry teacher will appreciate it!

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