Arithmetic Sequences for Fun and Profit Quiz

Answer: 1, 1, 2, 3, 5, 8, 13, 21, 34...

The only thing required for a sequence to be arithmetic is that the difference from one number to the next is always the same. Whether this difference is positive, negative, zero, integer or fractional does not matter. You could even have a sequence that starts with pi and decreases by the square root of two from each element to the next.

The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34... however is not arithmetic. It does use an arithmetic operation (addition) to arrive at its members and is typically discussed in arithmetics, but it does not meet the definition for an arithmetic sequence.

Answer: All odd numbers

Only the set of odd numbers would qualify as an infinite arithmetic sequence. Neither the primes nor the perfect squares meet the definition of an arithmetic sequence and the set of digits is finite, having only the ten elements 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Answer: The middle one: 60 degrees

As the difference between the lowest two and the highest two numbers are equal, we can write the three numbers as x-c, x and x+c. Adding up these three numbers results in 3x (the +c and -c cancel out). The angles of a triangle always add to 180 degrees, so we now get 3x=180 or x=60. Thus the middle angle is 60 degrees, but the other two are not known (the smallest could be anywhere between 0 and 60, the largest from 60 to just under 120 degrees.)

Answer: It will be a perfect square.

You will always wind up with a perfect square when adding up this sequence. Beginning at 1, we get 1, 1+3=4, 4+5=9, 9+7=16, 16+9=25 and so on. You can visualize this on a sheet of graph paper by adding one row and column to a square at a time: The sizes of the additional l-shaped tiles from one square to the next will exactly be 1, 3, 5, 7 and so on.

Answer: It will be evenly divisible by the number of elements added up.

You can make this into a little magic trick if you know the secret. Prepare a sheet with an odd number of boxes and have the player fill in the first two squares any way they want, then make them complete the sequence and add everything up. Before they start, you prepare an envelope with the number of boxes written in (use 11 to 19 of them for good effect) and say you can influence their thoughts so that they will provide a number divisible by the two-digit number you have written down.

The proof for this sequence is a bit tricky to write down in formal language, so I will just sketch it (those of you who hate formal maths can skip it):

It is well-known that adding up the sequence 0, 1, 2, 3, 4, 5... for n elements will result in n(n-1)/2, which is evenly divisible by n if n is odd (since that makes n-1 even, (n-1)/2 is an integer).

Now for our arbitrary sequence, we call our starting value x and our difference c. Thus, we have x, x+c, x+2c, ..., x+(n-1)c. We can rewrite that as nx+(0+1+2+...+(n-1))c. Now the latter part is (n(n-1)/2)c as said before. Thus both nx and (n(n-1)/2)c are divisible by n - and so is the sum.

Answer: Its last two digits will be "01".

This question can either be solved by elimination or by a bit of arithmetic. If you simply try step values of 1 and 2, you can already eliminate three choices: For a step of 1, the sum is 701 (neither four nor five digits) and for a step of 2, it is 1401 (which is divisible by 3 and thus not prime).

However, by some thought, you can also say that the 701st element will be 700 times an integer (which clearly ends in "00") plus 1, so its last two digits will be "01".

Answer: Yes, exactly one of them.

By the way the step value was constructed (neither even nor divisible by 5), it is assured that the step and 1,000 will be mutually prime (their largest common denominator is 1). Any such step value will need a full one thousand elements before it can repeat any three-digit ending combination. Having a thousand such combinations which are all different also means that each possible combination occurs exactly once - in particular also the sought after "000" indicating divisibility by 1000.

By the way, the pseudo-random generators used to make up random numbers in computers use a different principle although the underlying sequence is not arithmetic. These generators will also cycle through all possible numbers of their range in a fixed, fully predetermined sequence, but the fact that they are well mixed up and that these sequences are extremely long (many millions of numbers in today's machines) make them appear fully random to humans.

Answer: Any two elements

If you know the sequence is arithmetic, any two elements are enough. You first calculate the difference between both the values and the positions, then you divide the value difference by the position difference. That number becomes your step size. If neither of the elements is the first, you can then go backwards to determine it.

As an example: You know the fifth element to be 13 and the ninth to be 19. The difference in values is 6, the difference in positions is 4. The step is thus 6/4 or 1.5. Subtract 4 times 1.5 from 13 (the fifth element) to find that the first must be 7. The sequence is thus 7, 8.5, 10, 11.5, 13, 14.5 and so on.

Answer: You can never say that

Without knowing that a sequence is arithmetic, no number of elements known will ever offer proof that it is.

Imagine a sequence where each element's value is equal to its position number, except when its position number is evenly divisible by 65 million in which case the value is 0. You could know the first 64,999,999 elements and it would all look like a perfect arithmetic sequence. However, the next element breaks the pattern.

Answer: 5

The sequence I am looking for is 5, 11, 17, 23, 29 and this is also the only possible 5-length sequence with a step of 6 as with that step, one of your five numbers will have to be divisible by 5. If you want a longer sequence, your step must at least be 30 (divisible by 2, 3 and 5). You could for example use 541, 571, 601, 631, 661 and 691 for a sequence of 6.

It has been proven in 2004 that you can find any length of prime arithmetic sequence, but as of 2017, the longest such sequences actually known are a mere 26 numbers long; the lowest one begins at 3486107472997423, with a step of 371891575525470.