What is the sum of an infinitely decreasing geometric progression? Formula for the sum of the first n terms of GP

Let's consider a certain series.

7 28 112 448 1792...

It is absolutely clear that the value of any of its elements is exactly four times greater than the previous one. This means that this series is a progression.

A geometric progression is an infinite sequence of numbers. main feature which is that the next number is obtained from the previous one by multiplying by some specific number. This is expressed by the following formula.

a z +1 =a z ·q, where z is the number of the selected element.

Accordingly, z ∈ N.

The period when geometric progression is studied at school is 9th grade. Examples will help you understand the concept:

0.25 0.125 0.0625...

Based on this formula, the denominator of the progression can be found as follows:

Neither q nor b z can be zero. Also, each of the elements of the progression should not be equal to zero.

Accordingly, to find out the next number in a series, you need to multiply the last one by q.

To set this progression, you must specify its first element and denominator. After this, it is possible to find any of the subsequent terms and their sum.

Varieties

Depending on q and a 1, this progression is divided into several types:

• If both a 1 and q are greater than one, then such a sequence is a geometric progression increasing with each subsequent element. An example of this is presented below.

Example: a 1 =3, q=2 - both parameters are greater than one.

Then the number sequence can be written like this:

3 6 12 24 48 ...

• If |q| is less than one, that is, multiplication by it is equivalent to division, then a progression with similar conditions is a decreasing geometric progression. An example of this is presented below.

Example: a 1 =6, q=1/3 - a 1 is greater than one, q is less.

Then the number sequence can be written as follows:

6 2 2/3 ... - any element is 3 times larger than the element following it.

• Alternating sign. If q<0, то знаки у чисел последовательности постоянно чередуются вне зависимости от a 1 , а элементы ни возрастают, ни убывают.

Example: a 1 = -3, q = -2 - both parameters are less than zero.

Then the number sequence can be written like this:

3, 6, -12, 24,...

Formulas

There are many formulas for convenient use of geometric progressions:

• Z-term formula. Allows you to calculate an element under a specific number without calculating previous numbers.

Example:q = 3, a 1 = 4. It is required to count the fourth element of the progression.

Solution:a 4 = 4 · 3 4-1 = 4 · 3 3 = 4 · 27 = 108.

• The sum of the first elements whose quantity is equal to z. Allows you to calculate the sum of all elements of a sequence up toa zinclusive.

Since (1-q) is in the denominator, then (1 - q)≠ 0, therefore q is not equal to 1.

Note: if q=1, then the progression would be a series of infinitely repeating numbers.

Sum of geometric progression, examples:a 1 = 2, q= -2. Calculate S5.

Solution:S 5 = 22 - calculation using the formula.

• Amount if |q| < 1 и если z стремится к бесконечности.

Example:a 1 = 2 , q= 0.5. Find the amount.

Solution:Sz = 2 · = 4

Sz = 2 + 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 3.9375 4

Some properties:

• Characteristic property. If the following condition works for anyz, then the given number series is a geometric progression:

a z 2 = a z -1 · az+1

• Also, the square of any number in a geometric progression is found by adding the squares of any two other numbers in a given series, if they are equidistant from this element.

a z 2 = a z - t 2 + a z + t 2 , Wheret- the distance between these numbers.

• Elementsdiffer in qonce.
• The logarithms of the elements of a progression also form a progression, but an arithmetic one, that is, each of them is greater than the previous one by a certain number.

Examples of some classic problems

To better understand what a geometric progression is, examples with solutions for class 9 can help.

• Conditions:a 1 = 3, a 3 = 48. Findq.

Solution: each subsequent element is greater than the previous one inq once.It is necessary to express some elements in terms of others using a denominator.

Hence,a 3 = q 2 · a 1

When substitutingq= 4

• Conditions:a 2 = 6, a 3 = 12. Calculate S 6.

Solution:To do this, just find q, the first element and substitute it into the formula.

a 3 = q· a 2 , hence,q= 2

a 2 = q · a 1 ,That's why a 1 = 3

S 6 = 189

• · a 1 = 10, q= -2. Find the fourth element of the progression.

Solution: to do this, it is enough to express the fourth element through the first and through the denominator.

a 4 = q 3· a 1 = -80

Application example:

• A bank client made a deposit in the amount of 10,000 rubles, under the terms of which every year the client will have 6% of it added to the principal amount. How much money will be in the account after 4 years?

Solution: The initial amount is 10 thousand rubles. This means that a year after the investment the account will have an amount equal to 10,000 + 10,000 · 0.06 = 10000 1.06

Accordingly, the amount in the account after another year will be expressed as follows:

(10000 · 1.06) · 0.06 + 10000 · 1.06 = 1.06 · 1.06 · 10000

That is, every year the amount increases by 1.06 times. This means that to find the amount of funds in the account after 4 years, it is enough to find the fourth element of the progression, which is given by the first element equal to 10 thousand and the denominator equal to 1.06.

S = 1.06 1.06 1.06 1.06 10000 = 12625

Examples of sum calculation problems:

Geometric progression is used in various problems. An example for finding the sum can be given as follows:

a 1 = 4, q= 2, calculateS 5.

Solution: all the data necessary for the calculation is known, you just need to substitute them into the formula.

S 5 = 124

• a 2 = 6, a 3 = 18. Calculate the sum of the first six elements.

Solution:

In geom. progression, each next element is q times greater than the previous one, that is, to calculate the sum you need to know the elementa 1 and denominatorq.

a 2 · q = a 3

q = 3

Similarly, you need to finda 1 , knowinga 2 Andq.

a 1 · q = a 2

a 1 =2

S 6 = 728.

For example, sequence \(3\); \(6\); \(12\); \(24\); \(48\)... is a geometric progression, because each next element differs from the previous one by a factor of two (in other words, it can be obtained from the previous one by multiplying it by two):

Like any sequence, a geometric progression is denoted by a small Latin letter. Numbers that form a progression are called members(or elements). They are denoted by the same letter as the geometric progression, but with a numerical index equal to the number of the element in order.

For example, the geometric progression \(b_n = \(3; 6; 12; 24; 48…\)\) consists of the elements \(b_1=3\); \(b_2=6\); \(b_3=12\) and so on. In other words:

If you understand the above information, you will already be able to solve most of the problems on this topic.

Example (OGE):
Solution:

Answer : \(-686\).

Example (OGE): The first three terms of the progression \(324\) are given; \(-108\); \(36\)…. Find \(b_5\).
Solution:

	To continue the sequence, we need to know the denominator. Let's find it from two neighboring elements: what do we need to multiply \(324\) by to get \(-108\)?
\(324·q=-108\)	From here we can easily calculate the denominator.
\(q=-\) \(\frac(108)(324)\) \(=-\) \(\frac(1)(3)\)	Now we can easily find the element we need.
	The answer is ready.

Answer : \(4\).

Example: The progression is specified by the condition \(b_n=0.8·5^n\). Which number is a member of this progression:

a) \(-5\) b) \(100\) c) \(25\) d) \(0.8\) ?

Solution: From the wording of the task it is obvious that one of these numbers is definitely in our progression. Therefore, we can simply calculate its terms one by one until we find the value we need. Since our progression is given by the formula, we calculate the values ​​of the elements by substituting different \(n\):
\(n=1\); \(b_1=0.8·5^1=0.8·5=4\) – there is no such number in the list. Let's continue.
\(n=2\); \(b_2=0.8·5^2=0.8·25=20\) - and this is not there either.
\(n=3\); \(b_3=0.8·5^3=0.8·125=100\) – and here is our champion!

Answer: \(100\).

Example (OGE): Given are several consecutive terms of the geometric progression...\(8\); \(x\); \(50\); \(-125\)…. Find the value of the element labeled \(x\).

Solution:

Answer: \(-20\).

Example (OGE): The progression is specified by the conditions \(b_1=7\), \(b_(n+1)=2b_n\). Find the sum of the first \(4\) terms of this progression.

Solution:

Answer: \(105\).

Example (OGE): It is known that in geometric progression \(b_6=-11\), \(b_9=704\). Find the denominator of \(q\).

Solution:

	From the diagram on the left you can see that in order to “get” from \(b_6\) to \(b_9\) we take three “steps”, that is, we multiply \(b_6\) three times by the denominator of the progression. In other words, \(b_9=b_6·q·q·q=b_6·q^3\).
\(b_9=b_6·q^3\)	Let's substitute the values ​​we know.
\(704=(-11)q^3\)	Let's turn the equation around and divide it by \((-11)\).
\(q^3=\) \(\frac(704)(-11)\) \(\:\:\: ⇔ \:\:\: \)\(q^3=-\) \(64 \)	What number cubed gives \(-64\)? Of course, \(-4\)!
	The answer has been found. It can be checked by restoring the chain of numbers from \(-11\) to \(704\).
	Everything came together - the answer is correct.

Answer: \(-4\).

The most important formulas

As you can see, most geometric progression problems can be solved using pure logic, simply by understanding the essence (this is generally typical for mathematics). But sometimes knowledge of certain formulas and patterns speeds up and significantly facilitates the solution. We will study two such formulas.

Formula of the \(n\)th term: \(b_n=b_1·q^(n-1)\), where \(b_1\) is the first term of the progression; \(n\) – number of the required element; \(q\) – progression denominator; \(b_n\) – term of the progression with number \(n\).

Using this formula, you can, for example, solve the problem from the very first example literally in one action.

Example (OGE): Geometric progression specified by the conditions \(b_1=-2\); \(q=7\). Find \(b_4\).
Solution:

Answer: \(-686\).

This example was simple, so the formula did not make the calculations too much easier for us. Let's look at the problem a little more complicated.

Example: The geometric progression is specified by the conditions \(b_1=20480\); \(q=\frac(1)(2)\). Find \(b_(12)\).
Solution:

Answer: \(10\).

Of course, raising \(\frac(1)(2)\) to the \(11\)th power is not very joyful, but it’s still easier than \(11\) dividing \(20480\) by two.

Sum \(n\) of the first terms: \(S_n=\)\(\frac(b_1·(q^n-1))(q-1)\) , where \(b_1\) is the first term of the progression; \(n\) – number of summed elements; \(q\) – progression denominator; \(S_n\) – the sum of \(n\) first terms of the progression.

Example (OGE): Given a geometric progression \(b_n\), the denominator of which is \(5\), and the first term is \(b_1=\frac(2)(5)\). Find the sum of the first six terms of this progression.
Solution:

Answer: \(1562,4\).

And again, we could solve the problem head-on - find all six elements in turn, and then add the results. However, the number of calculations, and hence the chance of random error, would increase sharply.

For geometric progression, there are several more formulas that we did not consider here because of their low practical use. You can find these formulas.

Increasing and decreasing geometric progressions

For the progression \(b_n = \(3; 6; 12; 24; 48...\)\) considered at the very beginning of the article, the denominator \(q\) is greater than one and therefore each next term is greater than the previous one. Such progressions are called increasing.

If \(q\) is less than one, but is positive (that is, lies in the range from zero to one), then each next element will be less than the previous one. For example, in the progression \(4\); \(2\); \(1\); \(0.5\); \(0.25\)... the denominator of \(q\) is equal to \(\frac(1)(2)\).

These progressions are called decreasing. Note that none of the elements of such a progression will be negative, they just get smaller and smaller with each step. That is, we will gradually approach zero, but we will never reach it and will not go beyond it. In such cases, mathematicians say “tend to zero.”

Note that with a negative denominator, the elements of the geometric progression will necessarily change sign. For example, y progression \(5\); \(-15\); \(45\); \(-135\); \(675\)... the denominator of \(q\) is \(-3\), and because of this, the signs of the elements “blink”.

The formula for the nth term of a geometric progression is very simple. Both in meaning and in general appearance. But there are all kinds of problems on the formula of the nth term - from very primitive to quite serious. And in the process of our acquaintance, we will definitely consider both. Well, let's get acquainted?)

So, to begin with, actually formulan

Here she is:

b n = b 1 · qn -1

The formula is just a formula, nothing supernatural. It looks even simpler and more compact than a similar formula for. The meaning of the formula is also as simple as felt boots.

This formula allows you to find ANY member of a geometric progression BY ITS NUMBER " n".

As you can see, the meaning is complete analogy with an arithmetic progression. We know the number n - we can also count the term under this number. Whichever one we want. Without repeatedly multiplying by "q" many, many times. That's the whole point.)

I understand that at this level of working with progressions, all the quantities included in the formula should already be clear to you, but I still consider it my duty to decipher each one. Just in case.

So, here we go:

b 1 – first term of geometric progression;

q – ;

n– member number;

b n – nth (nth) term of a geometric progression.

This formula connects the four main parameters of any geometric progression - bn, b 1 , q And n. And all the progression problems revolve around these four key figures.

“How is it removed?”– I hear a curious question... Elementary! Look!

What is equal to second member of the progression? No problem! We write directly:

b 2 = b 1 ·q

What about the third member? Not a problem either! We multiply the second term once again onq.

Like this:

B 3 = b 2 q

Let us now remember that the second term, in turn, is equal to b 1 ·q and substitute this expression into our equality:

B 3 = b 2 q = (b 1 q) q = b 1 q q = b 1 q 2

We get:

B 3 = b 1 ·q 2

Now let’s read our entry in Russian: third term is equal to the first term multiplied by q in second degrees. Do you get it? Not yet? Okay, one more step.

What is the fourth term? All the same! Multiply previous(i.e. the third term) on q:

B 4 = b 3 q = (b 1 q 2) q = b 1 q 2 q = b 1 q 3

Total:

B 4 = b 1 ·q 3

And again we translate into Russian: fourth term is equal to the first term multiplied by q in third degrees.

And so on. So how is it? Did you catch the pattern? Yes! For any term with any number, the number of identical factors q (i.e., the degree of the denominator) will always be one less than the number of the desired membern.

Therefore, our formula will be, without options:

b n =b 1 · qn -1

That's all.)

Well, let's solve the problems, I guess?)

Solving formula problemsnth term of a geometric progression.

Let's start, as usual, with the direct application of the formula. Here's a typical problem:

In geometric progression, it is known that b 1 = 512 and q = -1/2. Find the tenth term of the progression.

Of course, this problem can be solved without any formulas at all. Directly in the sense of geometric progression. But we need to warm up with the formula for the nth term, right? Here we are warming up.

Our data for applying the formula is as follows.

The first member is known. This is 512.

b 1 = 512.

The denominator of the progression is also known: q = -1/2.

All that remains is to figure out what the number of member n is. No problem! Are we interested in the tenth term? So we substitute ten instead of n into the general formula.

And carefully calculate the arithmetic:

Answer: -1

As you can see, the tenth term of the progression turned out to be minus. Nothing surprising: our progression denominator is -1/2, i.e. negative number. And this tells us that the signs of our progression alternate, yes.)

Everything is simple here. Here is a similar problem, but a little more complicated in terms of calculations.

In geometric progression, it is known that:

b 1 = 3

Find the thirteenth term of the progression.

Everything is the same, only this time the denominator of the progression is irrational. Root of two. Well, that's okay. The formula is a universal thing; it can handle any numbers.

We work directly according to the formula:

The formula, of course, worked as it should, but... this is where some people get stuck. What to do next with the root? How to raise a root to the twelfth power?

How-how... You must understand that any formula, of course, is a good thing, but knowledge of all previous mathematics is not canceled! How to build? Yes, remember the properties of degrees! Let's turn the root into fractional degree and – according to the formula for raising a degree to a degree.

Like this:

Answer: 192

And that's all.)

What is the main difficulty in directly applying the nth term formula? Yes! The main difficulty is working with degrees! Namely, raising negative numbers, fractions, roots and similar constructions to powers. So those who have problems with this, please repeat the degrees and their properties! Otherwise, you will slow down this topic too, yes...)

Now let’s solve typical search problems one of the elements of the formula, if all others are given. To successfully solve such problems, the recipe is uniform and terribly simple - write the formulan-th member in general! Right in the notebook next to the condition. And then from the condition we figure out what is given to us and what is missing. And we express the desired value from the formula. All!

For example, such a harmless problem.

The fifth term of a geometric progression with denominator 3 is 567. Find the first term of this progression.

Nothing complicated. We work directly according to the spell.

Let's write the formula for the nth term!

b n = b 1 · qn -1

What have we been given? First, the denominator of the progression is given: q = 3.

Moreover, we are given fifth member: b 5 = 567 .

All? No! We have also been given number n! This is five: n = 5.

I hope you already understand what is in the recording b 5 = 567 two parameters are hidden at once - this is the fifth term itself (567) and its number (5). I already talked about this in a similar lesson, but I think it’s worth mentioning here too.)

Now we substitute our data into the formula:

567 = b 1 ·3 5-1

We do the arithmetic, simplify and get a simple linear equation:

81 b 1 = 567

We solve and get:

b 1 = 7

As you can see, there are no problems with finding the first term. But when searching for the denominator q and numbers n There may also be surprises. And you also need to be prepared for them (surprises), yes.)

For example, this problem:

The fifth term of a geometric progression with a positive denominator is 162, and the first term of this progression is 2. Find the denominator of the progression.

This time we are given the first and fifth terms, and are asked to find the denominator of the progression. Here we go.

We write the formulanth member!

b n = b 1 · qn -1

Our initial data will be as follows:

b 5 = 162

b 1 = 2

n = 5

Missing value q. No problem! Let’s find it now.) We substitute everything we know into the formula.

We get:

162 = 2q 5-1

2 q 4 = 162

q 4 = 81

A simple equation of the fourth degree. And now - carefully! At this stage of the solution, many students immediately joyfully extract the root (of the fourth degree) and get the answer q=3 .

Like this:

q4 = 81

q = 3

But actually, this is an unfinished answer. More precisely, incomplete. Why? The point is that the answer q = -3 also suitable: (-3) 4 will also be 81!

This is because the power equation x n = a always has two opposite roots at evenn . With plus and minus:

Both are suitable.

For example, when deciding (i.e. second degrees)

x 2 = 9

For some reason you are not surprised by the appearance two roots x=±3? It's the same here. And with any other even degree (fourth, sixth, tenth, etc.) will be the same. Details are in the topic about

Therefore, the correct solution would be:

q 4 = 81

q= ±3

Okay, we've sorted out the signs. Which one is correct - plus or minus? Well, let’s read the problem statement again in search of additional information. Of course, it may not exist, but in this problem such information available. Our condition states in plain text that a progression is given with positive denominator.

Therefore the answer is obvious:

q = 3

Everything is simple here. What do you think would happen if the problem statement were like this:

The fifth term of a geometric progression is 162, and the first term of this progression is 2. Find the denominator of the progression.

What is the difference? Yes! In condition Nothing no mention is made of the sign of the denominator. Neither directly nor indirectly. And here the problem would already have two solutions!

q = 3 And q = -3

Yes Yes! Both with a plus and with a minus.) Mathematically, this fact would mean that there are two progressions, which fit the conditions of the problem. And each has its own denominator. Just for fun, practice and write out the first five terms of each.)

Now let’s practice finding the member’s number. This problem is the most difficult, yes. But also more creative.)

Given a geometric progression:

3; 6; 12; 24; …

What number in this progression is the number 768?

The first step is still the same: write the formulanth member!

b n = b 1 · qn -1

And now, as usual, we substitute the data we know into it. Hm... it doesn't work! Where is the first term, where is the denominator, where is everything else?!

Where, where... Why do we need eyes? Flapping your eyelashes? This time the progression is given to us directly in the form sequences. Can we see the first member? We see! This is a triple (b 1 = 3). What about the denominator? We don’t see it yet, but it’s very easy to count. If, of course, you understand...

So we count. Directly according to the meaning of a geometric progression: we take any of its terms (except the first) and divide by the previous one.

At least like this:

q = 24/12 = 2

What else do we know? We also know some term of this progression, equal to 768. Under some number n:

b n = 768

We don’t know his number, but our task is precisely to find him.) So we are looking. We have already downloaded all the necessary data for substitution into the formula. Unbeknownst to yourself.)

Here we substitute:

768 = 3 2n -1

Let's do the elementary ones - divide both sides by three and rewrite the equation in the usual form: the unknown is on the left, the known is on the right.

We get:

2 n -1 = 256

This is an interesting equation. We need to find "n". What, unusual? Yes, I don't argue. Actually, this is the simplest thing. It is so called because the unknown (in this case it is the number n) costs in indicator degrees.

At the stage of learning about geometric progression (this is ninth grade), they don’t teach you how to solve exponential equations, yes... This is a topic for high school. But there's nothing scary. Even if you don’t know how such equations are solved, let’s try to find our n, guided by simple logic and common sense.

Let's start talking. On the left we have a deuce to a certain degree. We don’t yet know what exactly this degree is, but that’s not scary. But we know for sure that this degree is equal to 256! So we remember to what extent a two gives us 256. Do you remember? Yes! IN eighth degrees!

256 = 2 8

If you don’t remember or have problems recognizing the degrees, then that’s okay too: just successively square two, cube, fourth, fifth, and so on. Selection, in fact, but at this level will work quite well.

One way or another, we get:

2 n -1 = 2 8

n-1 = 8

n = 9

So 768 is ninth member of our progression. That's it, problem solved.)

Answer: 9

What? Boring? Tired of elementary stuff? Agree. And me too. Let's move to the next level.)

More complex tasks.

Now let’s solve more challenging problems. Not exactly super cool, but ones that require a little work to get to the answer.

For example, this one.

Find the second term of a geometric progression if its fourth term is -24 and its seventh term is 192.

This is a classic of the genre. Some two different terms of the progression are known, but another term needs to be found. Moreover, all members are NOT neighboring. Which is confusing at first, yes...

As in, to solve such problems we will consider two methods. The first method is universal. Algebraic. Works flawlessly with any source data. So that’s where we’ll start.)

We describe each term according to the formula nth member!

Everything is exactly the same as with an arithmetic progression. Only this time we are working with another general formula. That's all.) But the essence is the same: we take and one by one We substitute our initial data into the formula for the nth term. For each member - their own.

For the fourth term we write:

b 4 = b 1 · q 3

-24 = b 1 · q 3

Eat. One equation is ready.

For the seventh term we write:

b 7 = b 1 · q 6

192 = b 1 · q 6

In total, we got two equations for the same progression .

We assemble a system from them:

Despite its menacing appearance, the system is quite simple. The most obvious solution is simple substitution. We express b 1 from the upper equation and substitute it into the lower one:

After fiddling around with the bottom equation a bit (reducing the powers and dividing by -24), we get:

q 3 = -8

By the way, this same equation can be arrived at in a simpler way! Which one? Now I will show you another secret, but very beautiful, powerful and useful way to solve such systems. Such systems, the equations of which include only works. At least in one. Called division method one equation to another.

So, we have a system before us:

In both equations on the left - work, and on the right is just a number. This is a very good sign.) Let's take it and... divide, say, the lower equation by the upper one! What means, let's divide one equation by another? Very simple. Let's take it left side one equation (lower) and divide her on left side another equation (upper). The right side is similar: right side one equation divide on right side another.

The whole division process looks like this:

Now, reducing everything that can be reduced, we get:

q 3 = -8

What's good about this method? Yes, because in the process of such division everything bad and inconvenient can be safely reduced and a completely harmless equation remains! This is why it is so important to have multiplication only in at least one of the equations of the system. There is no multiplication - there is nothing to reduce, yes...

In general, this method (like many other non-trivial methods of solving systems) even deserves a separate lesson. I'll definitely look into it in more detail. Some day…

However, it doesn’t matter how exactly you solve the system, in any case, now we need to solve the resulting equation:

q 3 = -8

No problem: extract the cube root and you’re done!

Please note that there is no need to put a plus/minus here when extracting. Our root is of odd (third) degree. And the answer is also the same, yes.)

So, the denominator of the progression has been found. Minus two. Great! The process is ongoing.)

For the first term (say, from the upper equation) we get:

Great! We know the first term, we know the denominator. And now we have the opportunity to find any member of the progression. Including the second one.)

For the second term everything is quite simple:

b 2 = b 1 · q= 3·(-2) = -6

Answer: -6

So, we have broken down the algebraic method for solving the problem. Difficult? Not really, I agree. Long and tedious? Yes, definitely. But sometimes you can significantly reduce the amount of work. For this there is graphic method. Good old and familiar to us.)

Let's draw a problem!

Yes! Exactly. Again we depict our progression on the number axis. It’s not necessary to follow a ruler, it’s not necessary to maintain equal intervals between the terms (which, by the way, will not be the same, since the progression is geometric!), but simply schematically Let's draw our sequence.

I got it like this:

Now look at the picture and figure it out. How many identical factors "q" separate fourth And seventh members? That's right, three!

Therefore, we have every right to write:

-24·q 3 = 192

From here it is now easy to find q:

q 3 = -8

q = -2

That’s great, we already have the denominator in our pocket. Now let’s look at the picture again: how many such denominators sit between second And fourth members? Two! Therefore, to record the connection between these terms, we will construct the denominator squared.

So we write:

b 2 · q 2 = -24 , where b 2 = -24/ q 2

We substitute our found denominator into the expression for b 2, count and get:

Answer: -6

As you can see, everything is much simpler and faster than through the system. Moreover, here we didn’t even need to count the first term at all! At all.)

Here is such a simple and visual way-light. But it also has a serious drawback. Did you guess it? Yes! It is only good for very short pieces of progression. Those where the distances between the members of interest to us are not very large. But in all other cases it’s already difficult to draw a picture, yes... Then we solve the problem analytically, through the system.) And systems are universal things. They can handle any numbers.

Another epic challenge:

The second term of the geometric progression is 10 more than the first, and the third term is 30 more than the second. Find the denominator of the progression.

What, cool? Not at all! All the same. Again we translate the problem statement into pure algebra.

1) We describe each term according to the formula nth member!

Second term: b 2 = b 1 q

Third term: b 3 = b 1 q 2

2) We write down the connection between the members from the problem statement.

We read the condition: "The second term of the geometric progression is 10 greater than the first." Stop, this is valuable!

So we write:

b 2 = b 1 +10

And we translate this phrase into pure mathematics:

b 3 = b 2 +30

We got two equations. Let's combine them into a system:

The system looks simple. But there are too many different indices for the letters. Let's substitute instead of the second and third terms their expressions through the first term and the denominator! Was it in vain that we painted them?

We get:

But such a system is no longer a gift, yes... How to solve this? Unfortunately, there is no universal secret spell for solving complex nonlinear There are no systems in mathematics and there cannot be. It is fantastic! But the first thing that should come to your mind when trying to crack such a tough nut is to figure out But isn’t one of the equations of the system reduced to a beautiful form that allows, for example, to easily express one of the variables in terms of another?

Let's figure it out. The first equation of the system is clearly simpler than the second. We'll torture him.) Shouldn't we try from the first equation something express through something? Since we want to find the denominator q, then it would be most advantageous for us to express b 1 through q.

So let’s try to do this procedure with the first equation, using the good old ones:

b 1 q = b 1 +10

b 1 q – b 1 = 10

b 1 (q-1) = 10

All! So we expressed unnecessary give us the variable (b 1) through necessary(q). Yes, it’s not the simplest expression we got. Some kind of fraction... But our system is of a decent level, yes.)

Typical. We know what to do.

We write ODZ (Necessarily!) :

q ≠ 1

We multiply everything by the denominator (q-1) and cancel all fractions:

10 q 2 = 10 q + 30(q-1)

We divide everything by ten, open the brackets, and collect everything from the left:

q 2 – 4 q + 3 = 0

We solve the result and get two roots:

q 1 = 1

q 2 = 3

There is only one final answer: q = 3 .

Answer: 3

As you can see, the path to solving most problems involving the formula of the nth term of a geometric progression is always the same: read attentively condition of the problem and using the formula of the nth term we translate all useful information into pure algebra.

Namely:

1) We describe separately each term given in the problem according to the formulanth member.

2) From the conditions of the problem we translate the connection between the members into mathematical form. We compose an equation or system of equations.

3) We solve the resulting equation or system of equations, find the unknown parameters of the progression.

4) In case of an ambiguous answer, carefully read the problem statement in search of additional information (if any). We also check the received response with the terms of the DL (if any).

Now let’s list the main problems that most often lead to errors in the process of solving geometric progression problems.

1. Elementary arithmetic. Operations with fractions and negative numbers.

2. If there are problems with at least one of these three points, then you will inevitably make mistakes in this topic. Unfortunately... So don't be lazy and repeat what was mentioned above. And follow the links - go. Sometimes it helps.)

Modified and recurrent formulas.

Now let’s look at a couple of typical exam problems with a less familiar presentation of the condition. Yes, yes, you guessed it! This modified And recurrent nth term formulas. We have already encountered such formulas and worked on arithmetic progression. Everything is similar here. The essence is the same.

For example, this problem from the OGE:

The geometric progression is given by the formula b n = 3 2 n . Find the sum of its first and fourth terms.

This time the progression is not quite as usual for us. In the form of some kind of formula. So what? This formula is also a formulanth member! You and I know that the formula for the nth term can be written both in general form, using letters, and for specific progression. WITH specific first term and denominator.

In our case, we are, in fact, given a general term formula for a geometric progression with the following parameters:

b 1 = 6

q = 2

Let's check?) Let's write down the formula for the nth term in general form and substitute it into b 1 And q. We get:

b n = b 1 · qn -1

b n= 6 2n -1

We simplify using factorization and properties of powers, and we get:

b n= 6 2n -1 = 3·2·2n -1 = 3 2n -1+1 = 3 2n

As you can see, everything is fair. But our goal is not to demonstrate the derivation of a specific formula. This is so, a lyrical digression. Purely for understanding.) Our goal is to solve the problem according to the formula given to us in the condition. Do you get it?) So we work with the modified formula directly.

We count the first term. Let's substitute n=1 into the general formula:

b 1 = 3 2 1 = 3 2 = 6

Like this. By the way, I won’t be lazy and once again draw your attention to a typical mistake with the calculation of the first term. DO NOT, looking at the formula b n= 3 2n, immediately rush to write that the first term is a three! This is a gross mistake, yes...)

Let's continue. Let's substitute n=4 and count the fourth term:

b 4 = 3 2 4 = 3 16 = 48

And finally, we calculate the required amount:

b 1 + b 4 = 6+48 = 54

Answer: 54

Another problem.

The geometric progression is specified by the conditions:

b 1 = -7;

b n +1 = 3 b n

Find the fourth term of the progression.

Here the progression is given by a recurrent formula. Well, okay.) How to work with this formula – we know too.

So we act. Step by step.

1) Count two consecutive member of the progression.

The first term has already been given to us. Minus seven. But the next, second term, can be easily calculated using the recurrence formula. If you understand the principle of its operation, of course.)

So we count the second term according to the well-known first:

b 2 = 3 b 1 = 3·(-7) = -21

2) Calculate the denominator of the progression

No problem either. Straight, let's divide second dick on first.

We get:

q = -21/(-7) = 3

3) Write the formulanth member in the usual form and calculate the required member.

So, we know the first term, and so do the denominator. So we write:

b n= -7·3n -1

b 4 = -7·3 3 = -7·27 = -189

Answer: -189

As you can see, working with such formulas for a geometric progression is essentially no different from that for an arithmetic progression. It is only important to understand the general essence and meaning of these formulas. Well, you also need to understand the meaning of geometric progression, yes.) And then there will be no stupid mistakes.

Well, let's decide on our own?)

Very basic tasks for warming up:

1. Given a geometric progression in which b 1 = 243, a q = -2/3. Find the sixth term of the progression.

2. The general term of the geometric progression is given by the formula b n = 5∙2 n +1 . Find the number of the last three-digit term of this progression.

3. Geometric progression is given by the conditions:

b 1 = -3;

b n +1 = 6 b n

Find the fifth term of the progression.

A little more complicated:

4. Given a geometric progression:

b 1 =2048; q =-0,5

What is the sixth negative term equal to?

What seems super difficult? Not at all. Logic and understanding of the meaning of geometric progression will save you. Well, the formula for the nth term, of course.

5. The third term of the geometric progression is -14, and the eighth term is 112. Find the denominator of the progression.

6. The sum of the first and second terms of the geometric progression is 75, and the sum of the second and third terms is 150. Find the sixth term of the progression.

Answers (in disarray): 6; -3888; -1; 800; -32; 448.

That's almost all. All we have to do is learn to count the sum of the first n terms of a geometric progression yes discover infinitely decreasing geometric progression and its amount. A very interesting and unusual thing, by the way! More on this in the next lessons.)

A geometric progression is a numerical sequence, the first term of which is non-zero, and each subsequent term is equal to the previous term multiplied by the same non-zero number. Geometric progression is denoted b1,b2,b3, …, bn, …

Properties of geometric progression

The ratio of any term of the geometric error to its previous term is equal to the same number, that is, b2/b1 = b3/b2 = b4/b3 = ... = bn/b(n-1) = b(n+1)/bn = … . This follows directly from the definition of an arithmetic progression. This number is called the denominator of a geometric progression. Usually the denominator of a geometric progression is denoted by the letter q.

One of the ways to specify a geometric progression is to specify its first term b1 and the denominator of the geometric error q. For example, b1=4, q=-2. These two conditions define the geometric progression 4, -8, 16, -32, ….

If q>0 (q is not equal to 1), then the progression is a monotonic sequence. For example, the sequence, 2, 4,8,16,32, ... is a monotonically increasing sequence (b1=2, q=2).

If the denominator in the geometric error is q=1, then all terms of the geometric progression will be equal to each other. In such cases, the progression is said to be a constant sequence.

Formula for the nth term of the progression

In order for a number sequence (bn) to be a geometric progression, it is necessary that each of its members, starting from the second, be the geometric mean of neighboring members. That is, it is necessary to fulfill the following equation - (b(n+1))^2 = bn * b(n+2), for any n>0, where n belongs to the set of natural numbers N.

The formula for the nth term of the geometric progression is:

bn=b1*q^(n-1), where n belongs to the set of natural numbers N.

Let's look at a simple example:

In geometric progression b1=6, q=3, n=8 find bn.

Let's use the formula for the nth term of a geometric progression.

Arithmetic and geometric progressions

Theoretical information

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a n is a sequence in which each member, starting from the second, is equal to the previous member added to the same number d (d- progression difference)	Geometric progression b n is a sequence of non-zero numbers, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression)
Recurrence formula	For any natural n a n + 1 = a n + d	For any natural n b n + 1 = b n ∙ q, b n ≠ 0
Formula nth term	a n = a 1 + d (n – 1)	b n = b 1 ∙ q n - 1 , b n ≠ 0
Characteristic property
Sum of the first n terms

Examples of tasks with comments

Exercise 1

In arithmetic progression ( a n) a 1 = -6, a 2

According to the formula of the nth term:

a 22 = a 1+ d (22 - 1) = a 1+ 21 d

By condition:

a 1= -6, then a 22= -6 + 21 d .

It is necessary to find the difference of progressions:

d = a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = - 48.

Answer : a 22 = -48.

Task 2

Find the fifth term of the geometric progression: -3; 6;....

1st method (using the n-term formula)

According to the formula for the nth term of a geometric progression:

b 5 = b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

Because b 1 = -3,

2nd method (using recurrent formula)

Since the denominator of the progression is -2 (q = -2), then:

b 3 = 6 ∙ (-2) = -12;

b 4 = -12 ∙ (-2) = 24;

b 5 = 24 ∙ (-2) = -48.

Answer : b 5 = -48.

Task 3

In arithmetic progression ( a n ) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.

For an arithmetic progression, the characteristic property has the form .

Therefore:

.

Let's substitute the data into the formula:

Answer: 95.

Task 4

In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.

To find the sum of the first n terms of an arithmetic progression, two formulas are used:

.

Which of them is more convenient to use in this case?

By condition, the formula for the nth term of the original progression is known ( a n) a n= 3n - 4. You can find immediately and a 1, And a 16 without finding d. Therefore, we will use the first formula.

Answer: 368.

Task 5

In arithmetic progression( a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.

According to the formula of the nth term:

a 22 = a 1 + d (22 – 1) = a 1+ 21d.

By condition, if a 1= -6, then a 22= -6 + 21d . It is necessary to find the difference of progressions:

d = a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = -48.

Answer : a 22 = -48.

Task 6

Several consecutive terms of the geometric progression are written:

Find the term of the progression indicated by x.

When solving, we will use the formula for the nth term b n = b 1 ∙ q n - 1 for geometric progressions. The first term of the progression. To find the denominator of the progression q, you need to take any of the given terms of the progression and divide by the previous one. In our example, we can take and divide by. We obtain that q = 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.

Substituting the found values ​​into the formula, we get:

.

Answer : .

Task 7

From the arithmetic progressions given by the formula of the nth term, select the one for which the condition is satisfied a 27 > 9:

Since the given condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:

.

Answer: 4.

Task 8

In arithmetic progression a 1= 3, d = -1.5. Specify highest value n for which the inequality holds a n > -6.