**Arithmetic and geometric series**

When we add the terms of a sequence together, we form a series. We use the symbol *S _{n}* to show the sum of the first

*n*terms of a series.

*S*=

_{n}*a*

_{1}+

*a*

_{2}+

*a*

_{3}+⋯+

*a*

_{(n-1)}+

*a*

_{n}📌 Solutionample 1: **Recognizing Arithmetic and Geometric Sequences**

Which of the following can be the first four terms of an arithmetic sequence? Of a geometric sequence?

(A) 1, 2, 3, 5, … (B) -1, 3, -9, 27, …

(C) 3, 3, 3, 3, … (D) 10, 8.5, 7, 5.5, …

✍ Solution:

(A) Since 2-1≠5-3, there is no common difference, so the sequence is not an arithmetic sequence. Since 2/1≠3/2, there is no common ratio, so the sequence is not geometric either.

(B) The sequence is geometric with common ratio -3, but it is not arithmetic.

(C) The sequence is arithmetic with common difference 0 and it is also geometric with common ratio 1.

(D) The sequence is arithmetic with common difference -1.5, but it is not geometric.

📌 Solution2. Consider the sequence: ½, 4, ¼, 7.⅛, 10, …

a) If the pattern continues in the same way, write down the next TWO terms in the sequence. (Hint: Look for two different sequences in the pattern and separate them.)

b) Calculate the sum of the first 50 terms of the sequence.

✍ Solution:

a) *a*_{1}, *a*_{3} and *a*_{5} form a sequence with a common ratio of ½, so *a*_{7} is 1/16.

*a*_{2}, *a*_{4} and *a*_{6} form a sequence with a common difference of 3, so *a*_{8} is 13.

b) *S*_{50}=25 terms of 1st sequence +25 terms of 2nd sequence

*S*

_{50}=(½+¼+⅛+… to 25 terms)+(4+7+10+13+… to 25 terms).

📌 Solutionample 3 ▼

In an arithmetic sequence, the sum of the first term and the third term is 15. The first, third and seventh terms of the arithmetic sequence are the first three terms of a geometric sequence.

(i) Find the first term and the common difference of the arithmetic sequence, where the common difference is positive.

(ii) Find the first three terms and the common ratio of the geometric sequence.

✍ Solution:

(i) For the arithmetic sequence, *u _{n}*=

*a*+(

*n*-1)

*d*.

*u*

_{1}=

*a*;

*u*

_{3}=

*a*+2

*d*;

*u*

_{7}=

*a*+6

*d*

Given:

*u*

_{1}+

*u*

_{3}=15

*a*)+(

*a*+2

*d*)=15

*a*+

*a*+2

*d*=15

2

*a*+2

*d*=15… (1)

Given:

*u*

_{1},

*u*

_{3}and

*u*

_{7}, are the first three terms in a geometric sequence.

[common ratio]

[multiply both sides by

*a*(

*a*+2

*d*)]

*a*(

*a*+6

*d*)=(

*a*+2

*d*)(

*a*+2

*d*)

*a*

^{2}+6

*ad*=

*a*

^{2}+4

*ad*+4

*d*

^{2}

6

*ad*=4

*ad*+4

*d*

^{2}

2

*ad*-4

*d*

^{2}=0

*ad*-2

*d*

^{2}=0

*d*(

*a*-2

*d*)=0

*d*=0 or

*a*-2

*d*=0

*d*=0 or

*a*=2

*d*

Thus

*a*=2

*d*… (2) (we are given

*d*>0)

We now solve between the simultaneous equations (1) and (2).

*a*+2

*d*=15;(

*a*=2

*d*)

2

*a*+

*a*=15

3

*a*=15

*a*=5

*d*=

*a*

2

*d*=5

*d*=5/2

(ii) For the geometric sequence:

*u*

_{1}=

*a*=5

*u*

_{2}=

*a*+2

*d*=5+2(5/2)=5+5=10.

Thus, the first three terms of the geometric sequence are 5, 10 and 20 and the common ratio is 2.