MathCalculusConceptsArithmetic and Geometric Sequences

# Arithmetic and Geometric Sequences

## Using Arithmetic and Geometric Sequences in the Real World

Consider the two examples below:

(A) Bob is a fitness fanatic who runs 50 minutes a day to maintain his health, but after an unfortunate accident, he undergoes a knee surgery. During his recovery phase, his trainer tells him that he can return to his running program, but at a slower pace. The trainer suggests that he start with 10 minutes of running. Thereafter, he suggests that the run time can be increased by 5 minutes per day. Assuming this pattern, how many days will it be before Bob is back to running 50 minutes per day?

(B) George has $2000 in savings. He decides to invest it with a bank. He plans to leave the money in the bank for a period of three years, for which the bank agrees to pay him a compounded interest of 6%. How much money will George have at the end of three years under the given criteria? These are real life examples. In order to answer them, we need to decode the pattern according to the data provided in the question. Solution for Example A: Adding 5 minutes every week to an initial value of 10 minutes will result in a pattern of numbers which looks like: If we continue this sequence of numbers until we reach 50, we will obtain 9 terms. Hence, the answer for Example A is 9 days. Solution for Example B: For Year 1: 6% interest on$2000 =

$\frac{6}{100}×2000=120$

The amount at the end of Year 1 is

$2000+120=2120$

For Year 2: 6% interest on $2120 = $\frac{6}{100}×2120=127.2$ Therefore, the amount at the end of Year 2 is >$2120+127.2=2247.2$ For Year 3: 6% interest on$2247.5 =

$\frac{6}{100}×2247.2=134.8$

Therefore, the amount at the end of Year 3 is

$2247.2+134.8=2382$

## Arithmetic and Geometric Progressions

Getting the answer for the examples above was simple, because there were only a few terms to compute. But what if in Example A, the increase in number of minutes per day was restricted to 2 minutes. What if in Example B, the number of compounding years was given to be 12 years? In such cases, it would be time consuming to calculate every term till we get the desired value. A simpler way to solve such examples would be to decode the sequence of numbers and devise a formula for them.

If we identify the pattern of numbers in Example A, we see that there is a constant difference of 5 between any two successive numbers.

When the difference between any two consecutive terms is always the same in a given sequence of numbers, the numbers are said to be in
Arithmetic Progression (AP). This means that the next number in a given sequence is calculated by adding a fixed number to the previous number in the sequence.

To express this in a mathematical formula, if we call the “
first term of the sequencea, and the fixed term to be added d (commonly referred to as the common difference), then, the sequence of numbers can be written as:

To write the formula for n terms, observe that there will be (n – 1) number of differences between the successive terms.

To find the number of terms till a given value of n, that is to find the nth term of arithmetic progression (AP) denoted as Tn , the sequence can be generalized as:

${T}_{n}=a+\left(n-1\right)d$

This formula can be used to calculate

• A certain term of the arithmetic sequence Tn
• Number of terms in a given arithmetic sequence n
• First number of the arithmetic sequence a
• Common difference of the arithmetic sequence d

However, the concept and the formula mentioned above fails for Example B. The numbers for Example B are

The difference between any two successive given numbers is not a constant, but observe that the ratio of the second number to the first number is a constant.

$\frac{2382}{2247.2}=\frac{2247.2}{2120}=\frac{2120}{2000}=1.06$

Example B
is a case of Geometric Progression (GP). In a GP, the ratio of any two consecutive terms is always the same. Therefore, the next number in the sequence is calculated by multiplying a fixed number to the previous number in the sequence.

To express this in a mathematical formula, call the first term of the sequence a, and the fixed term to be multiplied r (commonly referred to as the “common ratio”). Then the sequence of numbers can be written as

To find the number of terms until a given value of n, that is to find the nth term of a GP denoted by ${T}_{n}$ we can generalize the sequence as

${T}_{n}=a{r}^{\left(n-1\right)}$

This formula can be used to calculate

• A certain term of the geometric sequence Tn
• Number of terms in a given geometric sequence n

• First number of the geometric sequence a

• Common ratio of the geometric sequence r

## Arithmetic and Geometric Series

When a sequence of numbers is added, the result is known as a series. When we add a finite number of terms in an arithmetic sequence, we get a finite arithmetic sequence, for example, sum of first 50 whole numbers.

Consider a sequence of terms in AP
given as

To obtain an arithmetic series, we need to add these. The result will be

It will take a long time to arrive at the final answer by using the term-by-term addition. Therefore, a generalized formula to find the sum of first
n terms of arithmetic progression denoted as Sn is given as

${S}_{n}=\frac{n}{2}\left\{2a+\left(n-1\right)d\right\}$

If the first term a and the last term l of a given arithmetic sequence is known, then the formula is written as

${S}_{n}=\frac{n}{2}\left\{a+l\right\}$

If there are an infinite number of terms, we call it an infinite series. The sum of the first n terms Sn is called a “partial sum”. As n tends to infinity, Sn tends to ∞.

The sum to infinity for an arithmetic series is undefined.

In the case of a geometric progression, the sum of the terms will be

$a+ar+a{r}^{2}+a{r}^{3}+...$

A generalized formula to find the sum of first n terms of geometric progression denoted as Sn is given as

${S}_{n}=a\frac{\left(1-{r}^{n}\right)}{1-r}$

If the sum to infinity of a geometric series is desired, we rewrite the equation as

${S}_{\infty }=\frac{a}{1-r}$ where $-1

Examples on Bartleby:

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