Sanctions against the Russian energy sector by the United States can lead to critical consequences - up to the collapse of the European energy system. So says Robert, the head of the British oil and gas company BP.

“I don't think it will happen. If you impose sanctions on Rosneft, or impose sanctions like those applied to Rusal, then you will actually turn off the energy systems of Europe, and this is already a little too much, ”

- said Dudley, speaking at the Oil & Money 2018 conference in London (quoted from).

The provision of debt and equity capital to enterprises from Russia was limited, as well as the supply of equipment for the exploration and production of oil on the shelf at a depth of more than 150 meters and for the development of shale rocks.

In August 2017, the United States tightened financial sanctions, introduced additional bans on the supply of goods and technologies for production, and also legislated the possibility of imposing restrictions on export pipelines. Due to the sanctions, almost all joint projects with foreigners for the development of offshore and shale oil were also suspended.

Experts have repeatedly noted that in the future these restrictions may lead to a decrease in the level of production in the Russian Federation if the country does not pay more attention to geological exploration and the development of its own technologies.

Obviously, if the toughest package of restrictions is adopted in November, the interaction may be complicated, but it is unlikely that it will go into the category of a complete stop,

Zharsky thinks.

If the expectations were different, then the same disturbing news would begin to come from the other interested party, but the oilmen do not stutter about such forecasts, the expert draws attention.

The imposition of tough sanctions is not only a problem for Russia, but also a headache for our foreign counterparties, which include the closest US allies, agrees BCS Premier investment strategist.

According to the analyst, in the event of strengthening sanctions, restrictive measures may rather be selective in nature and are unlikely to be directed to the entire industry.

Russia occupies more than 10% of the world oil market, the abrupt departure of such a major player will mean a rapid growth in oil quotes: potentially this is not only a blow to European, but also to all other oil consumers.

Thus, in September, oil production in Russia amounted to 11.35 million barrels per day (b / d). According to the CDU of the Fuel and Energy Complex of the Ministry of Energy, in January-September 2018, Russia supplied 190.212 million tons of oil to non-CIS countries.

As for the gas market, the situation for the EU is even more serious: Russia accounts for about 34% of all gas supplies to Europe. At the same time, last year Gazprom delivered about 195 billion cubic meters of gas to non-CIS countries (the EU plus Turkey). This year, according to the forecasts of experts and the monopolist himself, this figure will exceed 200 billion cubic meters.

It is very difficult to quickly replace such volumes. Not to mention the fact that economically gas from the Russian Federation is more profitable for European countries than the same liquefied natural gas (LNG).

Earlier I reported that sanctions against Russia cannot be imposed according to the tough scenario of Iran or North Korea, the country is too deeply integrated into the world economy. In November, an embargo on the supply of oil from Iran will be introduced, and the market will lose about 1-2 million barrels. Only the expectation of this brought the quotes to the level of $80-85 per barrel of Brent.

However, the administration does not consider risks, unleashing trade wars with the EU and China. US Secretary of the Interior Ryan Zinke recently said that the US could impose a naval blockade of Russia. So not a single, even the most improbable scenario, can be ruled out.

Among all the sequences of numbers, the geometric progression, which is considered in the 9th grade algebra course, is one of the most famous. What it is and how to solve a geometric progression - these questions are answered in this article.

A sequence of numbers that obeys a mathematical law

The title of this paragraph is a general definition of a geometric progression. The law by which it is described is quite simple: each next number differs from the previous one by a factor, which is called the "denominator". You can designate it with the letter r. Then we can write the following equality:

Here an is the member of the progression with number n.

If r is greater than 1, then the progression will increase in absolute value (it can decrease if its first term has a negative sign). If r is less than one, then the whole progression will tend to zero or from below (a1<0), либо сверху (a1>0). In the case of a negative denominator (r<0) иметь место будет чередующаяся числовая последовательность (каждый положительный член будет окружен двумя отрицательными). Наконец, при равенстве r единице получится простой набор чисел, который, как правило, не называют прогрессией.

An example of the type of progression under consideration is given below:

2, 3, 4, 5, 6, 75, …

Here the first term is 2 and the denominator is 1.5.

Important formulas

How to solve a geometric progression in grade 9? To do this, you should know not only its definition and understand what it is about, but also remember two important formulas. The first of these is shown below:

The expression allows you to easily find an arbitrary element of the sequence, but for this you need to know two numbers: the denominator and the first element. It is easy to prove this formula, you just need to remember the definition of a geometric progression: the second element is obtained by multiplying the first by the denominator to the first degree, the third element by multiplying the first by the denominator to the second degree, and so on. The usefulness of this expression is obvious: there is no need to sequentially restore the entire number series in order to find out what value its nth element will take.

The following formula is also useful in answering the question of how to solve a geometric progression. We are talking about the sum of its elements, starting with the first and ending with the nth. The corresponding expression is given below:

Sn = a1*(rn-1)/(r-1).

It is worth paying attention to its peculiarity: as in the formula for finding the nth element, here it is also enough to know the same two numbers (a1 and r). This result is not surprising, because each term of the progression is associated with the marked numbers.

Restoring progression

The first example, how to solve a geometric progression, has the following condition: it is known that the two numbers 10 and 20 form the kind of progression under consideration. In this case, the numbers are the eighth and fifteenth elements of the series. It is necessary to restore the entire series, knowing that it must be decreasing.

This somewhat confusing condition of the problem should be analyzed carefully: since we are talking about a decreasing series, the number 10 should be in position 15, and 20 in 8. Starting to solve, write out the corresponding equalities for each of the numbers:

a8 = a1*r7 and a15 = a1*r14.

You have two equalities with two unknowns. Solve them by expressing from the first a1 and substituting it into the second. Get:

a1 = a8*r-7 and a15 = a8*r-7 *r14=a8*r7 => r=7√(a15/a8).

Now it remains to substitute the appropriate values ​​from the condition and calculate the seventh root. Get:

r=7√(a15/a8) = 7√(10/20) ≈ 0.9057.

Substituting the resulting denominator into any of the expressions for the known nth element, a1 is obtained:

a1 = a8*r-7 = 20*(0.9057)-7 ≈ 40.0073.

This way you will find the first term and the denominator, which means you will restore the entire progression. First few members:

40,0073, 36,2346, 32,8177, 29,7230, …

It should be noted that when performing calculations, rounding to 4 decimal places was used.

Finding an unknown member of a series

Now it is worth considering another example: it is known that the seventh element of the series is 27, which is the thirteenth term if the denominator r \u003d -2. How to solve a geometric progression using this data? Very simple, you need to write out the formula for the 7th element:

Since only the number a1 is unknown in this equality, express it:

Use the last equation by substituting it into the formula for the 13th term you want to find. Get:

a13 = a1*r12 = a7*r-6*r12 = a7*r6.

It remains to substitute the numbers and write the answer:

a13 = a7*r6 = 27*(-2)6 = 1728.

The resulting number shows how fast the geometric progression grows.

Task for the sum

The last task, which reveals the question of how to solve a geometric progression, is related to finding the sum of several elements. Let a1 = 1.5, r = 2. You should calculate the sum of the terms of this series, starting from the 5th and ending with the 10th.

To get the answer to the question posed, you should apply the formula:

S510 = S10 - S4.

That is, first you need to find the sum of 10 elements, then the sum of the first 4 and subtract them among themselves. Following the specified algorithm, it will turn out:

S10 = a1*(rn-1)/(r-1) = 1.5*(210-1)/(2-1) = 1534.5;

S4 = a1*(rn-1)/(r-1) = 1.5*(24-1)/(2-1) = 22.5;

S510 = 1534.5 - 22.5 = 1512.

It is worth noting that in the final formula, the sum of exactly 4 terms was subtracted, since the fifth, according to the condition of the problem, should participate in the sum.

October 9, 2018

The geometric progression is one of the most interesting number series that is considered in the school algebra course. This article is devoted to a special case of the mentioned series: a decreasing infinite geometric progression and the sum of its terms.

What series of numbers are we talking about?

A geometric progression is a one-dimensional sequence of real numbers that are related to each other by the following relationship:

a 2 = a 1 *r, a 3 = a 2 *r, a 4 = a 3 *r, ...., a n = a n-1 *r

Generalizing the expressions above, we can write the following equality:

a n = a 1 *r n-1

As is clear from the above entries, a n is the element of the progression with the number n. The parameter r, by which n-1 elements should be multiplied to get the n-th element, is called the denominator.

What are the properties of the described sequence? The answer to the question depends on the value and sign of r. The following options are possible:

• The denominator r is positive and greater than 1. In this case, the progression will always increase in absolute value, while the absolute value of its members may also decrease if a 1 is negative.
• The denominator r is negative and greater than 1. In this case, the terms of the progression will appear with alternating sign (+ and -). Such series are of little practical interest.
• The modulus of the denominator r is less than 1. This series is called decreasing, regardless of the sign of r. It is this progression that is of great practical interest, and it will be discussed in this article.

Formula for sum

First, let's get an expression that will allow us to calculate the sum of an arbitrary number of elements of a given progression. Let's start solving this problem head-on. We have:

S n = a 1 +a 2 +a 3 +..+a n

The above equality can be used if it is necessary to calculate the result for a small number of terms (3-4 terms), each of which is determined by the formula for the nth term (see the previous paragraph). However, if there are a lot of terms, then it is inconvenient to count on the forehead and you can make a mistake, so they use a special formula.

We multiply both parts of the equality above by r, we get:

r*S n = r*a 1 +r*a 2 +r*a 3 +..+r*a n = a 2 +a 3 +a 4 +...+a n+1

Now we subtract the left and right parts of these two expressions in pairs, we have:

r*S n - S n = a 2 +a 3 +a 4 +...+a n+1 - (a 1 +a 2 +a 3 +..+a n) = a n+1 - a 1

Expressing the sum S n and using the formula for the term a n+1 , we get:

S n \u003d (a n+1 - a 1) / (r-1) \u003d a 1 * (r n - 1) / (r-1)

Thus, we have obtained a general formula for the sum of the first n terms of the considered type of the number series. Note that the formula is valid if r≠1. In the latter case, there is a simple series of identical numbers, the sum of which is calculated as the product of one number and their number.

Related videos

How to find the sum of an infinite decreasing geometric progression?

To answer this question, we should recall that the series will be decreasing when |r|<1. Воспользуемся полученной в предыдущем пункте формулой для S n:

S n \u003d a 1 * (r n - 1) / (r-1)

Note that any number whose modulus is less than 1 tends to zero when raised to a large power, that is, r ∞ -> 0. You can check this fact on any example:

r = -1/2, then (-1/2)**10 ≈ 9.7*10 -4, (-1/2)**20 ≈ 9.5*10 -7 and so on.

Having established this fact, let's pay attention to the expression for the sum: for n->∞ it will be rewritten as follows:

S ∞ = a 1 *(r ∞ - 1)/(r-1) = a 1 /(1-r)

An interesting result was obtained: the sum of an infinite progression of a decreasing geometric tends to a finite number, which does not depend on the number of terms. It is determined only by the first term and the denominator. Note that the sign of the sum is uniquely determined by the sign of a 1 , since the denominator is always a positive number (1-r>0).

The sum of squares of an infinite decreasing geometric progression

The title of the item defines the problem to be solved. To do this, we use a technique that is completely similar to that used to derive the general formula for S n . We have the first expression:

M n = a 1 2 + a 2 2 + a 3 2 + ... + a n 2

Multiply both sides of the equality by r 2, write the second expression:

r 2 *M n = r 2 *a 1 2 + r 2 *a 2 2 + r 2 *a 3 2 + ... + r 2 *a n 2 = a 2 2 + a 3 2 + a 4 2 .. .+a n+1 2

Now we find the difference between these two equalities:

r 2 *M n - M n = a 2 2 + a 3 2 + a 4 2 ... + a n+1 2 - (a 1 2 + a 2 2 + a 3 2 + ... + a n 2) = a n+1 2 - a 1 2

We express M n and use the formula for the nth element, we get the equality:

M n \u003d (a n+1 2 - a 1 2) / (r 2 -1) \u003d a 1 2 * (r 2n -1) / (r 2 -1)

In the previous paragraph, it was shown that r ∞ -> 0, then the final formula will take the form:

M ∞ = a 1 2 */(1-r 2)

Comparison of two received sums

Let's compare two formulas: for an infinite sum and an infinite sum of squares using the example of the following problem: the sum of an infinite geometric progression is 2, it is known that we are talking about a decreasing sequence for which the denominator is 1/3. It is necessary to find the infinite sum of the squares of this series of numbers.

Let's use the formula for the sum. Express a 1:

S ∞ = a 1 /(1-r) => a 1 = S ∞ *(1-r)

We substitute this expression into the formula for the sum of squares, we have:

M ∞ = a 1 2 */(1-r 2) = S ∞ 2 *(1-r) 2 /(1-r 2) = S ∞ 2 *(1-r)/(1+r)

We have obtained the desired formula, now we can substitute the data known from the condition:

M ∞ = S ∞ 2 *(1-r)/(1+r) = 2 2 *(1-1/3)/(1+1/3) = 2

Thus, we have obtained the same value for the infinite sum of squares as for the simple sum. Note that this result is valid only for this problem. In general, M ∞ ≠ S ∞ .

The task of calculating the area of ​​a rectangle

Every student knows the formula S = a * b, which determines the area of ​​a rectangle in terms of its sides. Few people know that the problem of finding the area of ​​this figure can be easily solved using the sum of an infinite geometric progression. Let's show how it's done.

Let's mentally divide the rectangle in half. The area of ​​one half is taken as unity. Now we divide the other half in half again. We get two halves, one of which we will divide in half. We will continue this procedure indefinitely (see the figure below).

As a result, the area of ​​the rectangle in the units we have chosen will be equal to:

S ∞ = 1+1/2+1/4+1/8+...

It can be seen that these terms are elements of a decreasing series, in which a 1 = 1 and r = 1/2. Using the formula for an infinite sum, we get:

S∞ = 1 /(1-1/2) = 2

In the scale we have chosen, half of the rectangle (one unit) corresponds to the area a*b/2. This means that the area of ​​the entire rectangle is:

S ∞ = 2*a*b/2 = a*b

The result obtained is obvious, nevertheless, it showed how a decreasing progression can be applied to solve problems in geometry.

The geometric progression is one of the most interesting number series that is considered in the school algebra course. This article is devoted to a particular case of the mentioned series: and the sum of its terms.

What series of numbers are we talking about?

A geometric progression is a one-dimensional sequence of real numbers that are related to each other by the following relationship:

a 2 = a 1 *r, a 3 = a 2 *r, a 4 = a 3 *r, ...., a n = a n-1 *r

Generalizing the expressions above, we can write the following equality:

a n = a 1 *r n-1

As is clear from the above entries, a n is the element of the progression with the number n. The parameter r, by which n-1 elements should be multiplied to get the n-th element, is called the denominator.

What are the properties of the described sequence? The answer to the question depends on the value and sign of r. The following options are possible:

• The denominator r is positive and greater than 1. In this case, the progression will always increase in absolute value, while the absolute value of its members may also decrease if a 1 is negative.
• The denominator r is negative and greater than 1. In this case, the terms of the progression will appear with alternating sign (+ and -). Such series are of little practical interest.
• The modulus of the denominator r is less than 1. This series is called decreasing, regardless of the sign of r. It is this progression that is of great practical interest, and it will be discussed in this article.

Formula for sum

First, let's get an expression that will allow us to calculate the sum of an arbitrary number of elements of a given progression. Let's start solving this problem head-on. We have:

S n = a 1 +a 2 +a 3 +..+a n

The above equality can be used if it is necessary to calculate the result for a small number of terms (3-4 terms), each of which is determined by the formula for the nth term (see the previous paragraph). However, if there are a lot of terms, then it is inconvenient to count on the forehead and you can make a mistake, so they use a special formula.

We multiply both parts of the equality above by r, we get:

r*S n = r*a 1 +r*a 2 +r*a 3 +..+r*a n = a 2 +a 3 +a 4 +...+a n+1

Now we subtract the left and right parts of these two expressions in pairs, we have:

r*S n - S n = a 2 +a 3 +a 4 +...+a n+1 - (a 1 +a 2 +a 3 +..+a n) = a n+1 - a 1

Expressing the sum S n and using the formula for the term a n+1 , we get:

S n \u003d (a n+1 - a 1) / (r-1) \u003d a 1 * (r n - 1) / (r-1)

Thus, we have obtained a general formula for the sum of the first n terms of the considered type of the number series. Note that the formula is valid if r≠1. In the latter case, there is a simple series of identical numbers, the sum of which is calculated as the product of one number and their number.

How to find the sum of an infinite decreasing geometric progression?

To answer this question, we should recall that the series will be decreasing when |r|<1. Воспользуемся полученной в предыдущем пункте формулой для S n:

S n \u003d a 1 * (r n - 1) / (r-1)

Note that any number whose modulus is less than 1 tends to zero when raised to a large power, that is, r ∞ -> 0. You can check this fact on any example:

r = -1/2, then (-1/2)**10 ≈ 9.7*10 -4, (-1/2)**20 ≈ 9.5*10 -7 and so on.

Having established this fact, let's pay attention to the expression for the sum: for n->∞ it will be rewritten as follows:

S ∞ = a 1 *(r ∞ - 1)/(r-1) = a 1 /(1-r)

An interesting result was obtained: the sum of an infinite progression of a decreasing geometric tends to a finite number, which does not depend on the number of terms. It is determined only by the first term and the denominator. Note that the sign of the sum is uniquely determined by the sign of a 1 , since the denominator is always a positive number (1-r>0).

The sum of squares of an infinite decreasing geometric progression

The title of the item defines the problem to be solved. To do this, we use a technique that is completely similar to that used to derive the general formula for S n . We have the first expression:

M n = a 1 2 + a 2 2 + a 3 2 + ... + a n 2

Multiply both sides of the equality by r 2, write the second expression:

r 2 *M n = r 2 *a 1 2 + r 2 *a 2 2 + r 2 *a 3 2 + ... + r 2 *a n 2 = a 2 2 + a 3 2 + a 4 2 .. .+a n+1 2

Now we find the difference between these two equalities:

r 2 *M n - M n = a 2 2 + a 3 2 + a 4 2 ... + a n+1 2 - (a 1 2 + a 2 2 + a 3 2 + ... + a n 2) = a n+1 2 - a 1 2

We express M n and use the formula for the nth element, we get the equality:

M n \u003d (a n+1 2 - a 1 2) / (r 2 -1) \u003d a 1 2 * (r 2n -1) / (r 2 -1)

In the previous paragraph, it was shown that r ∞ -> 0, then the final formula will take the form:

M ∞ = a 1 2 */(1-r 2)

Comparison of two received sums

Let's compare two formulas: for an infinite sum and an infinite sum of squares using the example of the following problem: the sum of an infinite geometric progression is 2, it is known that we are talking about a decreasing sequence for which the denominator is 1/3. It is necessary to find the infinite sum of the squares of this series of numbers.

Let's use the formula for the sum. Express a 1:

S ∞ = a 1 /(1-r) => a 1 = S ∞ *(1-r)

We substitute this expression into the formula for the sum of squares, we have:

M ∞ = a 1 2 */(1-r 2) = S ∞ 2 *(1-r) 2 /(1-r 2) = S ∞ 2 *(1-r)/(1+r)

We have obtained the desired formula, now we can substitute the data known from the condition:

M ∞ = S ∞ 2 *(1-r)/(1+r) = 2 2 *(1-1/3)/(1+1/3) = 2

Thus, we have obtained the same value for the infinite sum of squares as for the simple sum. Note that this result is valid only for this problem. In general, M ∞ ≠ S ∞ .

The task of calculating the area of ​​a rectangle

Every student knows the formula S = a * b, which determines the area of ​​a rectangle in terms of its sides. Few people know that the problem of finding the area of ​​this figure can be easily solved using the sum of an infinite geometric progression. Let's show how it's done.

Let's mentally divide the rectangle in half. The area of ​​one half is taken as unity. Now we divide the other half in half again. We get two halves, one of which we will divide in half. We will continue this procedure indefinitely (see the figure below).

As a result, the area of ​​the rectangle in the units we have chosen will be equal to:

S ∞ = 1+1/2+1/4+1/8+...

It can be seen that these terms are elements of a decreasing series, in which a 1 = 1 and r = 1/2. Using the formula for an infinite sum, we get:

S∞ = 1 /(1-1/2) = 2

In the scale we have chosen, half of the rectangle (one unit) corresponds to the area a*b/2. This means that the area of ​​the entire rectangle is:

S ∞ = 2*a*b/2 = a*b

The result obtained is obvious, nevertheless, it showed how a decreasing progression can be applied to solve problems in geometry.

Among all the sequences of numbers, the geometric progression, which is considered in the 9th grade algebra course, is one of the most famous. What it is and how to solve a geometric progression - these questions are answered in this article.

A sequence of numbers that obeys a mathematical law

The title of this paragraph is a general definition of a geometric progression. The law by which it is described is quite simple: each next number differs from the previous one by a factor, which is called the "denominator". You can designate it with the letter r. Then we can write the following equality:

Here a n is a member of the progression with number n.

If r is greater than 1, then the progression will increase in absolute value (it can decrease if its first term has a negative sign). If r is less than one, then the whole progression will tend to zero or from below (a 1<0), либо сверху (a 1 >0). In the case of a negative denominator (r<0) иметь место будет чередующаяся числовая последовательность (каждый положительный член будет окружен двумя отрицательными). Наконец, при равенстве r единице получится простой набор чисел, который, как правило, не называют прогрессией.

An example of the type of progression under consideration is given below:

2, 3, 4, 5, 6, 75, ...

Here the first term is 2 and the denominator is 1.5.

Important Formulas

How to solve a geometric progression in grade 9? To do this, you should know not only its definition and understand what it is about, but also remember two important formulas. The first of these is shown below:

The expression allows you to easily find an arbitrary element of the sequence, but for this you need to know two numbers: the denominator and the first element. It is easy to prove this formula, you just need to remember the definition of a geometric progression: the second element is obtained by multiplying the first by the denominator to the first degree, the third element by multiplying the first by the denominator to the second degree, and so on. The usefulness of this expression is obvious: there is no need to sequentially restore the entire number series in order to find out what value its nth element will take.

The following formula is also useful in answering the question of how to solve a geometric progression. We are talking about the sum of its elements, starting with the first and ending with the nth. The corresponding expression is given below:

S n \u003d a 1 * (r n -1) / (r-1).

It is worth paying attention to its peculiarity: as in the formula for finding the nth element, here it is also enough to know the same two numbers (a 1 and r). This result is not surprising, because each term of the progression is associated with the marked numbers.

Restoring progression

The first example, how to solve a geometric progression, has the following condition: it is known that the two numbers 10 and 20 form the kind of progression under consideration. In this case, the numbers are the eighth and fifteenth elements of the series. It is necessary to restore the entire series, knowing that it must be decreasing.

This somewhat confusing condition of the problem should be analyzed carefully: since we are talking about a decreasing series, the number 10 should be in position 15, and 20 in 8. Starting to solve, write out the corresponding equalities for each of the numbers:

a 8 = a 1 *r 7 and a 15 = a 1 *r 14 .

You have two equalities with two unknowns. Solve them by expressing from the first a 1 and substituting it into the second. Get:

a 1 = a 8 *r -7 and a 15 = a 8 *r -7 *r 14 = a 8 *r 7 => r= 7 √ (a 15 / a 8).

Now it remains to substitute the appropriate values ​​from the condition and calculate the seventh root. Get:

r \u003d 7 √ (a 15 / a 8) \u003d 7 √ (10 / 20) ≈ 0.9057.

Substituting the resulting denominator into any of the expressions for the known nth element, we get a 1:

a 1 \u003d a 8 * r -7 \u003d 20 * (0.9057) -7 ≈ 40.0073.

This way you will find the first term and the denominator, which means you will restore the entire progression. First few members:

40,0073, 36,2346, 32,8177, 29,7230, ...

It should be noted that when performing calculations, rounding to 4 decimal places was used.

Finding an unknown member of a series

Now it is worth considering another example: it is known that the seventh element of the series is 27, which is the thirteenth term if the denominator r \u003d -2. How to solve a geometric progression using this data? Very simple, you need to write out the formula for the 7th element:

Since only the number a 1 is unknown in this equality, express it:

Use the last equation by substituting it into the formula for the 13th term you want to find. Get:

a 13 = a 1 *r 12 = a 7 *r -6 *r 12 = a 7 *r 6 .

It remains to substitute the numbers and write the answer:

a 13 \u003d a 7 * r 6 \u003d 27 * (-2) 6 \u003d 1728.

The resulting number shows how fast the geometric progression grows.

Task for the sum

The last task, which reveals the question of how to solve a geometric progression, is related to finding the sum of several elements. Let a 1 \u003d 1.5, r \u003d 2. The sum of the terms of this series should be calculated, starting from the 5th and ending with the 10th.

To get the answer to the question posed, you should apply the formula:

That is, first you need to find the sum of 10 elements, then the sum of the first 4 and subtract them among themselves. Following the specified algorithm, it will turn out:

S 10 \u003d a 1 * (r n -1) / (r-1) \u003d 1.5 * (2 10 -1) / (2-1) \u003d 1534.5;

S 4 \u003d a 1 * (r n -1) / (r-1) \u003d 1.5 * (2 4 -1) / (2-1) \u003d 22.5;

S 5 10 \u003d 1534.5 - 22.5 \u003d 1512.

It is worth noting that in the final formula, the sum of exactly 4 terms was subtracted, since the fifth, according to the condition of the problem, should participate in the sum.