Number Series - A Summary

Arithmetic Progressions

These are series of numbers e.g. 2, 6, 10, 14, 18 ... or 100, 97.5, 95, 92.5, 90, . . . . where a constant number is added to or subtracted from each term to form the next.

Assuming that the first term is a and that the number added/subtracted (the "common difference") is d, then the series is: a, a + d, a + 2d, a + 3d .... The nth term of the series is a + (n - 1)d.

The sum of the first n terms of the series is: 0.5n(2a + [n - 1]d).

Geometric Progressions

These are series of numbers, e.g. 2, 6, 18, 54, 162, ... or 100, -50, 25, -12.5, 6.25 ... where each term is multiplied/divided by a constant number (the "common ratio") to form the next. The number can be positive (in which case all the terms are the same sign) or negative (in which case alternate terms have alternate signs). The first term is usually called a. The common ratio is usually called r.

If r is outside the range -1 to 1, the terms of the series get bigger and bigger (even if they change sign), and the series diverges. If r is within the range -1 to 1, the terms get smaller and smaller (closer to 0) and the series converges.

The sum of the first n terms of a geometric progression is
a(1 - r n)
1 - r

If r lies within the range -1 < r < 1, then the series has a sum to infinity (i.e. if you added up an infinite number of terms of the series, you would still get a finite number). An example is the series 1, 1/2, 1/4, 1/8, 1/16 ... etc. where a = 1 and r = 1/2. In this case, if you add the terms of the series, you get closer and closer to 2 without ever reaching it. If you could add up an infinite number of terms, then the sum would be 2.

To find the sum to infinity of the series, simply put n = ¥ in the equation above. Providing -1 < r < 1, the term r¥ will be 0, and the whole formula reduces to this:

a
1 - r

Some worked examples of number series questions:

  1. "The seventh term of an arithmetic progression is 56 and the eleventh term is 88. Calculate the twentieth term, and the sum of the first 100 terms."

    The seventh term of the series is a + 6d, and the eleventh term is a + 10d. This gives us two simultaneous equations as shown.

    a + 6d = 56

    a + 10d = 88

    It is an easy matter to see that the values that fit these equations are a = 8 and d = 8 (It is just coincidence that they happen to be the same).

    The twentieth term of the series must be 8 + 19 x 8 = 160. The sum of the first 100 terms is 0.5 x 100(2 x 8 + (8 - 1) x 8) = 50(16 + 7 x 8) = 50(16 + 56) = 50 x 72 = 3600.

  2. "A geometric progression has a common ratio of 0.75 and a sum to infinity of 30. Calculate the sum of the first 10 terms."

    a / (1 - 0.75) = a / 0.25 = 30. This means that a = 30 x 0.25 = 6.5.

    The sum of the first 10 terms is 6.5 x (1 - 0.7510) / (1 - 0.75) = 6.5 x (1 - 0.056314) / 0.25 = 24.53585

  3. "Billy has 56p in his piggy bank and puts in an amount of money equal to the amount he put in the week before + 33p every week until he has at least £10.00, which he then blows on sweets (Greedy pig!) How long does it take him to accumulate this?"

    The sums of money going into the piggy bank every week form an arithmetic progression where the first term is 56, the common difference is 33 and the sum of the first n terms is 1000, where n is unknown. Using the formula for the sum of the first n terms of an arithmetic progression,

    0.5n (2 x 56 + (n - 1) x 33) = 1000

    n (112 + (n - 1) x 33) = 2000

    Multiplying this equation out and rearranging it gives: 33n2 + 79n - 2000 = 0

    This is a quadratic equation which solves to give two possible values for n:
    -9.07 and 6.68.

    Since n represents a number of weeks, -9.07 is clearly the wrong value. The value for n must be 6.68, which we round up to the nearest week to give 7 weeks.

    A double check: By the end of week 6 the piggy bank holds 56p + 89p + £1.22 + £1.55 + £1.88 + £2.21 = £8.31. By the end of week 7, the piggy bank holds an extra £2.54, which puts it over £10.00.

  4. "An investor starts with £500 in an investment account, and each month it earns a constant £32 interest. After how many years will the sum in the account exceed £700?"

    This is a lot simpler than the previous example. In this case, we know that a + (n - 1)d = 700 where a = 500 and d = 32. The reason that we don’t use the sum formula is that the amount in the account after each successive year represents each term of the series, not those terms added up.

    The equation becomes 500 + (n - 1) x 32 = 700. This solves to give n = 7.25, which we round up to 8. After eight months the account will hold in excess of £700.


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