Section of mathematics called progression is one of the most interesting sections of mathematics. But before we discuss arithmetic progression questions let’s remember some historical moments and figure out what are arithmetic and geometric progressions.

Here is a vivid example of the use of progressions by English economist and Bishop Malthus, who used geometric and arithmetic progressions to justify wars: consumer products (food, clothing) evolve under the laws of arithmetic progression, and people multiply by the laws of geometric progression. To get rid of excess population wars are a necessity. Now we understand that this statement is incorrect and does not correspond to the mentality of modern society.

There is also known an interesting story about the famous German mathematician Gauss, who as a child discovered outstanding abilities in mathematics. The teacher offered to students to add up all the natural numbers from 1 to 100. Little Gauss solved the problem in one minute, realizing that the sum of 1 + 100, 99 + 2, etc. are equal, he multiplied 101 by 50, or on the number of such sums. In other words, he noticed a pattern inherent in arithmetic progressions.

So, we turn to the definition: arithmetic progression - is a sequence of numbers in which the difference between successive terms is constant. For example, 7, 10, 13, 16 - is an arithmetic progression, in which the difference between adjacent members is three.

Probably you have a question: why his progressions are needed? In fact, its use is wide and varied. Let’s consider a few cases:

 

  1. Probably the first situation in which people had to deal with a geometric progression - herd population count conducted several times at regular intervals. If there is no emergency, the number of births and deaths of animals in proportion to the number of all animals. So, if at any time the number of sheep the shepherd has increased from 10 to 20 goals, then for the next same period it will grow back twice and will be equal to 40.
  2. In life geometrical progression first of all is present in calculation of so called „compound interest”. If you put money on a fixed deposit in a bank, then in a year the deposit will be increased by 3% from the starting amount, and new amount will be equal to the deposit multiplied by 1,03. And already in a year this amount will increase by 3%, then again multiplied on 1.03. In 20 years, the amount on the account will be multiplied in (1.03)20 =1.8 times.
    If the percentage will be bigger, then the result will change drastically. At 50% annual interest in 10 years the amount will multiply in (1.5)10 = 55.7 times. This was the interest rate British money-lenders issued credits in the XIII century. It caused a terrible unrest. Laws were made to limit interest rates. King Henry VII had even completely abolished the charging of interest, which resulted in decline in banking and production industries, were not able to get access to credit. In the end, the charging of interest has been allowed, but had to be not more than 10%.
  3. One more example of geometrical progression - change in mass of radioactive material over time. It is known that during the unit of time such material loses some of its mass (passes into another substance and energy). For each of the radioactive substance is determined by the value of T - time of half-life. Weights of non-decayed substance at time 0, T, 2T, 3T, ... will form an infinitely decreasing geometric progression.
  4. A growth of trees in the forest comes under the laws of geometric progression. At the same time every type of tree can have its own annual growth coefficient. Consideration of such changes allows planning tree cutting as a part of forests and at the same forest renovation.
  5. Under favourable conditions bacteria multiply in such a way that during one minute one of them divides into two.

 

By studying various examples one should come to a conclusion, that progressions are important in real life as well as in science. However we have only spoken in detail about arithmetic progressions, so let us discuss geometrical.

Lets start with a simple question: why geometrical progression is calls „geometrical”? The answer is really simple: because its every member is equal to average geometrical of its neighboring members. And such average number is called geometrical, because it is a part of a square, sides of which have lengths equal to measurements, from which the average is calculated.

Geometrical progressions surprise with their substantial growth. In real life growing geometrical progressions have to be treated with care. If in a geometrical progression increases a number of animals in a herd than soon it will run out of grass.

It should be concluded that the ability to deal with progressions - is indispensable for any modern man who wants to build his business. Knowledge of progression is extremely important for many occupations, so no matter what university apply to, the ability to work with progressions will be your trump card. Our team wants to help you with this, and offers to take the test progression online in which arithmetic progression questions and answers designed by our experts in the field of mathematics. Detailed answers to each question will help you to understand your mistakes and improve the result!

5.1

The sequence \(u_1,u_2,u_3,\dots\) is defined by \(u_n=3n-1\), for \(n\geqslant1\).

Find the value of \(u_{40}\).

\(u_{40} = 3 \times 40 - 1 = 119\)

5.2

A geometric progression \(u_1, u_2, u_3,\dots\) is defined by \(u_1 = 15\) and \(u_{n+1} = 0.8u_n\) for \(n \geqslant 1\).

Find \(\displaystyle\sum_{n=1}^{\infty}u_n\).

\(\displaystyle\sum_{n=1}^{\infty}u_n = \frac{15}{1-0.8} = \frac{15}{0.2} = 75\).

5.3

In an arithmetic progression, the first term is \(12\) and the sum of the first \(70\) terms is \(12\ 915\).

 Find the common difference.

The sum of the first \(n\) members of an arithmetic progression is \(\frac{n(2u_1+(n-1)d)}{2}\), where \(u_1\) is the first term and \(d\) is the common difference. Therefore,

\(\frac{70(2\times12+69d)}{2} = 12\ 915\\ 70(24+69d) = 25\ 830\\ 24+69d = 369\\ 69d = 345\\ d = 5\)

5.4

In a geometric progression, the second term is \(-4\) and the sum to infinity is \(9\).

Find the common ratio.

Sum of an infinite geometric progression is \(\frac{b_1}{1-q}\), where \(b_1\) is the first term and \(q\) is the common ratio. \(b_2 = qb_1\), so \(b_1 = \frac{b_2}{q} = \frac{-4}{q}\). Therefore,

\(\frac{b_1}{1-q} = 9,\\ \frac{\frac{-4}{q}}{1-q} = 9,\\ -4 = 9q(1-q),\\ 9q^2-9q-4 = 0.\)

The solution to the quadratic equation is:

\(D = b^2-4ac = 81-4\times9\times(-4) = 81+144 = 225\),

\(q = {-b \pm \sqrt{D} \over 2a}={9 \pm 25 \over 18}\). So \(q = \frac{17}{9}\), or \(q = -\frac{8}{9}\). Since \(|q|\) must be less than \(1\)\(q = -\frac89\).

5.5

The tenth term of an arithmetic progression is equal to twice the fourth term. The twentieth term of the progression is \(44\).

Find the first term.

\(\left\{ \begin{aligned} &u_{10} = 2u_{4}\\ &u_{20} = 44 \end{aligned} \right. \\ \left\{ \begin{aligned} &u_1 + 9d = 2(u_1 + 3d)\\ &u_1+19d = 44 \end{aligned} \right. \\ \left\{ \begin{aligned} &u_1 = 3d\\ &u_1+19d = 44 \end{aligned} \right. \\ u_1 + \frac{19}{3}u_1 = 44\\ \frac{22}{3}u_1 = 44\\ u_1 = 6\)

5.6

A sequence \( u_1, u_2, u_3 ,\dots\) is defined by \( u_1 = 8 \) and \(u_{n+1} = u_n + 3\).

State what type of sequence it is. 

Since \(u_{n+1}-u_n=3\), the difference between the consecutive terms of the sequence is constant, so it's an arithmetic progression.

5.7

In an arithmetic progression the ninth term is \(18\) and the sum of the first nine terms is \(72\).

Find the first term.

Since \(S_n = \frac{n(u_1+u_n)}{2}\)\(S_9 = \frac{9(u_1+u_9)}{2} = \frac{9(u_1+18)}{2} = 72 \), so,

\(9(u_1+18) = 144\\ u_1+18 =16\\ u_1 = -2\)

5.8

The first three terms of an arithmetic progression are \(2x\)\(x+4\) and \(2x-7\) respectively.

Find the value of \(x\).

Since consecutive terms of an arithmetic progression have a common difference, we obtain

 \((x+4)-2x = (2x-7)-(x+4)\\ -x+4=x-11\\ 2x=15\\ x=7.5\)

5.9

The sequence \(u_1,u_2,u_3,\dots\) is defined by \(u_n=85-5n\), for \(n\geqslant1\). Given that \(u_1, u_5 \) and \(u_p\) are, respectively, the first, second and third terms of a geometric progression.

Find the value of \(p\).

Since \(u_1 = 85-5 = 80\) and \(u_5 = 85 - 5\times 5 = 60\), the common ratio is \(\frac{60}{80} = \frac{3}{4}=0.75\). Therefore, 

\(\frac{u_p}{u_5} = \frac{85-5p}{60} = 0.75\\ 85-5p=45\\ 5p=40\\ p=8\)

5.10

The first term of a geometric progression is \(7\) and the common ratio is \(-2\).

Find the ninth term.

\(u_n = u_1q^{n-1}\), where \(u_1\) is the first term and \(q\) is the common ratio. So, \(u_9 = u_1q^8 = 7\times(-2)^8 = 1792\).

7

The sequence \(u_1,u_2,u_3,\dots\) is defined by \(u_n=3n-1\), for \(n\geqslant1\).

Find \(\displaystyle\sum_{n = 1}^{40}u_n\).

\(\displaystyle\sum_{n = 1}^{40}u_n = \displaystyle\sum_{n = 1}^{40}(3n-1) = 3\displaystyle\sum_{n = 1}^{40}n - \displaystyle\sum_{n = 1}^{40}1 = 3\frac{40\times41}{2} - 40 = 2460 - 40 = 2420\).

8

The tenth term of an arithmetic progression is equal to twice the fourth term. The twentieth term of the progression is \(44\).

Find the common difference.

\(\begin{cases} a10=2a4\\ a20=44\\ \end{cases}> \begin{cases} a+9d=2(a+3d)\\ a+19d=44\\ \end{cases}> \begin{cases} a+9d=2a+6d\\ a+19d=44\\ \end{cases}> \begin{cases} a=3d\\ a+19d=44\\ \end{cases}> \begin{cases} a=3d\\ 22d=44\\ \end{cases}> \begin{cases} a=6\\ d=2\\ \end{cases}\)

d – the common difference.

d=2

9

A sequence \( u_1, u_2, u_3 ,\dots\) is defined by \( u_1 = 8 \) and \(u_{n+1} = u_n + 3\).

Express \(u_n\) in the form \(u_n=pn+q\), where \(p\) and \(q\) are integers.

Since \(u_1 = 8\) and the common difference \(d = 3\), we obtain \(u_n = u_1 + (n-1)d = 8+3(n-1)= 3n+5\).

10

In an arithmetic progression the ninth term is \(18\) and the sum of the first nine terms is \(72\).

Find the common difference.

\(u_9 = u_1+8d = -2+8d=18\\ 8d = 20\\ d = 2.5\)

11

The sequence \(u_1,u_2,u_3,\dots\) is defined by \(u_n=85-5n\), for \(n\geqslant1\). Given that \(u_1, u_5 \) and \(u_p\) are, respectively, the first, second and third terms of a geometric progression.

Find the sum to infinity of the geometric progression.

\(S = \frac{u_1}{1-q} = \frac{80}{1-0.75} = \frac{80}{0.25}=320\).

12

The first term of a geometric progression is \(7\) and the common ratio is \(-2\).

Find the sum of the first \(10\) terms.

\(S_n = u_1\frac{q^n-1}{q-1}\)\(S_{10} = 7\frac{(-2)^{10}-1}{-2-1} = 7\frac{1023}{-3} = 7\times(-341) = -2387\).