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SPI Mathematics · Formula sheet

Sequence and Series formulas

12 formulas and 26 common traps for SPI Mathematics Sequence and Series, grouped by subtopic.

Full notes with worked examples

The nth Term of an AP

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What makes a list an AP

Common difference

d=a2−a1=a3−a2d = a_2 - a_1 = a_3 - a_2

The nth term formula

The nth term and the number of terms

an=a+(n−1)dn=l−ad+1a_n = a + (n - 1)d \qquad n = \frac{l - a}{d} + 1
  • aathe first term
  • ddthe common difference
  • llthe last term

The gap between two terms

Difference of two terms

am−ak=(m−k)da_m - a_k = (m - k)d

Terms counted from the end

nth term from the end

l−(n−1)dl - (n - 1)d

Common traps

Subtracting the wrong way round

The common difference is a term minus the term before it. For 20, 17, 14 it is 17−20=−317 - 20 = -3, not 3. A list that falls has a negative common difference.

Judging surds by their look

2,4,6\sqrt{2}, \sqrt{4}, \sqrt{6} looks evenly spaced but is not an AP: the gaps are about 0.59 and 0.45. Rewrite each surd as a multiple of one root before deciding.

Squares of an AP are not an AP

2, 4, 6 is an AP, but their squares 4, 16, 36 have gaps 12 and 20. Squaring the terms breaks the equal steps.

Using n instead of n minus 1

The 10th term of 3, 7, 11, ... is 3+9×4=393 + 9 \times 4 = 39. Adding 10×410 \times 4 gives 43, one step too far.

Forgetting the +1 when counting terms

From 12 to 96 in steps of 12 there are 96−1212+1=8\frac{96 - 12}{12} + 1 = 8 terms, not 7. Dividing the distance counts the steps; the terms are one more.

Starting the count at the wrong number

To count the three-digit multiples of 8, start at 104 (the first multiple of 8 above 99), not at 100, and end at 992, the last one below 1000.

Counting the terms instead of the steps

From the 9th term to the 15th there are 6 steps, not 7. The gap is (m−k)d(m - k)d, not (m−k+1)d(m - k + 1)d.

Losing the sign

a10−a14a_{10} - a_{14} is −4d-4d, because the later term is subtracted. With d=3d = 3 it is −12-12.

Getting the direction wrong in a falling AP

For 50, 46, 42, ..., 2 the common difference is −4-4, so the terms just before 2 are 6, 10, 14. Going back from the end you subtract dd, which here adds 4.

Taking n steps back instead of n minus 1

The 5th term from the end is 4 steps back from the last term, so it is l−4dl - 4d, not l−5dl - 5d.

Three Terms in an AP

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The middle term is the average of its neighbours

Three terms in AP

2b=a+cb=a+c22b = a + c \qquad b = \frac{a + c}{2}

Three unknown terms written around the middle one

Three terms centred on the middle one

(a−d)+a+(a+d)=3a(a−d)(a+d)=a2−d2(a - d) + a + (a + d) = 3a \qquad (a - d)(a + d) = a^2 - d^2

Common traps

Writing the middle term as the sum

The middle term is the average of its neighbours, not their sum. For 4, 10, 16: 2×10=4+162 \times 10 = 4 + 16, while 4+16=204 + 16 = 20 is not 10.

Skipping the check

After solving for the unknown, put it back into all three terms. If the gaps are not equal, an algebra slip has crept in.

Calling the terms a, a + d, a + 2d

That also works, but the sum becomes 3a+3d3a + 3d and leaves two unknowns. Centre the terms on the middle one and the sum gives it at once.

Choosing the wrong sign for d

d2=9d^2 = 9 gives d=3d = 3 or d=−3d = -3. The two signs list the same numbers in opposite orders, so the first and third terms swap. Read 'increasing' or 'decreasing' before you answer.

Sum of n Terms of an AP

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The two sum formulas

Sum of n terms

Sn=n2(a+l)=n2[2a+(n−1)d]S_n = \frac{n}{2}(a + l) = \frac{n}{2}\left[2a + (n - 1)d\right]
  • aathe first term
  • llthe last term
  • ddthe common difference

Standard sums of natural, odd and even numbers

Standard sums

1+2+⋯+n=n(n+1)21+3+⋯+(2n−1)=n22+4+⋯+2n=n(n+1)1 + 2 + \cdots + n = \frac{n(n + 1)}{2} \qquad 1 + 3 + \cdots + (2n - 1) = n^2 \qquad 2 + 4 + \cdots + 2n = n(n + 1)

Finding the number of terms from the sum

Number of terms from the sum

n=2Sa+ln = \frac{2S}{a + l}

Terms from a sum written in n

Terms from the sum

a1=S1an=Sn−Sn−1Sn=pn2+qn⇒d=2pa_1 = S_1 \qquad a_n = S_n - S_{n-1} \qquad S_n = pn^2 + qn \Rightarrow d = 2p

Common traps

Giving the last term when the sum is asked

For 1, 4, 7, ... the 10th term is 28 but the sum of 10 terms is 145. An option that matches the nth term formula is a common wrong answer.

Halving twice

The n2\frac{n}{2} already does the halving. Do not also divide (a+l)(a + l) by 2, or the answer comes out half the right size.

Mixing up the odd and the even sums

The first 10 odd numbers add up to 100 and the first 10 even numbers to 110. Each even number is one more than the matching odd number, so the even sum is larger by nn.

Using the last number as n

1+3+⋯+411 + 3 + \cdots + 41 has 21 terms, because 2n−1=412n - 1 = 41 gives n=21n = 21. Its sum is 212=44121^2 = 441, not 41241^2.

Keeping the negative root

The quadratic for nn usually has a negative or fractional second root. A number of terms is a positive whole number, so that root is thrown away.

Forgetting the half

Writing n(a+l)=Sn(a + l) = S without the 12\frac{1}{2} gives half the right number of terms. The formula is n2(a+l)\frac{n}{2}(a + l).

Reading the sum as a term

If Sn=n2+5nS_n = n^2 + 5n, the 3rd term is not S3=24S_3 = 24. It is S3−S2=24−14=10S_3 - S_2 = 24 - 14 = 10.

Expanding the bracket carelessly

Sn−1S_{n-1} means n−1n - 1 in place of every nn, so (n−1)2=n2−2n+1(n - 1)^2 = n^2 - 2n + 1. The minus sign in front of Sn−1S_{n-1} then changes every sign inside it.

Geometric Progressions and Means

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Geometric progression: ratio, nth term and sum

Geometric progression

an=arn−1Sn=a(rn−1)r−1b2=aca_n = ar^{n-1} \qquad S_n = \frac{a(r^n - 1)}{r - 1} \qquad b^2 = ac
  • aathe first term
  • rrthe common ratio

Arithmetic and geometric means of two numbers

The two means

AM=a+b2GM=abt2−2 AM t+GM2=0\text{AM} = \frac{a + b}{2} \qquad \text{GM} = \sqrt{ab} \qquad t^2 - 2\,\text{AM}\,t + \text{GM}^2 = 0

Common traps

Using n instead of n minus 1 in the power

The 4th term of 5, 10, 20, ... is 5×23=405 \times 2^3 = 40, not 5×24=805 \times 2^4 = 80.

Testing a GP by subtracting

A GP is tested by dividing. 2, 4, 8 has equal ratios but unequal gaps, so it is a GP and not an AP.

Treating the geometric mean as the product

A geometric mean of 6 means the product is 36, not 6. Square the geometric mean before you use it.

Applying a ratio of means to the numbers

A ratio such as AM : GM = 5 : 4 compares the two means, not the two numbers. Find the AM from the sum first, then scale it to get the GM.

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