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Problems (arithmetic and geometric series)

Solutions

Online resources

Problems (arithmetic and geometric series)

- What is the sum of all odd numbers between 50 and 300?
- What is the sum of all all multiples of 7 between 1 and 200?
- What is the average (mean) of all odd numbers from 1 to 300?
- Suppose you deposit \($1000\) a month into a savings account with \(0.2\%\) interest compounded at the end of each month. After 10 years how much money will there be in your account?
- The price of gold is \($59.19\) per gram (as of this writing). Suppose you buy \($100\) worth of gold at one month intervals for the next 10 years. Assuming a monthly inflation rate of \(0.5\%\) (this means that the “purchasing power” of \($1\) today equals equals that of \($1.005\) a month from now). How much gold will you accumulate? Compare with an inflation rate of 0. (Assume the future price of gold in dollars is affected by inflation only.)
- Consider the infinitely repeating decimal fraction \(0.555…\). Express it as a ratio of two integers. (Hint: Begin by expressing it as a geometric series.)
- Consider the infinitely repeating decimal fraction \(0.123123123….\). Express it as a ratio of two integers.
- Consider \(f(x) = \frac{1}{1-x}\). Re-express this function as a series. How many terms does the series have? Do the same for \(g(x) = \frac{1}{1+x}\).

Solutions

- The odd numbers form an arithmetic sequence: \(a_n = a_n-1 +2\). The sum of any n consecutive terms from an arithmetic sequence is \(\frac{n}{2}(b_1 + b_n)\), where \(b_1\) and \(b_n\) are the first and last of the terms respectively. Thus the sum of all odd numbers between 50 and 300 is \(\frac{125}{2}(51+299) ~=~ 21875\).
- Multiples of 7 form an arithmetic sequence: \(a_n = a_n-1 + 7\). The biggest multiple of 7 smaller than 200 is \(7*28 = 196\). Thus the sum in question is, \(\frac{28}{2}(7+196) ~=~ 2842\).
- As noted previously, The sum of any n consecutive terms from an arithmetic sequence is \(\frac{n}{2}(b_1 + b_n)\), where \(b_1\) and \(b_n\) are the first and last of the terms. Thus the average of n consecutive terms is \(\frac{b_1+b_n}{2}\). Thus the average in question is \(\frac{1+299}{2} = 150\).
- A thousand dollars deposited at the beginning of the ten year period will grow to \(1.002^{120} \times 1000\) at 10 years. A thousand dollars deposited k months hence will grow to \((1.002)^{120-k}\). Thus the total amount in the account after 10 years equals a geometric series: \(1.002^{120} \times 1000 ~+~ 1.002^{119} \times 1000 ~+~ \cdots 1.002\times 1000\) with a common ratio of \(\frac{1}{1.002}\). The sum of n consecutive terms of a geometric series is \(a_0\frac{1-r^n}{1-r}\). Thus the total amount after 10 years is \(1.002^{120}~\frac{1-1.002^{-120}}{1-\frac{1}{1.002}}\times 1000 ~\sim~136 \times 1000\).
- Today \($100\) will buy \(\frac{100}{59.19} ~\sim~ 1.7\) grams of gold. One month from now, \($100\) will buy \(\frac{1}{1.005} \times \frac{100}{59.19} ~\sim~ 1.68\) grams of gold. \(k\) months from now, you will buy \(\big(\frac{1}{1.005}\big)^k \times \frac{100}{59.19}\) grams. Thus the total amount purchased over 10 years is a geometric series, \(\frac{100}{59.19}\big(1+ \frac{1}{1.005} + \big(\frac{1}{1.005}\big)^2 + \cdots + \big(\frac{1}{1.005}\big)^{119}\big) = \frac{100}{59.19}\frac{1-\big(\frac{1}{1.005}\big)^{120}}{1-\frac{1}{1.005}} ~\sim~ 153\) grams. With no inflation, or if you had bought \($12000\) worth of gold on day 1, you would have accumulated \(120 \times \frac{100}{55.19} ~\sim~ 203\) grams!
- \(0.555… = 5~\frac{1}{10} ~+~ 5~\big(\frac{1}{10}\big)^2 ~+~ \cdots\). This is an infinite geometric series with common ratio \(\frac{1}{10}\) and initial term \(\frac{5}{10}\). Summing the series we get \(0.555… ~=~ \frac{5}{10}\frac{1}{1-\frac{1}{10}} ~=~ \frac{5}{9}\).
- \(0.123123… = 123~\frac{1}{10^3} ~+~ 123~\big(\frac{1}{10^3}\big)^2 ~+~ \cdots\). This is an infinite geometric series with common ratio \(\frac{1}{10}\) and initial term \(\frac{123}{10^3}\). Summing the series we get \(0.123123… ~=~ \frac{123}{1000}\frac{1}{1-\frac{1}{1000}} ~=~ \frac{123}{999}\).
- The sum of an infinite geometric series with initial term \(a\) and common ratio \(r\) is \(a~\frac{1}{1-r}\). Thus \(\frac{1}{1-x}\) is the sum of a series with initial term 1 and common ratio \(x\). Working backwards, \(f(x) ~=~ 1 + x + x^2 + x^3 + \cdots\). Similarly, \(g(x) ~=~ 1 + (-x) + (-x)^2 + (-x)^3 + \cdots\).

Online resources

This short video from MIT derives the formula for the sum of an arithmetic sequence.

This short video from MIT derives the formula for the sum of an infinite geometric sequence.