Everything Maths Grade 12

CHAPTER 3. SEQUENCES AND SERIES 3.8

By simply adding together the first n terms, we are actually writing out the series

Sn= a 1 +a 1 r +a 1 r^2 +··· +a 1 rn−^2 +a 1 rn−^1 (3.28)

We may multiply the above equation by r on both sides, giving us

rSn= a 1 r +a 1 r^2 +a 1 r^3 +··· +a 1 rn−^1 +a 1 rn (3.29)

You may notice that allthe terms on the right side of (3.28) and (3.29)are the same, except the first
and last terms. If we subtract (3.28) from (3.29),we are left with just

rSn−Sn= a 1 rn−a 1

Sn(r− 1) = a 1 (rn− 1)

Dividing by (r− 1) on both sides, we arriveat the general form of ageometric series:

Sn=

�n

i=1

a 1 .ri−^1 =

a 1 (rn− 1)

r− 1

(3.30)

See video: VMgjq at http://www.everythingmaths.co.za

Exercise 3 - 4

1. Prove that
   a +ar +ar^2 +··· +arn−^1 =
   a (1−rn)
   (1−r)


2. Given the geometricsequence 1;−3; 9; ... determine:

(a) The 8 thterm of the sequence

(b) The sum of the firsteight terms of the sequence.

3. Determine:
   �^4

n=1

3. 2 n−^1

4. Find the sum of the first 11 terms of the geometricseries 6 + 3 +^32 +^34 +...


5. Show that the sum of the first n terms of the geometricseries

54 + 18 + 6 +··· + 5 (^13 )n−^1

is given by 81 − 34 −n.

6. The eighth term of ageometric sequence is 640. The third term is 20. Find the sum of the first 7
   terms.


7. Solve for n:

�n

t=1

8 (^12 )t= 15^34.

8. The ratio between the sum of the first three terms of a geometric series and the sum of the 4 th-,
   5 th- and 6 th-terms of the same seriesis 8 : 27. Determine the common ratio and the first 2 terms
   if the third term is 8.

More practice video solutions or help at http://www.everythingmaths.co.za

(1.) 01cr (2.) 01cs (3.) 01ct (4.) 01cu (5.) 01cv (6.) 01cw

(7.) 01cx (8.) 01cy