QUESTION 4 a) The numbers 3, x, (x + 6) form the first 3 terms of a positive geometric sequence. Find
i)
ii)
the possible value(s) of x
the tenth term of the sequence
5+6 2 marks
[2 marks]
[3 marks]
S
b) Let A =
6 1 1 2
2 3 -1 and B
1-3 1 – 2
-8
3
4.
e
3x1
i)
Write down the inverse of A V
[6 marks]
ii) Find a matrix X such that AX = B ✓
[4 marks]
c) Sketch the graph of the following polynomial and state the domain and the range.
22
f(x) = -(x + 3)(x + 2)(x - 1)3
0
3x3 ARV
ferse
7
15 marks]

3, x, x+6 are the first 3 terms of G.P

common ratio is x/3=x+6/x

so we cross multiply x×x=3×x+6

ײ= 3x+18 so we have x²-3x-18

(x-6)(x+3)=0

the possible value of x are 6 and -3

b) tenth term of the seauence

first of all the possible sequence are 3, 6, 12 and 3, -3, 3

but we need only the positive GP

so the common ratio of.the GP r is 6/3=2

the 10th term of a GP sequence is ar^10-1=ar^9

where a = first time

3×2^9=3×512=1536

the 10th term is 1536

d no solution

step-by-step explanation:

i believe it’s 165 because if u add 137 and 58 u get 195 and then subtract that from 360 and there’s ur answer