Annuity Scheme

Under an annuity repayment scheme, the borrower pays the same amount in each period.

However, the structure of the payment changes:

• at the beginning of the term, most of the payment goes to interest;
• toward the end of the term, most goes to repaying the principal.

The payment amount is chosen so that the loan is fully repaid by the end of the term.

Example

Loan amount: 331 000
Term: 3 months
Interest rate: 10% per month
Monthly payment: 133 100

Period	Remaining debt	Accrued interest	Principal repaid	Payment for the period
1	331 000	33 100	100 000	133 100
2	231 000	23 100	110 000	133 100
3	121 000	12 100	121 000	133 100

Note that the numbers in the “Principal repaid” column form a geometric progression:

\[ 100\,000,\quad 110\,000,\quad 121\,000. \]

Payment Amount

Let the loan amount be \( S \), the interest rate per period be \( r \), the number of periods be \( n \), and the payment amount be \( A \).

For convenience in calculations, introduce the growth factor:

\[ k = 1 + \frac{r}{100}. \]

After interest is applied in the first period, the debt becomes \( kS \), and after the first payment it becomes \( kS - A \).

In the second period, the debt again increases by a factor of \( k \): \[ k(kS - A) = k^2S - kA, \] and after the second payment becomes \[ k^2S - kA - A. \]

Continuing this process, we find that before the last payment the debt equals \[ k^n S - k^{n-1}A - k^{n-2}A - \ldots - kA, \]

and after the final payment the loan must be fully repaid. Therefore, \[ k^n S - A\left(k^{n-1} + k^{n-2} + \ldots + 1\right) = 0. \]

The sum in parentheses is the sum of a geometric progression: \[ k^{n-1} + k^{n-2} + \ldots + 1 = \frac{k^n - 1}{k - 1}. \]

Substituting this expression, we obtain the equation: \[ k^n S - A \cdot \frac{k^n - 1}{k - 1} = 0. \]

Hence, the annuity payment amount is: \[ \boxed{ A = S \cdot \frac{k^n (k - 1)}{k^n - 1} } \]

Characteristic property of a geometric progression

Three consecutive terms of a geometric progression have an important property: the square of the middle term is equal to the product of the two neighboring terms.

\[ a_k^2 = a_{k-1} \cdot a_{k+1}. \]

This property makes it possible to set up equations and reconstruct elements of a geometric progression from incomplete data.

Subtraction method

To derive the formula for the sum of the first \(n\) terms of a geometric progression, the subtraction method is used.

Consider the sum

\[ S_n = a_1 + a_1q + a_1q^2 + \ldots + a_1q^{n-1}. \]

Multiply this equality by the common ratio \(q\):

\[ qS_n = a_1q + a_1q^2 + \ldots + a_1q^{n}. \]

Subtract the second equality from the first one. Most terms cancel out, and only the extreme terms remain.

Formula for the sum of a geometric progression

As a result of the subtraction, we obtain:

\[ S_n (1 - q) = a_1 (1 - q^n). \]

If \( q \neq 1 \), the sum of the first \(n\) terms of a geometric progression is computed using the formula:

\[ S_n = \frac{a_1 (1 - q^n)}{1 - q}. \]

This formula is especially convenient when working with large powers and allows one to quickly find the sums of long geometric series.

Example

Let us find the sum of the first eight terms of a geometric progression, if it is known that \( a_1 = 2 \) and \( q = 3 \).

Use the sum formula:

\[ S_8 = \frac{2(1 - 3^8)}{1 - 3}. \]

Compute the power: \( 3^8 = 6561 \).

\[ S_8 = \frac{2(1 - 6561)}{-2} = \frac{-13120}{-2} = 6560. \]

Answer: \( S_8 = 6560 \).

Concept of a Geometric Progression

A numerical sequence is an ordered set of numbers in which each element is assigned its own index. The elements of a sequence are usually denoted by lowercase Latin letters with subscripts: \( a_1, a_2, a_3, \ldots \).

Let us consider several examples of numerical sequences:

• \( 2, 4, 8, 16, 32, \ldots \) — the sequence of powers of 2;
• \( 3, 6, 12, 24, 48, \ldots \) — each next element is twice the previous one;
• \( 1, -2, 4, -8, 16, \ldots \) — a sequence with alternating signs.

Task: Find the seventh element in each of the given sequences.

A geometric progression is a numerical sequence in which each next element is obtained from the previous one by multiplying by the same nonzero number. This number is called the common ratio of the geometric progression and is denoted by the letter q.

For example, in the sequence \[ 2, 6, 18, 54, \ldots \] each next element is obtained by multiplying the previous one by 3, therefore \( q = 3 \). The notation \( a_4 \) corresponds to the fourth element of the sequence, that is, \( a_4 = 54 \).

Task: Determine the common ratio of the geometric progression \( 5, 10, 20, 40, \ldots \) and find its seventh element.

The general term of a geometric progression is given by the formula:
\( a_n = a_1 \cdot q^{\,n-1} \)

Using this formula, one can find any term of the progression if its first term and common ratio are known.

Task: Using the formula for the general term, find the seventh element of a geometric progression with first term \( a_1 = 3 \) and common ratio \( q = 2 \).

The type of monotonicity of a geometric progression whose first term is positive depends on the common ratio as follows:

• if \( q > 1 \), the progression is increasing (each subsequent term is greater than the previous one);
• if \( 0 < q < 1 \), the progression is decreasing (each subsequent term is less than the previous one);
• if \( q < 0 \), the signs of the terms alternate (there is no monotonicity).

Task: Determine the type of monotonicity of a geometric progression whose first term is negative.

1. Loan Agreement and Repayment Scheme

Loan Agreement Parameters

When taking out a mortgage, the parties agree on the following main parameters:

• Loan Amount — the size of the loan;
• Term — the number of repayment periods (months);
• Interest Rate — the cost of using the loan;
• Repayment Scheme — the method of returning the debt to the bank.

The interest rate is usually stated as an annual rate. However, interest accrual and payment calculations are made per repayment period, most often monthly. Therefore, the annual rate must be converted to a rate per period. For example, if the annual rate is \(12\%\), the monthly rate is \( \dfrac{12\%}{12} = 1\% \). This means that each month interest is charged at \(1\%\) of the remaining principal.

Payment Structure

Each loan payment consists of two parts:

• interest charged on the remaining debt;
• the portion of principal repayment.

Interest decreases as the remaining debt decreases, and principal repayment depends on the chosen repayment scheme.

Consider an example. Let the annual interest rate be \(24\%\), the remaining principal at the beginning of the period be \(5000\) dollars, and the principal repayment for this period be \(1000\) dollars.

Convert the annual rate to a monthly rate:

\[ \frac{24\%}{12} = 2\% \text{ per month}. \]

Then the interest for the period will be:

\[ 5000 \cdot 0{,}02 = 100 \text{ dollars}. \]

The total payment for this period is the sum of the interest and the principal repayment:

\[ 100 + 1000 = 1100 \text{ dollars}. \]

Thus, under a differentiated repayment scheme, the payment amount in each period is determined by the current remaining debt and the fixed principal repayment amount.

It is convenient to organize calculations for the entire loan term in a table:

Period	Remaining Debt	Interest Charged	Repayment Amount	Payment for Period
1	6000	120	1000	1120
2	5000	100	2000	2100
3	3000	60	3000	3060

Differentiated Payments

Under a differentiated scheme, the parties agree that in each period the same amount of principal is repaid.

As a result:

• the remaining debt decreases evenly;
• interest decreases each period;
• total payment gradually decreases.

Consider an example.

Loan: $10,000
Term: 4 months
Interest rate: 6% per year

Monthly rate: \[ \frac{6\%}{12} = 0{,}5\% = 0{,}005 \]

Principal repayment each month: \[ \frac{10\,000}{4} = 2\,500 \]

Period	Remaining Debt	Interest Charged	Repayment Amount	Payment for Period
1	10 000	50	2 500	2 550
2	7 500	37,5	2 500	2 537,5
3	5 000	25	2 500	2 525
4	2 500	12,5	2 500	2 512,5

Note an important fact: the values in all columns form arithmetic progressions! Therefore, the total of all payments can be found:

• by direct addition;
• using the formula for the sum of an arithmetic progression.

\[ S = \frac{n(a_1 + a_n)}{2} = \frac{4(2550 + 2512{,}5)}{2} = 10\,125 \]

The overpayment for the loan will be only 1.25%:

\[ 10\,125 - 10\,000 = 125 \]

Characteristic Property of an Arithmetic Progression

Three consecutive terms of an arithmetic progression have an important property: the middle term equals the arithmetic mean of its two neighbors.

\[ a_k = \frac{a_{k-1} + a_{k+1}}{2}. \]

This property makes it possible to form equations and reconstruct the terms of a progression from incomplete data.

Pairwise Summation Method

To quickly compute the sum of the first \(n\) terms of an arithmetic progression, the pairwise summation method is used.

The first and last terms, the second and the second-to-last terms, and so on, form pairs with the same sum:

\[ a_1 + a_n = a_2 + a_{n-1} = \ldots \]

There are exactly \( \dfrac{n}{2} \) such pairs if \(n\) is even, and \( \dfrac{n-1}{2} \) pairs plus one middle term if \(n\) is odd.

This method is the basis of the formula for the sum of an arithmetic progression.

Formula for the Sum of an Arithmetic Progression

The sum of the first \(n\) terms of an arithmetic progression equals the product of the number of terms and the arithmetic mean of the first and the last terms:

\[ S_n = \frac{n(a_1 + a_n)}{2}. \]

This formula is especially convenient when the extreme terms of the progression are known.

Example

Find the sum of the first twelve terms of an arithmetic progression if it is known that \( a_2 = 8 \) and \( a_{10} = 40 \).

We use the formula for the general term of an arithmetic progression:

\[ a_n = a_1 + (n - 1)d. \]

From the given conditions we obtain the system:

\[ \begin{cases} a_1 + d = 8, \\ a_1 + 9d = 40. \end{cases} \]

Subtract the first equation from the second:

\[ 8d = 32 \quad \Rightarrow \quad d = 4. \]

Substituting the value of the common difference into the first equation, we get:

\[ a_1 = 8 - 4 = 4. \]

Now find the twelfth term of the progression:

\[ a_{12} = a_1 + 11d = 4 + 44 = 48. \]

Use the formula for the sum of the first \(n\) terms of an arithmetic progression:

\[ S_{12} = \frac{12(a_1 + a_{12})}{2} = 6 \cdot (4 + 48) = 312. \]

Answer: \( S_{12} = 312 \).

Concept of Arithmetic Progression

Numerical sequence — an ordered set of numbers, in which a rule for generating all elements is specified.

Examples of different numerical sequences:

• \(1, 2, 3, 4, 5, \ldots\) — sequence of natural numbers;
• \(1, 1, 2, 3, 5, 8, \ldots\) — Fibonacci sequence, where each element from the third onward is the sum of the two previous elements;
• \(1, 4, 1, 5, 9, \ldots\) — digits after the decimal point of \( \pi \), with no obvious pattern.

Task: Find the fifteenth element in each of the sequences listed above.

Arithmetic progression — a sequence in which each next element differs from the previous one by the same number. Elements of a sequence are usually denoted by lowercase Latin letters with subscripts indicating the position of the number in the list. For example, in the progression \[1, 4, 7, 10, 13, \ldots\] the notation \(a_3\) refers to the third number in this list, that is \(a_3 = 7\). The number by which consecutive elements of a progression differ is called its common difference and is denoted by d. In our example, \(d = 3\).
Task: Find the fifteenth element of this progression.

The general term of an arithmetic progression can be described by the following formula:
\( a_n = a_1 + (n - 1)d \)

Task: Use this formula to find the fifteenth element of the progression above.