Adding the Terms of a Sequence

Expert reviewed 21 July 2024 7 minute read


HSC Maths Advanced Syllabus

  • establish and use the formulae for the sum of the first nn terms of an arithmetic sequence: Sn=n2(a+l)S_n = \frac{n}{2}(a + l) where ll is the last term in the sequence and Sn=n2(2a+(n1)d)S_n = \frac{n}{2}(2a + (n - 1)d)
  • establish and use the formula for the sum of the first 𝑛 terms of a geometric sequence: Sn=a(rn1)r1=a(1rn)1rS_n=\frac{a(r^n-1)}{r-1}=\frac{a(1-r^n)}{1-r}

How to sum a Sequence

Adding the terms of a sequence is known as finding the sum of a series. This is denoted by the term SnS_n, where:

Sn=T1+T2+T3+...+TnS_n=T_1+T_2+T_3+...+ T_n

In this case nn **represents the number of terms you are adding in the sequence. For example,

S2=T1+T2S_2=T_1+T_2

Sigma notation

Sigma notation is just an alternative way of displaying a sequence:

n=39Tn=T3+T4+...+T9\sum_{n=3}^{9}T_n=T_3+T_4+...+T_9

To read this correctly, we must know that the number on-top of the greek letter sigma (\sum) is the term of the sequence the calculation is finishing at, while the term at the bottom of (\sum) is the term at which the calculation begins.

Summing an AP

If you are finding the summation of an arithmetic series, where each term is obtained by adding a constant difference dd to the previous term, the sum of the first n terms (SnS_n) is given by:

Sn=n2(a+l)orSn=n2(2a+(n1)d)S_n = \frac{n}{2}(a+l)\qquad or \qquad S_n=\frac{n}{2}(2a+(n-1)d)

where aa is the first term of the sequence, ll is the last term, and dd is the common difference. Both formulas can be used depending on the scenario.

Practice Question 1

If the sum of the first 10 terms of an AP is 275, and the first term in the sequence is 5, what is the 10th term ?

From the information given we know that S10=275S_{10}=275 and a=5a=5, we can use the summing formula to find the 10th term, which in this case is the final term.

S10=102(5+l)275=5(5+l)275=25+5l5l=250l=50S_{10}=\frac{10}{2}(5+l)\\275=5(5+l)\\275=25+5l\\5l=250\\l=50

\therefore The last term in the sequence is 50.

Summing a GP

For a geometric sequence, where each term is found by multiplying the previous term by a constant ratio, the sum of the first n terms is found using:

Sn=a(rn1)r1(whenr>1)orSn=a(1rn)1r(whenr<1)S_n=\frac{a(r^n-1)}{r-1}\qquad(when\quad r>1)\\or\\S_n=\frac{a(1-r^n)}{1-r}\qquad(when\quad r<1)

where aa is the first term of the sequence and rr is the common ratio.

Practice Question 2

What is the sum of the first 9 terms of the following GP: 2,4,8,16,...2,4,8,16,...

From the information given, we can see the first term and common ratio of the progression are both 2. Thus, we can implement the formula for summing a GP.

Because the common ratio rr is greater than 1 we use the following equation:

S9=2(291)21=2(5121)1=2×511=1022S_9=\frac{2(2^9-1)}{2-1}\\=\frac{2(512-1)}{1}\\=2\times511\\=1022

\therefore The sum of the first 9 times of the given GP is 1022

Return to Module 1: Sequences and Series